Questions as text
Q1 · By expanding sin(2x + x) and using double-angle formulae, show that sin 3x = 3 sin x −4… 9709/21 May/June 2005
7 (i) By expanding sin(2x + x) and using double-angle formulae, show that sin 3x = 3 sin x −4 sin3x. [5] (ii) Hence show that 13π sin3x dx = 24.5 [5] 0
10 marks
Mark scheme: 7 (i) Make relevant use of the sin(A + B) formula B1 Make relevant use of sin2A and cos2A formulae M1 Obtain a correct expression in terms of sin x and cos x A1 Use cos2 x = 1 – sin2 x to obtain an expression in terms of sin x M1(dep*) Obtain given answer correctly A1 5 3 1 (ii) Replace integrand by sin x – sin 3x, or equivalent B1 4 4 3 1 Integrate, obtaining − cos x + cos 3 x, or equivalent B1√ + B1√ 4 12 Use limits correctly M1 Obtain given answer correctly A1 5
This question in 9709/21 May/June 2005
Q2 · By sketching a suitable pair of graphs, show that the equation secx = 3 −x, where x is in… 9709/21 May/June 2007
5 (i) By sketching a suitable pair of graphs, show that the equation secx = 3 −x, where x is in radians, has only one root in the interval 0 < x < 12π. [2] (ii) Verify by calculation that this root lies between 1.0 and 1.2. [2] (iii) Show that this root also satisfies the equation 1 x = cos−1 . [1] 3 −x (iv) Use the iterative formula 1 xn+1 = cos−1 , 3 −xn with initial value x1 = 1.1, to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
8 marks
Mark scheme: 5 (i) Make recognisable sketch of a relevant graph, e.g. y = secx B1 Sketch an appropriate second graph, e.g. y = 3 –x, correctly and justify the given statement B1 [2] (ii) Consider sign of secx – (3 – x) at x = 1 and x = 1.2, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (iii) Show that the given equation is equivalent to secx = 3 –x, or vice versa B1 [1] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.04 A1 Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.035, 1.045) B1 [3] 1 1
This question in 9709/21 May/June 2007
Q3 · Express cos2x in terms of cos 2x 9709/21 May/June 2007
6 (i) Express cos2x in terms of cos 2x. [1] (ii) Hence show that 13π cos2x dx = 1 + 1 √3. [4] 6π 8 0 (iii) By using an appropriate trigonometrical identity, deduce the exact value of 3π! sin2x dx. 0 [3]
8 marks
Mark scheme: 1 1 6 (i) State correct expression + cos 2 x , or equivalent B1 [1] 2 2 (ii) Integrate an expression of the form a + b cos 2 x , where ab ≠ 0 , correctly M1 1 1 State correct integral x + sin 2 x , or equivalent A1 2 4 Use correct limits correctly M1 Obtain given answer correctly A1 [4] (iii) Use identity sin 2 x = 1 − cos 2 x and attempt indefinite integration M1 1 1 Obtain integral x − x − sin 2 x , or equivalent A1 2 4 1 3 Use limits and obtain answer π − A1 [3] 6 8 [Solutions that use the result of part (ii), score M1A1 for integrating 1 and A1 for the final answer.] GCE A/AS LEVEL – May/June 2007 9709 02
This question in 9709/21 May/June 2007
Q4 · Express 8 sin θ −15 cos θ in the form R sin(θ −α), where R > 0 and 0◦< α < 90◦, giving… 9709/21 Oct/Nov 2007
6 (i) Express 8 sin θ −15 cos θ in the form R sin(θ −α), where R > 0 and 0◦< α < 90◦, giving the exact value of R and the value of α correct to 2 decimal places. [3] (ii) Hence solve the equation 8 sin θ −15 cos θ = 14, giving all solutions in the interval 0◦≤θ ≤360◦. [4]
7 marks
Mark scheme: 6 (i) State answer R = 17, allow 289 B1 Use trig formula to find α M1 Obtain α = 61.93°, (1.08 radians) A1 [3] (ii) Carry out evaluation of sin −(14/17)1 ≈55.44°, or equivalent M1 Obtain answer 117.4°, (2.06 radians) A1 Carry out correct method for second answer M1 Obtain answer 186.5° and no others in the range (3.255 radians) A1√ [4] [Ignore answers outside the given range.]
This question in 9709/21 Oct/Nov 2007
Q5 · Prove the identity (cos x + 3 sin x)2 ≡5 −4 cos 2x + 3 sin 2x 9709/21 Oct/Nov 2007
7 (i) Prove the identity (cos x + 3 sin x)2 ≡5 −4 cos 2x + 3 sin 2x. [4] (ii) Using the identity, or otherwise, find the exact value of 14π (cos x + 3 sin x)2 dx. [4] 0
8 marks
Mark scheme: 7 (i) Expand and use sin 2A formula M1 Use cos 2A formula at least once M1 Obtain any correct expression in terms of cos 2x and sin 2x only – can be implied A1 Obtain given answer correctly A1 [4] 3 (ii) State indefinite integral 5x – 2sin 2x – cos 2x B2 2 [Award B1 if one error in one term] Substitute limits correctly – must be correct limits M1 1 Obtain answer (5π – 2), or exact simplified equivalent A1 [4] 4
This question in 9709/21 Oct/Nov 2007
Q6 · Express 5 cos θ −sin θ in the form R cos(θ + α), where R > 0 and 0◦< α < 90◦, giving the… 9709/21 May/June 2008
5 (i) Express 5 cos θ −sin θ in the form R cos(θ + α), where R > 0 and 0◦< α < 90◦, giving the exact value of R and the value of α correct to 2 decimal places. [3] (ii) Hence solve the equation 5 cos θ −sin θ = 4, giving all solutions in the interval 0◦≤θ ≤360◦. [4]
7 marks
Mark scheme: 5 (i) State R = 26 B1 Use trig formula to find a M1 Obtain α = 11.31° with no errors seen A1 [3] 4 (ii) Carry out evaluation of cos–1 ( ) (≈ 38.3288…°) M1 26 Obtain answer 27.0° A1 Carry out correct method for second answer M1 Obtain answer 310.4° and no others in the range A1√ [4] [Ignore answers outside the given range.]
This question in 9709/21 May/June 2008
Q7 · Show that the equation sin(x + 30◦) = 2 cos(x + 60◦) can be written in the form (3 √3)… 9709/21 Oct/Nov 2008
4 (i) Show that the equation sin(x + 30◦) = 2 cos(x + 60◦) can be written in the form (3 √3) sin x = cos x. [3] (ii) Hence solve the equation sin(x + 30◦) = 2 cos(x + 60◦), for −180◦≤x ≤180◦. [3] 2
6 marks
Mark scheme: 4 (i) Use correct sin(A + B) and cos(A + B) formulae M1 Substitute exact values for sin 30° etc. M1 Obtain given answer correctly A1 [3] (ii) Solve for x M1 Obtain answer x = 10.9° A1 Obtain second answer x = –169.1° and no others in the range A1 [3] [Ignore answers outside the given range.]
This question in 9709/21 Oct/Nov 2008
Q8 · By sketching a suitable pair of graphs, show that the equation cos x = 2 −2x, where x is… 9709/21 Oct/Nov 2008
7 (i) By sketching a suitable pair of graphs, show that the equation cos x = 2 −2x, where x is in radians, has only one root for 0 ≤x ≤12π. [2] (ii) Verify by calculation that this root lies between 0.5 and 1. [2] (iii) Show that, if a sequence of values given by the iterative formula xn+1 = 1 −12 cos xn converges, then it converges to the root of the equation in part (i). [1] (iv) Use this iterative formula, with initial value x1 = 0.6, to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
8 marks
Mark scheme: 7 (i) Make a recognizable sketch of a relevant graph, e.g. y = cos x or y = 2 – 2x B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of cos x – (2 – 2x) at x = 0.5 and x = 1, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (iii) Show that the given equation is equivalent to x = 1 – 1 cos x, or vice versa B1 [1] 2 (iv) Use the iterative formula correctly at least once Ml Obtain final answer 0.58 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.575, 0.585) B1 [3]
This question in 9709/21 Oct/Nov 2008
Q9 · Prove the identity + sin x sec2x + sec x tan x ≡1 9709/21 Oct/Nov 2008
8 (i) (a) Prove the identity + sin x sec2x + sec x tan x ≡1 . cos2x (b) Hence prove that 1 sec2x + sec x tan x ≡ 1 −sin x. [3] 1 dy (ii) By differentiating cos x, show that if y = sec x then dx = secx tan x. [3] (iii) Using the results of parts (i) and (ii), find the exact value of 14π 1 dx. [3] 1 −sin x 0
9 marks
Mark scheme: 8 (i) (a) Use trig formulae and justify given result B1 (b) Use 1 – sin2 x = cos2 x M1 Obtain given result correctly A1 [3] (ii) Use quotient or chain rule M1 Obtain correct derivative in any form A1 Obtain given result correctly A1 [3] (iii) Obtain integral tan x + sec x B1 Substitute limits correctly M1 Obtain exact answer 2 , or equivalent A1 [3]
This question in 9709/21 Oct/Nov 2008
Q10 · Solve the equation sec x 4 tan2x, giving all solutions in the interval [6] = −2 0◦≤x ≤180◦ 9709/21 May/June 2009
5 Solve the equation sec x 4 tan2x, giving all solutions in the interval [6] = −2 0◦≤x ≤180◦.
6 marks
Mark scheme: 5 Use tan2 x = sec2 x –1 or sin2 x = 1 – cos2 x M1 Obtain 3-term quadratic in sec x or cos x, e.g. 2sec2 x + sec x – 6 = 0 A1 Make reasonable solution attempt at a 3-term quadratic M1 Obtain sec x = 3 and sec x = –2, or equivalent A1 2 [or 6cos2 x – cos x – 2 = 0 cos x = 2 3 , −1 2 ] Obtain answer x = 48.2° A1 Obtain answer x = 120° and no others in the range A1 [6] [Ignore answers outside the given range.] GCE A/AS LEVEL – May/June 2009 9709 02
This question in 9709/21 May/June 2009
Q11 · Show that the equation 2 sin x can be written in the form tan x k, where k is a… 9709/21 Oct/Nov 2009
4 (i) Show that the equation 2 sin x can be written in the form tan x k, where k is a sin(60◦−x) = = constant. [4] (ii) Hence solve the equation [2] 2 sin x, for x 0◦< sin(60◦−x) = < 360◦.
6 marks
Mark scheme: 4 (i) Use trig formulae to express equation in terms of sin x and cos x M1 Use cos 60° = 1 and sin 60° = 3 , or equivalent M1 2 2 Obtain equation in sin x and cos x in any correct form A1 Obtain tan x = 3 / 5 , or 0.3464…, or equivalent A1 [4] (ii) Obtain answer x = 19.1° B1 Obtain answer x = 199.1° and no others in the range B1√ [2] [ignore answers outside the given range.]
This question in 9709/21 Oct/Nov 2009
Q12 · Express cos2 2x in terms of cos 4x 9709/21 Oct/Nov 2009
5 (i) Express cos2 2x in terms of cos 4x. [2] 18π (ii) Hence find the exact value of cos2 2x dx. [4] ã 0
6 marks
Mark scheme: 5 (i) Use double angle formulae and obtain a + bcos 4x M1 Obtain answer 1 + 1 cos 4x, or equivalent A1 [2] 2 2 (ii) Integrate and obtain 1 x + 1 sin 4x A1√ + A1√ 2 8 Substitute limits correctly M1 Obtain answer 1 π + 1 , or exact equivalent A1 [4] 16 8 GCE A/AS LEVEL – October/November 2009 9709 21
This question in 9709/21 Oct/Nov 2009
Q13 · Express 3 cos x 4 sin x in the form R where R 0 and α 90◦, stating the exact value of R… 9709/22 Oct/Nov 2009
6 (i) Express 3 cos x 4 sin x in the form R where R 0 and α 90◦, stating the exact value of R and giving+ the value of α correctcos(x −α),to 2 decimal places.> 0◦< < [3] (ii) Hence solve the equation 3 cos x 4 sin x 4.5, + = giving all solutions in the interval [4] x 0◦< < 360◦.
7 marks
Mark scheme: 6 (i) State answer R = 5 B1 Use trig formula to find a M1 Obtain a = 53.13° A1 [3] (ii) Evaluate cos–1 (4.5/5) ≈ 25.84° M1 Obtain answer 79.0° A1 Carry out correct method for second answer M1 Obtain answer 27.3° and no others in the given range A1√ [4] [Treat the giving of answers in radians as a misread. Ignore answers outside the given range.]
This question in 9709/22 Oct/Nov 2009
Q14 · Show that the equation 6 tan x can be written in the form tan(x + 45◦) = 6 tan2x tan x 1 0 9709/21 May/June 2010
3 (i) Show that the equation 6 tan x can be written in the form tan(x + 45◦) = 6 tan2x tan x 1 0. −5 + = [3] (ii) Hence solve the equation [3] 6 tan x, for x 0◦< tan(x + 45◦) = < 180◦.
6 marks
Mark scheme: 3 (i) Use tan(A ± B) formula to obtain an equation in tan x M1 Use tan 45° = 1 and obtain a correct equation in any form A1 Obtain the given equation correctly A1 [3] (ii) Solve the given quadratic in tan x and evaluate an inverse tangent M1 Obtain a correct answer, e.g. 18.4° A1 Obtain second answer, e.g. 26.6°, and no others in the given interval A1 [3] [Treat the giving of answers in radians as a misread. Ignore answers outside the given interval.] 2
This question in 9709/21 May/June 2010
Q15 · Prove the identity sin x 9709/23 May/June 2010
8 (i) Prove the identity sin x. sin(x −30◦) + cos(x −60◦) ≡(√3) [3] (ii) Hence solve the equation 1 2 sec x, sin(x −30◦) + cos(x −60◦) = for [6] x 0◦< < 360◦.
9 marks
Mark scheme: 8 (i) Use correct sin(A – B) and cos(A – B) formulae M1 Substitute exact values for sin 30° etc. M1 Obtain given answer correctly A1 [3] 1 (ii) State 3 sin x = sec x B1 2 Rearrange to sin 2x = k, where k is a non-zero constant M1 1 −1 1 Carry out evaluation of sin M1 2 3 Obtain answer 17.6° A1 Carry out correct method for second answer M1 Obtain remaining 3 answers from 17.6°, 72.4°, 197.6°, 252.4° and no others in the range A1 [6] [Ignore answers outside the given range]
This question in 9709/23 May/June 2010
Q16 · Solve the equation 8 cot θ 2 cosec2θ, giving all solutions in the interval [6] + = 0◦≤θ… 9709/21 Oct/Nov 2010
5 Solve the equation 8 cot θ 2 cosec2θ, giving all solutions in the interval [6] + = 0◦≤θ ≤360◦.
6 marks
Mark scheme: 5 Use correct trig identity to obtain a quadratic in cot θ or tan θ M1 Solve the quadratic correctly A1 1 2 Obtain tan θ = or − A1√ 2 3 Obtain answer 26.6° or 146.3° A1 Carry out correct method for second answer from either root M1 Obtain remaining 3 answers from 26.6°, 146.3°, 206.6°, 326.3° and no others in the range A1 [6] [Ignore answers outside the given range] GCE A/AS LEVEL – October/November 2010 9709 21 6
This question in 9709/21 Oct/Nov 2010
Q17 · Solve the equation 8 cot θ 2 cosec2θ, giving all solutions in the interval [6] + = 0◦≤θ… 9709/22 Oct/Nov 2010
5 Solve the equation 8 cot θ 2 cosec2θ, giving all solutions in the interval [6] + = 0◦≤θ ≤360◦.
6 marks
Mark scheme: 5 Use correct trig identity to obtain a quadratic in cot θ or tan θ M1 Solve the quadratic correctly A1 1 2 Obtain tan θ = or − A1√ 2 3 Obtain answer 26.6° or 146.3° A1 Carry out correct method for second answer from either root M1 Obtain remaining 3 answers from 26.6°, 146.3°, 206.6°, 326.3° and no others in the range A1 [6] [Ignore answers outside the given range] GCE A/AS LEVEL – October/November 2010 9709 22 6
This question in 9709/22 Oct/Nov 2010
Question 18 9709/23 Oct/Nov 2010
4 (a) Find dx. [2] ã e1−2x (b) Express sin23x in terms of cos 6x and hence find sin23x dx. [4] ã
6 marks
Mark scheme: 4 (a) Obtain integral of the form ke1 – 2x with any non-zero k M1 Correct integral A1 [2] (b) Attempt to use double angle formula to expand cos (3x + 3x) M1 1 1 State correct expression − cos6x or equivalent A1 2 2 Integrate an expression of the form a + b cos6x, where ab ≠ 0, correctly M1 1 1 State correct integral x – sin6x, or equivalent A1 [4] 2 12
This question in 9709/23 Oct/Nov 2010
Q19 · Express 2 sin θ θ in the form R where R 0 and α 90◦, giving the exact value of R and… 9709/23 Oct/Nov 2010
6 (i) Express 2 sin θ θ in the form R where R 0 and α 90◦, giving the exact value of R and the−cosvalue of α correct sin(θto 2 decimal−α), places. > 0◦< < [3] (ii) Hence solve the equation 2 sin θ θ −cos = −0.4, giving all solutions in the interval [4] 0◦≤θ ≤360◦.
7 marks
Mark scheme: 6 (i) State R = 5 B1 Use trig formula to find α M1 Obtain α = 26.57° with no errors seen A1 [3] ± 4.0 ≈ ±10.3048)° M1 (ii) Carry out evaluation of sin–1 ( 5 Obtain answer 16.3° A1 Carry out correct method for second answer M1 Obtain answer 216.9° and no others in the range A1 [4]
This question in 9709/23 Oct/Nov 2010
Q20 · Α8 (i) Express 4 sin θ 0 and Give the exact cos θ in the form R where R 0◦< −6 sin(θ −α)… 9709/21 May/June 2011
α8 (i) Express 4 sin θ 0 and Give the exact cos θ in the form R where R 0◦< −6 sin(θ −α), > < 90◦. value of R and the value of α correct to 2 decimal places. [3] (ii) Solve the equation 4 sin θ cos θ 3 for [4] −6 = 0◦≤θ ≤360◦. (iii) Find the greatest and least possible values of sin θ cos 8 as θ varies. [2] (4 −6 θ)2 +
9 marks
Mark scheme: 8 (i) State or imply R = 52 or 2 13 B1 Use appropriate formula to find α M1 Obtain 56.31° A1 [3] (ii) Attempt to find at least one value of θ – α M1 Obtain one correct value 80.9° of θ A1 Carry out correct method to find second answer M1 Obtain 211.7° and no others in range A1 [4] (iii) Obtain 60, following their value of R B1 √ Obtain 8. Allow quoted solution B1 [2]
This question in 9709/21 May/June 2011
Q21 · Prove that cos 2θ 9709/22 May/June 2011
8 (i) Prove that cos 2θ. [3] sin22θ(cosec2θ −sec2θ) ≡4 (ii) Hence (a) solve for equation 3, [4] 0◦≤θ ≤180◦the sin22θ(cosec2θ −sec2θ) = (b) find the exact value of [2] cosec215◦−sec215◦.
9 marks
Mark scheme: 1 1 8 (i) Use cosecθ = and secθ = B1 sin θ cos θ Attempt to simplify left-hand side M1 Confirm given right-hand side 4cos2θ with no errors seen A1 [3] 3 (ii) (a) State or imply cos2θ = B1 4 Attempt correct process to find at least one angle M1 Obtain 20.7° A1 Obtain 159.3° and no others in range A1 [4] 4 cos 30 o (b) Recognise as 2 o B1 sin 30 Obtain 8 3 B1 [2]
This question in 9709/22 May/June 2011
Q22 · Prove that cos 2θ 9709/23 May/June 2011
8 (i) Prove that cos 2θ. [3] sin22θ(cosec2θ −sec2θ) ≡4 (ii) Hence (a) solve for equation 3, [4] 0◦≤θ ≤180◦the sin22θ(cosec2θ −sec2θ) = (b) find the exact value of [2] cosec215◦−sec215◦.
9 marks
Mark scheme: 1 1 8 (i) Use cosecθ = and secθ = B1 sin θ cos θ Attempt to simplify left-hand side M1 Confirm given right-hand side 4cos2θ with no errors seen A1 [3] 3 (ii) (a) State or imply cos2θ = B1 4 Attempt correct process to find at least one angle M1 Obtain 20.7° A1 Obtain 159.3° and no others in range A1 [4] 4 cos 30 o (b) Recognise as 2 o B1 sin 30 Obtain 8 3 B1 [2]
This question in 9709/23 May/June 2011
Q23 · By first expanding show that cos(2x + x), cos 3x cos3x cos x 9709/21 Oct/Nov 2011
8 (i) By first expanding show that cos(2x + x), cos 3x cos3x cos x. ≡4 −3 [5] (ii) Hence show that 16π cos3x dx 12.5 ã 0 (2 −cos x) = [5]
10 marks
Mark scheme: 8 (i) Make relevant use of the cos(A + B) formula M1* Make relevant use of the cos 2A and sin 2A formulae M1* Obtain a correct expression in terms of cos x and sin x A1 Use sin2 x = 1 – cos2 x to obtain an expression in terms of cos x M1(dep*) Obtain given answer correctly A1 [5] 1 1 (ii) Replace integrand by cos 3 x + cos x , or equivalent B1 2 2 1 1 Integrate, obtaining sin 3 x + sin x , or equivalent B1 + B1√ 6 2 Use limits correctly M1 Obtain given answer A1 [5]
This question in 9709/21 Oct/Nov 2011
Q24 · By sketching a suitable pair of graphs, show that the equation 1 sin x, x = [2] where x… 9709/22 Oct/Nov 2011
5 (i) By sketching a suitable pair of graphs, show that the equation 1 sin x, x = [2] where x is in radians, has only one root for 0 x < ≤12π. (ii) Verify by calculation that this root lies between x 1.1 and x 1.2. [2] = = 1 (iii) Use the iterative formula xn+1 = sin xn to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
7 marks
Mark scheme: 1 5 (i) Make a recognisable sketch of a relevant graph, e.g. y = sin x or y = B1 x Sketch a second relevant graph and justify the given statement B1 [2] 1 (ii) Consider sign of − sin x at x = 1.1 and x = 1.2, or equivalent M1 x Complete the argument correctly with appropriate calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.11 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.105, 1.115) B1 [3] GCE AS/A LEVEL – October/November 2011 9709 22 dx dy 2
This question in 9709/22 Oct/Nov 2011
Q25 · Α8 (i) Express 5 cos θ 0 and giving the exact sin θ in the form R where R −3 cos(θ + α)… 9709/22 Oct/Nov 2011
α8 (i) Express 5 cos θ 0 and giving the exact sin θ in the form R where R −3 cos(θ + α), > < 90◦, 0◦< value of R and the value of α correct to 2 decimal places. [3] (ii) Hence solve the equation 5 cos θ sin θ 4, −3 = giving all solutions in the interval [4] 0◦≤θ ≤360◦. (iii) Write down the least value of 15 cos θ sin θ as θ varies. [1] −9
8 marks
Mark scheme: 8 (i) State R = 34 B1 Use trig formula to find α M1 Obtain α = 30.96º with no errors seen A1 [3] ± 4 (ii) Carry out evaluation of cos–1 (≈ 46.6861º or 313.3139º) M1 R Obtain answer 15 .7º A1 Carry out correct method for second answer M1 Obtain answer 282.3º or 282.4º and no others in the range A1 [4] (iii) State − 3 34 (= –3R) B1√ [1]
This question in 9709/22 Oct/Nov 2011
Q26 · Express cos2x in terms of cos 2x 9709/23 Oct/Nov 2011
4 (i) Express cos2x in terms of cos 2x. [1] (ii) Hence show that 16π 1 1 sin dx 1 √3 8 12π 4. + + [5] ã 0 (cos2x + 2x) =
6 marks
Mark scheme: 1 1 4 (i) State correct expression + cos 2 x , or equivalent B1 [1] 2 2 (ii) Integrate an expression of the form a + b cos 2x, where ab ≠ 0, correctly M1 1 1 State correct integral x + sin 2 x , or equivalent A1 2 4 1 Obtain correct integral (for sin 2x term) of − cos 2 x B1 2 Attempt to substitute limits, using exact values M1 Obtain given answer correctly A1 [5]
This question in 9709/23 Oct/Nov 2011
Q27 · Given that 35 sec2θ 12 tan θ, find the value of tan θ 9709/21 May/June 2012
4 (i) Given that 35 sec2θ 12 tan θ, find the value of tan θ. [3] + = (ii) Hence, showing the use of an appropriate formula in each case, find the exact value of (a) [2] tan(θ −45◦), (b) tan 2θ. [2]
7 marks
Mark scheme: 4 (i) Use sec2 θ = 1 + tan2 θ B1 Attempt solution of quadratic equation in tan θ M1 Obtain tan2 θ – 12 tanθ + 36 = 0 or equivalent and hence tan θ = 6 A1 [3] (ii) (a) Attempt use of tan(A – B) formula M1 Obtain 75 following their value of tan θ A1√ [2] (b) Attempt use of tan 2θ formula M1 Obtain − 1235 A1 [2] 1 x
This question in 9709/21 May/June 2012
Q28 · Show that sin x cos can be written in the form 5 2 sin 2x cos 2x 9709/21 May/June 2012
7 (i) Show that sin x cos can be written in the form 5 2 sin 2x cos 2x. [5] 2 2 (2 + x)2 + −3 14π (ii) Hence find the exact value of sin x cos dx. [4] ã 0 (2 + x)2
9 marks
Mark scheme: 7 (i) Expand to obtain 4 sin2 x + 4 sin x cos x + cos2 x B1 Use 2 sin x cos x = sin 2x B1 Attempt to express sin2 x or cos2 x (or both) in terms of cos 2x M1 Obtain correct 12 k 1( − cos 2 x ) for their k sin2 x or equivalent A1√ Confirm given answer 52 + 2 sin 2 x − 32 cos 2 x A1 [5] (ii) Integrate to obtain form px + q cos 2x + r sin 2x M1 Obtain 52 x − cos 2 x − 34 sin 2 x A1 Substitute limits in integral of form px + q cos 2x + r sin 2x and attempt simplification DM1 Obtain 85 π + 14 or exact equivalent A1 [4]
This question in 9709/21 May/June 2012
Q29 · Α4 (i) Express 9 sin θ 0 and Give the value cos θ in the form R where R −12 sin(θ −α), >… 9709/22 May/June 2012
α4 (i) Express 9 sin θ 0 and Give the value cos θ in the form R where R −12 sin(θ −α), > < 90◦. 0◦< of α correct to 2 decimal places. [3] Hence (ii) solve the equation 9 sin θ cos θ 4 for [4] −12 = 0◦≤θ ≤360◦, (iii) state the largest value of k for which the equation 9 sin θ cos θ k has any solutions. [1] −12 =
8 marks
Mark scheme: 4 (i) State or imply R = 15 B1 Use appropriate formula to find α M1 Obtain 53.13° A1 [3] (ii) Attempt to find at least one value of θ – α M1 Obtain one correct value 68.6° of θ A1 Carry out correct method to find second answer M1 Obtain 217.7° and no others in range A1 [4] (iii) State 15, following their value of R from part (i) B1√ [1] dx 1
This question in 9709/22 May/June 2012
Q30 · Show that tan2x cos2x 1 cos 2x and hence find the exact value of + ≡sec2x + 2 −12 14π dx 9709/22 May/June 2012
7 (i) Show that tan2x cos2x 1 cos 2x and hence find the exact value of + ≡sec2x + 2 −12 14π dx. ã 0 (tan2x + cos2x) [7] (ii) y 1 x O 4p 0 is shown in The region enclosed by the curve y tan x cos x and the lines x 0, x = + = = 14π and y = the diagram. Find the exact volume of the solid produced when this region is rotated completely about the x-axis. [4]
11 marks
Mark scheme: 7 (i) Replace tan2 x by sec2 x – 1 B1 Express cos2 x in the form ± 12 ± 12 cos 2 x M1 Obtain given answer sec 2 x + 12 cos 2 x − 12 correctly A1 Attempt integration of expression M1 Obtain tan x + 14 sin 2 x − 12 x A1 Use limits correctly for integral involving at least tan x and sin 2x M1 Obtain 54 − 18 π or exact equivalent A1 [7] (ii) State or imply volume is ∫ π (tan x + cos x 2) dx B1 Attempt expansion and simplification M1 Integrate to obtain one term of form k cos x M1 Obtain π ( 54 − 18 π ) + π ( 2 − 2 ) or equivalent A1 [4]
This question in 9709/22 May/June 2012
Q31 · Given that 35 sec2θ 12 tan θ, find the value of tan θ 9709/23 May/June 2012
4 (i) Given that 35 sec2θ 12 tan θ, find the value of tan θ. [3] + = (ii) Hence, showing the use of an appropriate formula in each case, find the exact value of (a) [2] tan(θ −45◦), (b) tan 2θ. [2]
7 marks
Mark scheme: 4 (i) Use sec2 θ = 1 + tan2 θ B1 Attempt solution of quadratic equation in tan θ M1 Obtain tan2 θ – 12 tanθ + 36 = 0 or equivalent and hence tan θ = 6 A1 [3] (ii) (a) Attempt use of tan(A – B) formula M1 Obtain 75 following their value of tan θ A1√ [2] (b) Attempt use of tan 2θ formula M1 Obtain − 1235 A1 [2] 1 x
This question in 9709/23 May/June 2012
Q32 · Show that sin x cos can be written in the form 5 2 sin 2x cos 2x 9709/23 May/June 2012
7 (i) Show that sin x cos can be written in the form 5 2 sin 2x cos 2x. [5] 2 2 (2 + x)2 + −3 14π (ii) Hence find the exact value of sin x cos dx. [4] ã 0 (2 + x)2
9 marks
Mark scheme: 7 (i) Expand to obtain 4 sin2 x + 4 sin x cos x + cos2 x B1 Use 2 sin x cos x = sin 2x B1 Attempt to express sin2 x or cos2 x (or both) in terms of cos 2x M1 Obtain correct 12 k 1( − cos 2 x ) for their k sin2 x or equivalent A1√ Confirm given answer 52 + 2 sin 2 x − 32 cos 2 x A1 [5] (ii) Integrate to obtain form px + q cos 2x + r sin 2x M1 Obtain 52 x − cos 2 x − 34 sin 2 x A1 Substitute limits in integral of form px + q cos 2x + r sin 2x and attempt simplification DM1 Obtain 85 π + 14 or exact equivalent A1 [4]
This question in 9709/23 May/June 2012
Q33 · Solve the equation 2 cos 2θ 4 cos θ = −3, for [4] 0◦≤θ ≤180◦ 9709/21 Oct/Nov 2012
3 Solve the equation 2 cos 2θ 4 cos θ = −3, for [4] 0◦≤θ ≤180◦.
4 marks
Mark scheme: 3 Make relevant use of the cos 2θ formula M1 Obtain a correct quadratic in cos θ A1 Solve a quadratic in cos θ M1 Obtain answer θ = 60 and no others in the range A1 [4] (Ignore answers outside the given range) dx − 2 dy 2
This question in 9709/21 Oct/Nov 2012
Q34 · Given that tan A t and 4, find tan B in terms of t 9709/22 Oct/Nov 2012
8 (a) Given that tan A t and 4, find tan B in terms of t. [3] = tan(A + B) = (b) Solve the equation 2 3 tan x, tan(45◦−x) = giving all solutions in the interval [6] 0◦≤x ≤360◦.
9 marks
Mark scheme: 8 (a) Use tan (A + B) formula to obtain an equation in tan B M1 t + tan B State equation = 4 , or equivalent A1 1 − t tan B 4 − t Solve to obtain tan B = A1 [3] 1 + 4t tan 45 − tan x (b) State equation 2 = 3 tan x , or equivalent B1 1 + tan 45 tan x Transform to a quadratic equation M1 Obtain 3tan2 x + 5tan x – 2 = 0 (or equivalent) A1 Solve the quadratic and calculate one angle, or establish that tan x = ⅓, –2 M1 Obtain one answer, e.g. x = 18.4o A1 Obtain other 3 answers 116.6o, 198.4o, 296.6o and no others in range A1 [6]
This question in 9709/22 Oct/Nov 2012
Q35 · Solve the equation 2 cos 2θ 4 cos θ = −3, for [4] 0◦≤θ ≤180◦ 9709/23 Oct/Nov 2012
3 Solve the equation 2 cos 2θ 4 cos θ = −3, for [4] 0◦≤θ ≤180◦.
4 marks
Mark scheme: 3 Make relevant use of the cos 2θ formula M1 Obtain a correct quadratic in cos θ A1 Solve a quadratic in cos θ M1 Obtain answer θ = 60 and no others in the range A1 [4] (Ignore answers outside the given range) dx − 2 dy 2
This question in 9709/23 Oct/Nov 2012
Q36 · Show that 12 sin2x cos2x [3] ≡32(1 −cos 4x) 9709/21 May/June 2013
3 (i) Show that 12 sin2x cos2x [3] ≡32(1 −cos 4x). (ii) Hence show that 13π π 3√3 12 sin2x cos2x dx . ã 1 = 8 + 16 [3] 4π
6 marks
Mark scheme: 3 (i) Either Use sin 2x = 2sin x cos x to convert integrand to k sin2 2x M1 Use cos 4x = 1 – 2 sin2 2x M1 1 1 State correct expression − cos 4 x or equivalent A1 2 2 Or 1 − cos 2 x 1 − cos 2 x Use cos2 x = and/or x = to obtain an equation in cos 2 x only M1 2 2 1 + cos 4 x Use cos2 2 x = M1 2 1 1 State correct expression − cos 4 x or equivalent A1 [3] 2 2 3 3 (ii) State correct integral x − sin 4 x , or equivalent B1 2 8 Attempt to substitute limits, using exact values M1 Obtain given answer correctly A1 [3] 3
This question in 9709/21 May/June 2013
Q37 · By sketching a suitable pair of graphs, show that the equation cot x 4x = −2, where x is… 9709/21 May/June 2013
6 (i) By sketching a suitable pair of graphs, show that the equation cot x 4x = −2, where x is in radians, has only one root for 0 2π. [2] ≤x ≤1 (ii) Verify by calculation that this root lies between x 0.7 and x 0.9. [2] = = (iii) Show that this root also satisfies the equation 1 2 tan x x + . = 4 tan x [1] 1 2 tan xn (iv) Use the iterative formula + to determine this root correct to 2 decimal places. xn+1 = 4 tan xn Give the result of each iteration to 4 decimal places. [3]
8 marks
Mark scheme: 6 (i) Make a recognisable sketch of a relevant graph, e.g. y = cot x or y = 4x – 2 B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of 4x – 2 – cot x at x = 0.7 and x = 0.9, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] 1+ 2 tan x (iii) Show that given equation is equivalent to x = , or vice versa B1 [1] 4 tan x (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.76 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.755, 0.765) B1 [3] GCE AS LEVEL – May/June 2013 9709 21
This question in 9709/21 May/June 2013
Q38 · Α7 (i) Express 5 sin 2θ 0 and giving the 2 cos 2θ in the form R where R 0◦< + sin(2θ +… 9709/21 May/June 2013
α7 (i) Express 5 sin 2θ 0 and giving the 2 cos 2θ in the form R where R 0◦< + sin(2θ + α), > < 90◦, exact value of R and the value of α correct to 2 decimal places. [3] Hence (ii) solve the equation 5 sin 2θ 2 cos 2θ 4, + = giving all solutions in the interval [5] 0◦≤θ ≤360◦, 1 (iii) determine the least value of as θ varies. [2] sin 2θ 4 cos (10 + 2θ)2
10 marks
Mark scheme: 7 (i) State R = 29 B1 Use trig formula to find α M1 Obtain α = 21.80 0 with no errors seen A1 [3] −1 4 0 (ii) Carry out evaluation of sin (≈ 47.97 ) M1 R Carry out correct method for one correct answer M1 Obtain one correct answer e.g. 13.1o A1 Carry out correct method for a further answer M1 Obtain remaining 3 answers 55.10, 193.10, 235.10 and no others in the range A1 [5] (iii) Greatest value of 10 sin 2θ + 4 cos 2θ = 2√29 M1 1 A1 [2] 116
This question in 9709/21 May/June 2013
Q39 · Show that 12 sin2x cos2x [3] ≡32(1 −cos 4x) 9709/23 May/June 2013
3 (i) Show that 12 sin2x cos2x [3] ≡32(1 −cos 4x). (ii) Hence show that 13π π 3√3 12 sin2x cos2x dx . ã 1 = 8 + 16 [3] 4π
6 marks
Mark scheme: 3 (i) Either Use sin 2x = 2sin x cos x to convert integrand to k sin2 2x M1 Use cos 4x = 1 – 2 sin2 2x M1 1 1 State correct expression − cos 4 x or equivalent A1 2 2 Or 1 − cos 2 x 1 − cos 2 x Use cos2 x = and/or x = to obtain an equation in cos 2 x only M1 2 2 1 + cos 4 x Use cos2 2 x = M1 2 1 1 State correct expression − cos 4 x or equivalent A1 [3] 2 2 3 3 (ii) State correct integral x − sin 4 x , or equivalent B1 2 8 Attempt to substitute limits, using exact values M1 Obtain given answer correctly A1 [3] 3
This question in 9709/23 May/June 2013
Q40 · By sketching a suitable pair of graphs, show that the equation cot x 4x = −2, where x is… 9709/23 May/June 2013
6 (i) By sketching a suitable pair of graphs, show that the equation cot x 4x = −2, where x is in radians, has only one root for 0 2π. [2] ≤x ≤1 (ii) Verify by calculation that this root lies between x 0.7 and x 0.9. [2] = = (iii) Show that this root also satisfies the equation 1 2 tan x x + . = 4 tan x [1] 1 2 tan xn (iv) Use the iterative formula + to determine this root correct to 2 decimal places. xn+1 = 4 tan xn Give the result of each iteration to 4 decimal places. [3]
8 marks
Mark scheme: 6 (i) Make a recognisable sketch of a relevant graph, e.g. y = cot x or y = 4x – 2 B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of 4x – 2 – cot x at x = 0.7 and x = 0.9, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] 1+ 2 tan x (iii) Show that given equation is equivalent to x = , or vice versa B1 [1] 4 tan x (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.76 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.755, 0.765) B1 [3] GCE AS LEVEL – May/June 2013 9709 23
This question in 9709/23 May/June 2013
Q41 · Α7 (i) Express 5 sin 2θ 0 and giving the 2 cos 2θ in the form R where R 0◦< + sin(2θ +… 9709/23 May/June 2013
α7 (i) Express 5 sin 2θ 0 and giving the 2 cos 2θ in the form R where R 0◦< + sin(2θ + α), > < 90◦, exact value of R and the value of α correct to 2 decimal places. [3] Hence (ii) solve the equation 5 sin 2θ 2 cos 2θ 4, + = giving all solutions in the interval [5] 0◦≤θ ≤360◦, 1 (iii) determine the least value of as θ varies. [2] sin 2θ 4 cos (10 + 2θ)2
10 marks
Mark scheme: 7 (i) State R = 29 B1 Use trig formula to find α M1 Obtain α = 21.80 0 with no errors seen A1 [3] −1 4 0 (ii) Carry out evaluation of sin (≈ 47.97 ) M1 R Carry out correct method for one correct answer M1 Obtain one correct answer e.g. 13.1o A1 Carry out correct method for a further answer M1 Obtain remaining 3 answers 55.10, 193.10, 235.10 and no others in the range A1 [5] (iii) Greatest value of 10 sin 2θ + 4 cos 2θ = 2√29 M1 1 A1 [2] 116
This question in 9709/23 May/June 2013
Q42 · Solve the equation 2 cosec 10, giving all solutions in the interval [6] cot21 −5 1 = 0Å… 9709/22 Oct/Nov 2013
3 Solve the equation 2 cosec 10, giving all solutions in the interval [6] cot21 −5 1 = 0Å ≤1 ≤360Å.
6 marks
Mark scheme: 3 Use trig identity correctly to obtain a quadratic in cosec θ or sin θ M1 Solve the quadratic correctly M1 Obtain sin θ = ¼ or − ⅔ A1 Obtain one correct answer A1 Carry out correct method for second answer from either root DM1 Obtain remaining 3 answers from 14.5, 165.5, 221.8, 318.2 and no others in the range A1 [Ignore answers outside the given range] [6]
This question in 9709/22 Oct/Nov 2013
Q43 · 5 (i) Prove that tan cot [3] sin 1 + 1 21 9709/21 May/June 2014
2 5 (i) Prove that tan cot [3] sin 1 + 1 21. (ii) Hence (a) find the exact value of tan 1 cot 1 [2] 80 + 80, 1 20 6 (b) evaluate [3] tan cot d1. Ô 0 1 + 1
8 marks
Mark scheme: 5 (i) Express left-hand side as a single fraction M1 Use sin 2θ = 2 sin θ cosθ at some point B1 Complete proof with no errors seen (AG) A1 [3] GCE AS LEVEL – May/June 2014 9709 21 2 (ii) (a) State or equivalent B1 sin 1 π 4 Obtain 2 2 or exact equivalent (dependent on first B1) B1 [2] (b) State or imply k sin 2θ for any k B1 Integrate to obtain − 32 cos 2θ B1 Substitute both limits correctly to obtain 3 B1 [3] ( )
This question in 9709/21 May/June 2014
Q44 · Solve the equation 3 sin tan 2 for [4] 21 1 = 0Å < 1 < 180Å 9709/22 May/June 2014
2 Solve the equation 3 sin tan 2 for [4] 21 1 = 0Å < 1 < 180Å.
4 marks
Mark scheme: 2 Use sin 2θ = 2 sin θ cos θ B1 Simplify to obtain form c1sin 2 θ = c 2 or equivalent M1 Find at least one value of θ from equation of form sin θ = k M1 Obtain 35.3° and 144.7° A1 [4]
This question in 9709/22 May/June 2014
Q45 · Solve the equation 3 sin tan 2 for [4] 21 1 = 0Å < 1 < 180Å 9709/23 May/June 2014
2 Solve the equation 3 sin tan 2 for [4] 21 1 = 0Å < 1 < 180Å.
4 marks
Mark scheme: 2 Use sin 2θ = 2 sin θ cos θ B1 Simplify to obtain form c1sin 2 θ = c 2 or equivalent M1 Find at least one value of θ from equation of form sin θ = k M1 Obtain 35.3° and 144.7° A1 [4]
This question in 9709/23 May/June 2014
Q46 · Find 4 cos2 1 [3] Ó 21 d1 9709/21 Oct/Nov 2014
3 (a) Find 4 cos2 1 [3] Ó 21 d1. 6 1 (b) Find the exact value of dx. [4] Ô 2x 3 −1 +
7 marks
Mark scheme: 3 (a) Express integrand in the form p cos θ + 2 M1 State correct 2 cos θ + 2 A1 Integrate to obtain 2 sin θ + 2θ (+ c) A1 [3] (b) Integrate to obtain form k ln (2 x + 3) M1 1 Obtain correct ln (2 x + 3) A1 2 Apply limits correctly DM1 1 Obtain ln 15 A1 [4] 2
This question in 9709/21 Oct/Nov 2014
Q47 · Prove that 2 cosec tan sec2 [3] 21 1 1 9709/21 May/June 2015
6 (i) Prove that 2 cosec tan sec2 [3] 21 1 1. (ii) Hence (a) solve the equation 2 cosec tan 5 for 0 [3] 21 1 = < 1 < 0, 160 (b) find the exact value of 2 cosec 4x tan2x dx. [4] Ó 0
10 marks
Mark scheme: 1 6 (i) State or imply cosec2θ = B1 sin2θ Express left-hand side in terms of sinθ and cosθ M1 Obtain given answer sec 2 θ correctly A1 [3] 1 (ii) (a) State or imply cos θ = or tan θ = 2 at least B1 5 Obtain 1.11 or awrt 1.11, allow 0.353π B1 Obtain 2.03 or awrt 2.03 , allow 0.648π and no other values between 0 and π B1 [3] (b) State integrand as sec 2 2x B1 Integrate to obtain expression of form k tan mx M1 Obtain correct 12 tan 2x A1 Obtain 12 3 or exact equivalent A1 [4] 2 dy 3
This question in 9709/21 May/June 2015
Q48 · Y P x O The diagram shows part of the curve with equation y = 4 sin2x + 8 sin x + 3 and… 9709/22 May/June 2015
6 y P x O The diagram shows part of the curve with equation y = 4 sin2x + 8 sin x + 3 and its point of intersection P with the x-axis. (i) Find the exact x-coordinate of P. [3] (ii) Show that the equation of the curve can be written y = 5 + 8 sin x −2 cos 2x, and use integration to find the exact area of the shaded region enclosed by the curve and the axes. [6]
9 marks
Mark scheme: 6 (i) Solve three-term quadratic equation for sin x M1 Obtain at least sin x = − 12 and no errors seen A1 Obtain x = 76 π A1 [3] (ii) State sin 2 x = 12 − 12 cos2 x B1 Obtain given 5 + 8sin x − 2cos2 x with necessary detail seen B1 Integrate to obtain expression of form ax + b cos x + c sin 2 x M1 Obtain correct 5 x − 8cos x − sin 2 x A1 Apply limits 0 and their x-value correctly M1 depM Obtain 356 π + 72 3 + 8 or exact equivalent A1 [6] 4 dy
This question in 9709/22 May/June 2015
Q49 · Y P x O The diagram shows part of the curve with equation y = 4 sin2x + 8 sin x + 3 and… 9709/23 May/June 2015
6 y P x O The diagram shows part of the curve with equation y = 4 sin2x + 8 sin x + 3 and its point of intersection P with the x-axis. (i) Find the exact x-coordinate of P. [3] (ii) Show that the equation of the curve can be written y = 5 + 8 sin x −2 cos 2x, and use integration to find the exact area of the shaded region enclosed by the curve and the axes. [6]
9 marks
Mark scheme: 6 (i) Solve three-term quadratic equation for sin x M1 Obtain at least sin x = − 12 and no errors seen A1 Obtain x = 76 π A1 [3] (ii) State sin 2 x = 12 − 12 cos2 x B1 Obtain given 5 + 8sin x − 2cos2 x with necessary detail seen B1 Integrate to obtain expression of form ax + b cos x + c sin 2 x M1 Obtain correct 5 x − 8cos x − sin 2 x A1 Apply limits 0 and their x-value correctly M1 depM Obtain 356 π + 72 3 + 8 or exact equivalent A1 [6] 4 dy
This question in 9709/23 May/June 2015
Q50 · Show that sin 2x cotx 2 cos2x 9709/22 Feb/March 2016
8 (i) Show that sin 2x cotx 2 cos2x. [2] (ii) Using the identity in part (i), (a) find the least possible value of 3 sin 2x cotx 5 cos 2x 8 + + as x varies, [4] 160 (b) find the exact value of cosec 4x tan 2x dx. [5] Ó 1 80
11 marks
Mark scheme: cos x 8 (i) State 2sin x cos x . B1 sin x Simplify to confirm 2cos 2 x B1 [2] (ii) (a) Use cos2 x = 2cos 2 x − 1 B1 Express in terms of cos x M1 Obtain 16cos 2 x + 3 or equivalent A1 State 3, following their expression of form a cos 2 x + b A1 [4] 1 2 (b) Obtain integrand as sec 2 x B1 2 Integrate to obtain form k tan2 x M1* 1 Obtain correct tan 2 x A1 4 Apply limits correctly dep M1* 1 1 Obtain 3 − or exact equivalent A1 [5] 4 4
This question in 9709/22 Feb/March 2016
Q51 · Solve the equation 5 tan 4 cot for [5] 21 = 1 0Å < 1 < 180Å 9709/21 May/June 2016
2 Solve the equation 5 tan 4 cot for [5] 21 = 1 0Å < 1 < 180Å.
5 marks
Mark scheme: 2 Use cot θ = 1 ÷ tan θ B1 Form equation involving tanθ only and with no denominators involving θ M1 Obtain tan 2 θ = 72 A1 Obtain 28.1 A1 Obtain 151.9 A1 [5] Allow other valid methods 2
This question in 9709/21 May/June 2016
Q52 · Solve the equation 3u 1 2u 9709/22 May/June 2016
3 (i) Solve the equation 3u 1 2u . [3] + = −5 (ii) Hence solve the equation 3 cotx 1 2 cotx for 0 x 1 giving your answer correct to 3 significant figures. + = −5 < < 20, [2]
5 marks
Mark scheme: 3 (i) State or imply non-modular equation (3u + 1) 2 = (2u − 5) 2 or corresponding pair of linear equations B1 Attempt solution of 3-term quadratic equation or of 2 linear equations M1 Obtain −6 and 54 A1 [3] (ii) Evaluate tan −1 1k for at least one of their solutions k from part (i) M1 Obtain 0.896 A1 [2]
This question in 9709/22 May/June 2016
Q53 · Cos 2x 9 cosx 5 5 (i) Show that 2 cos x 1 9709/21 Oct/Nov 2016
cos 2x 9 cosx 5 5 (i) Show that 2 cos x 1. [3] + + cosx 4 + + 0 cos 4x 9 cos 2x 5 (ii) Hence find the exact value of dx. [4] + + Ô cos 2x 4 −0 +
7 marks
Mark scheme: 5 (i) Use cos2 x = 2cos 2 x − 1 and attempt factorisation of numerator M1 Obtain (2cos x + 1)(cos x + 4) A1 Confirm given result 2cos x + 1 A1 [3] (ii) Express integrand as 2cos2 x + 1 B1 Integrate to obtain sin 2 x + x B1 Apply limits correctly to integral of form k1 sin 2 x + k 2 x M1 Obtain 2π A1 [4] dy
This question in 9709/21 Oct/Nov 2016
Q54 · Given that tan cot 8, show that 34 9709/22 Feb/March 2017
2 (i) Given that tan cot 8, show that 34. [3] 21 1 = tan21 = … … … … … … … … … … … … (ii) Hence solve the equation tan cot 8 for [2] 21 1 = 0Å < 1 < 180Å. … … … … … … … … … … … …
5 marks
Mark scheme: 2(i) 1 B1 Use identity cotθ= tanθ Attempt use of identity for tan 2θ M1 Confirm given tan 2 θ = 34 A1 Total: 3 2(ii) Obtain 40.9 B1 Obtain 139.1 B1 Total: 2
This question in 9709/22 Feb/March 2017
Q55 · Solve the equation 3 cosec for [5] sec21 = 1 0Å < 1 < 180Å 9709/21 Oct/Nov 2018
3 Solve the equation 3 cosec for [5] sec21 = 1 0Å < 1 < 180Å. … … … … … … … … … … … … … … … … … … … … … … … … …
5 marks
Mark scheme: 3 State 2 1 3 sin cos θ θ = or 2 3 1 tan sin θ θ + = B1 Produce quadratic equation in sinθ M1 Dependent on B1 Solve 3-term quadratic equation to find value between –1 and 1 for sinθ M1 Dependent on first M1 Obtain 1 6 sin ( 1 37) θ = −+ and hence 57.9 A1 Obtain 122.1 and no others between 0 and 180 A1 5
This question in 9709/21 Oct/Nov 2018
Q56 · Solve the equation 3 cosec for [5] sec21 = 1 0Å < 1 < 180Å 9709/23 Oct/Nov 2018
3 Solve the equation 3 cosec for [5] sec21 = 1 0Å < 1 < 180Å. … … … … … … … … … … … … … … … … … … … … … … … … …
5 marks
Mark scheme: 3 State 2 1 3 sin cos θ θ = or 2 3 1 tan sin θ θ + = B1 Produce quadratic equation in sinθ M1 Dependent on B1 Solve 3-term quadratic equation to find value between –1 and 1 for sinθ M1 Dependent on first M1 Obtain 1 6 sin ( 1 37) θ = −+ and hence 57.9 A1 Obtain 122.1 and no others between 0 and 180 A1 5
This question in 9709/23 Oct/Nov 2018
Q57 · Solve the equation sec2 tan2 5 tan 4 for Show all necessary working 9709/22 Feb/March 2019
1 Solve the equation sec2 tan2 5 tan 4 for Show all necessary working. [4] 1 + 1 = 1 + 0Å < 1 < 180Å. … … … … … … … … … … … … … … … … … … … … … … … … …
4 marks
Mark scheme: 1 Use identity 2 2 sec 1 tan θ θ = + B1 Attempt solution of quadratic equation to find two values of tanθ M1 Obtain 1 2 tan , 3 θ = − A1 Obtain 71.6 and 153.4 and no others between 0 and 180 A1 4
This question in 9709/22 Feb/March 2019
Q58 · Find tan2 3x dx 9709/21 May/June 2019
4 (a) Find tan2 3x dx. [3] Ó … … … … … … … … … 1 e3x 4 (b) Find the exact value dx. Show all necessary working. [4] + ex ofÔ0 … … … … … … … … … … … … …
7 marks
Mark scheme: 4(a) Use identity 2 2 tan 3 sec 3 1 x x = − B1 Integrate to obtain form 1 2 tan3 k x k x + M1 Obtain correct 1 3 tan3x x c − + A1 3 4(b) Express integrand as 2 e 4e x x − + B1 Integrate to obtain form 2 3 4 e e x x k k − + M1 Obtain correct 2 1 2 e 4e x x − − A1 Use limits to obtain 2 1 7 1 2 2 e 4e− − + or similarly simplified equivalent A1 4
This question in 9709/21 May/June 2019
Q59 · Express 4 sin 4 cos in the form R sin , where R 0 and [3] 1 + 1 1 + ! 9709/22 May/June 2019
7 (a) (i) Express 4 sin 4 cos in the form R sin , where R 0 and [3] 1 + 1 1 + ! > 0Å < ! < 90Å. … … … … … … … … … … … … … (ii) Hence find the smallest positive value of satisfying the equation 4 sin 4 cos 5. [2] 1 1 + 1 = … … … … … … … … … … … (b) Solve the equation 4 cot 2x 5 tan x = + for 0 x showing all necessary working and giving the answers correct to 2 decimal places. < < 0, [6] … … … … … … … … … … … … … … … … … … … … … … …
11 marks
Mark scheme: 7(a)(i) State 32 R = or equivalent or 5.657… Use appropriate trigonometry to find α M1 Obtain 45 α = A1 3 7(a)(ii) Carry out correct process to find one value of θ M1 Obtain 17.1 A1 Ignore other positive values greater than 17.1 2 Question Answer Marks Guidance 7(b) Use or imply 1 cot 2 tan2 x x = B1 Use identity of form 2 2tan tan 2 1 tan x x x ± = ± to obtain equation in tan x M1 Obtain 2 6tan 10tan 4 0 x x + − = or equivalent A1 Attempt solution of 3-term quadratic equation for tan x M1 Obtain 1 3 tan x = and hence 0.32 A1 Allow greater accuracy Obtain tan 2 x = − and hence 2.03 and no others between 0 and π A1 Allow greater accuracy 6
This question in 9709/22 May/June 2019
Q60 · Express 4 sin 4 cos in the form R sin , where R 0 and [3] 1 + 1 1 + ! 9709/23 May/June 2019
7 (a) (i) Express 4 sin 4 cos in the form R sin , where R 0 and [3] 1 + 1 1 + ! > 0Å < ! < 90Å. … … … … … … … … … … … … … (ii) Hence find the smallest positive value of satisfying the equation 4 sin 4 cos 5. [2] 1 1 + 1 = … … … … … … … … … … … (b) Solve the equation 4 cot 2x 5 tan x = + for 0 x showing all necessary working and giving the answers correct to 2 decimal places. < < 0, [6] … … … … … … … … … … … … … … … … … … … … … … …
11 marks
Mark scheme: 7(a)(i) State 32 R = or equivalent or 5.657… Use appropriate trigonometry to find α M1 Obtain 45 α = A1 3 7(a)(ii) Carry out correct process to find one value of θ M1 Obtain 17.1 A1 Ignore other positive values greater than 17.1 2 Question Answer Marks Guidance 7(b) Use or imply 1 cot 2 tan2 x x = B1 Use identity of form 2 2tan tan 2 1 tan x x x ± = ± to obtain equation in tan x M1 Obtain 2 6tan 10tan 4 0 x x + − = or equivalent A1 Attempt solution of 3-term quadratic equation for tan x M1 Obtain 1 3 tan x = and hence 0.32 A1 Allow greater accuracy Obtain tan 2 x = − and hence 2.03 and no others between 0 and π A1 Allow greater accuracy 6
This question in 9709/23 May/June 2019
Q61 · Show that 3 sin cot 6 [2] 21 1 cos21 9709/22 May/June 2020
8 (a) Show that 3 sin cot 6 [2] 21 1 cos21. … … … … … … … … … … … (b) Solve the equation 3 sin cot 5 for 0 [3] 21 1 = < 1 < π. … … … … … … … … … … … … … 1 2π 1(c) Find the exact value of 3 sin x cot 2x dx. [5] 1 Ó 4π … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 8(a) Use at least one of sin 2 2sin cos θ θ θ = and cos cot sin θ θ θ = B1 Use both and conclude 2 6cos θ AG B1 2 8(b) Attempt solution of 2 5 cos 6 θ = to find at least one value M1 Obtain 0.421 A1 Obtain 2.72 A1 3 Question Answer Marks 8(c) Express integrand in form cos + a b x M1 Obtain correct integrand 3 3cos + x A1 Integrate to obtain sin + px q x *M1 Apply limits correctly DM1 Obtain 3 3 3 4 2 π + − or exact equivalent A1 5
This question in 9709/22 May/June 2020
Q62 · Show that 3 sin cot 6 [2] 21 1 cos21 9709/23 May/June 2020
8 (a) Show that 3 sin cot 6 [2] 21 1 cos21. … … … … … … … … … … … (b) Solve the equation 3 sin cot 5 for 0 [3] 21 1 = < 1 < π. … … … … … … … … … … … … … 1 2π 1(c) Find the exact value of 3 sin x cot 2x dx. [5] 1 Ó 4π … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 8(a) Use at least one of sin 2 2sin cos θ θ θ = and cos cot sin θ θ θ = B1 Use both and conclude 2 6cos θ AG B1 2 8(b) Attempt solution of 2 5 cos 6 θ = to find at least one value M1 Obtain 0.421 A1 Obtain 2.72 A1 3 Question Answer Marks 8(c) Express integrand in form cos + a b x M1 Obtain correct integrand 3 3cos + x A1 Integrate to obtain sin + px q x *M1 Apply limits correctly DM1 Obtain 3 3 3 4 2 π + − or exact equivalent A1 5
This question in 9709/23 May/June 2020
Q63 · It is given that 3 sin cos where is an angle such that 21 = 1 1 0Å < 1 < 90Å 9709/21 Oct/Nov 2020
6 It is given that 3 sin cos where is an angle such that 21 = 1 1 0Å < 1 < 90Å. (a) Find the exact value of sin [2] 1. … … … … … … … (b) Find the exact value of sec [2] 1. … … … … … … … (c) Find the exact value of cos [2] 21. … … … … … … …
6 marks
Mark scheme: 6(a) Use sin2 2sin cos = θ θ θ Obtain 1 sin 6 θ = B1 2 6(b) Use correct identity or identities to find value of secθ M1 Obtain 6 35 or exact equivalent A1 2 6(c) Use correct identity or identities to find value of cos2θ M1 Obtain 17 18 or exact equivalent A1 2
This question in 9709/21 Oct/Nov 2020
Q64 · Solve the equation 7 cot 3 cosec for [3] 1 = 1 0Å < 1 < 90Å 9709/22 Oct/Nov 2020
1 Solve the equation 7 cot 3 cosec for [3] 1 = 1 0Å < 1 < 90Å. … … … … … … … … … … … … … … … … … … … … … … … … …
3 marks
Mark scheme: 1 Use cos cot sin θ θ θ = and 1 cosec sin θ θ = B1 SOI Simplify to obtain cos k θ = where 0 1 k < < M1 Obtain 3 cos 7 θ = and hence 64.6 θ = and no other solutions in the range A1 Alternative method for question 1 Use identity 2 2 cosec 1 cot = + θ θ B1 Simplify to obtain 1 tan k θ = or 2 sin k θ = where 2 0 1 k < < M1 Obtain 1 tan 40 3 θ = or 1 sin 40 7 θ = and hence 64.6 θ = and no other solutions in the range A1 3
This question in 9709/22 Oct/Nov 2020
Q65 · It is given that 3 sin cos where is an angle such that 21 = 1 1 0Å < 1 < 90Å 9709/23 Oct/Nov 2020
6 It is given that 3 sin cos where is an angle such that 21 = 1 1 0Å < 1 < 90Å. (a) Find the exact value of sin [2] 1. … … … … … … … (b) Find the exact value of sec [2] 1. … … … … … … … (c) Find the exact value of cos [2] 21. … … … … … … …
6 marks
Mark scheme: 6(a) Use sin2 2sin cos = θ θ θ B1 Obtain 1 sin 6 θ = B1 2 6(b) Use correct identity or identities to find value of secθ M1 Obtain 6 35 or exact equivalent A1 2 6(c) Use correct identity or identities to find value of cos2θ M1 Obtain 17 18 or exact equivalent A1 2
This question in 9709/23 Oct/Nov 2020
Q66 · Solve the equation cot 8 for 0 [5] sec21 1 = < 1 < π 9709/22 Feb/March 2021
2 Solve the equation cot 8 for 0 [5] sec21 1 = < 1 < π. … … … … … … … … … … … … … … … … … … … … … … … … …
5 marks
Mark scheme: 2 State 2 1 cos 8 sin cos θ θ θ × = B1 OE involving sinθ and cosθ only. Attempt use of sin 2θ identity to obtain sin 2 k θ = M1 Obtain 1 sin 2 4 θ = A1 Use correct process to find two values of θ between 0 and π M1 Allow if working in degrees. Obtain 0.126 and 1.44 A1 AWRT Alternative method for question 2 State 2 1 tan 8 tan θ θ + = B1 Attempt solution of 3-term quadratic equation to find values of tanθ M1 OE involving tanθ only. Obtain 8 60 tan 2 θ ± = A1 OE Solve tan ... θ = to find two values of θ between 0 and π M1 Allow if working in degrees. Obtain 0.126 and 1.44 A1 AWRT 5
This question in 9709/22 Feb/March 2021
Q67 · Express 5 3 cos x 5 sin x in the form R cos x , where R 0 and 0 1 [3] + −! 9709/22 Feb/March 2021
7 (a) Express 5 3 cos x 5 sin x in the form R cos x , where R 0 and 0 1 [3] + −! > < ! < 2π. … … … … … … … … … … … … … … … (b) As x varies, find the least possible value of 4 5 3 cos x 5 sin x, + + and determine the corresponding value of x where x [3] −π < < π. … … … … … … … … … … … … … 1 (c) Find [3] 2 Ô 5 3 cos 5 sin d1. 31 + 31 … … … … … … … … … … … … … … … … …
9 marks
Mark scheme: 7(a) State 10 R = B1 Use appropriate trigonometry to find α M1 Obtain 1 6 α = π A1 3 7(b) State 6 − B1 FT Following their value of R. Attempt to find x from their cos( ) 1 x α − = − M1 Obtain 1 6 x − π = −π and hence 5 π 6 − A1 3 7(c) State integrand of form 2 1 1 sec 3 π 6 k θ − *M1 Integrate to obtain form 2 1 tan 3 π 6 k θ − DM1 Obtain 1 1 tan 3 π 300 6 c θ − + A1 3
This question in 9709/22 Feb/March 2021
Q68 · Y P x O The diagram shows the curve with parametric equations x 4t e2t, y 6t sin 2t, = +… 9709/22 May/June 2021
7 y P x O The diagram shows the curve with parametric equations x 4t e2t, y 6t sin 2t, = + = for 0 The point P on the curve has parameter p and y-coordinate 3. ≤t ≤1. 1 (a) Show that p [1] = 2 sin 2p. … … … (b) Show by calculation that the value of p lies between 0.5 and 0.6. [2] … … … … … (c) Use an iterative formula, based on the equation in part (a), to find the value of p correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … (d) Find the gradient of the curve at P. [5] … … … … … … … … … … … … … … … … …
11 marks
Mark scheme: 7(a) Equate y to 3 and confirm 1 2sin 2 = p p 1 7(b) Consider sign of 1 2sin 2 − p p or equivalent for 0.5 and 0.6 M1 Obtain 0.09... − and 0.06... or equivalents and justify conclusion A1 AG 2 7(c) Use iteration process correctly at least once M1 Need to see 0.55494… Obtain final answer 0.557 only A1 Allow recovery. Allow if iterations are to 4sf Allow if insufficient iterations seen. Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval [0.5565, 0.5575] A1 If not starting at 0.55 then max marks M1A1A0 3 Question Answer Marks Guidance 7(d) Obtain 2 d 4 2e d = + t x t B1 Use product rule to find d d y t M1 Must be of the form sin 2 cos2 + p t qt t Obtain 6sin 2 12 cos2 + t t t A1 Allow unsimplified. Divide to obtain d d y x using their d d y t and d d x t correctly DM1 Must have either B1 or previous M1. Obtain 0.826 A1 AWRT 5
This question in 9709/22 May/June 2021
Q69 · The polynomials f x and g x are defined by f x 4x3 ax2 8x 15 and g x x2 bx 18, = + + + = +… 9709/23 Oct/Nov 2021
6 The polynomials f x and g x are defined by f x 4x3 ax2 8x 15 and g x x2 bx 18, = + + + = + + where a and b are constants. (a) Given that x 3 is a factor of f x , find the value of a. [2] + … … … … … … … … … … (b) Given that the remainder is 40 when g x is divided by x , find the value of b. [2] −2 … … … … … … … … … … … (c) When a and b have these values, factorise f x x completely. [3] −g … … … … … … … … … … … … (d) Hence solve the equation f cosec cosec 0 for 0 [3] −g = < < 2π. … … … … … … … … … … … …
10 marks
Mark scheme: 6(a) Substitute = − , equate to zero and attempt solution for a M1 Allow attempt at synthetic division, must be a complete method. Allow one sign error carried through. Allow attempt at algebraic long division, must be complete with the remainder equated to zero. Obtain 13 = a A1 2 6(b) Substitute 2 = x , equate to 40 and attempt solution for b M1 Obtain 9 = b A1 2 6(c) Identify 3 + x as factor of f ( ) g( ) − x x B1 May be implied by synthetic division. If working backwards from solutions from a calculator then B0 M0. Attempt, by division or equivalent, to find quadratic factor M1 ( )( )( ) 3 2 1 2 1 + + − k x x x where 1 ≠ k gets B1 M1. Obtain ( 3)(2 1)(2 1) + − + x x x A1 3 6(d) Attempt correct process to find at least 1 value from cosecθ = k where 1 < − k M1 Allow for o 199.5 or o 19.5 − . Obtain 3.48or 5.94 A1 Obtain a second correct solution A1 And no others within the range. 3
This question in 9709/23 Oct/Nov 2021
Q70 · Show that sin cot 1 9709/22 Feb/March 2022
4 (a) Show that sin cot 1. [3] 21 1 −cos 21 … … … … … … … … … (b) Hence find the exact value of sin 1 cot 1 [2] 6π 12π. … … … … … … … (c) Find the smallest positive value of (in radians) satisfying the equation 1 sin cot cos 1. [2] 21 1 −3 21 = … … … … …
7 marks
Mark scheme: 4(a) Use at least two of sin 2 2sin cos θ θ θ = , 2 2 cos2 cos sin θ θ θ = − , cos cot sin θ θ θ = B1 OE Express LHS in terms of sinθ and cosθ only and attempt valid simplification M1 Obtain 2 2 cos sin θ θ + or equivalent and hence 1 A1 AG – necessary detail needed 3 4(b) Substitute 1 12 θ = π and show or imply 1 1 1 6 12 6 sin cot 1 cos π π = + π M1 Obtain 1 2 1 3 + or exact equivalent A1 2 Question Answer Marks Guidance 4(c) Use the identity from part (a) to obtain 2cos2 0 θ − = or equivalent M1 Or alternative starting again, using valid simplification and reaching single trigonometric function. Obtain 1 4 θ = π A1 2
This question in 9709/22 Feb/March 2022
Q71 · Express the equation 7 tan 4 cot sec 0 in terms of sin only 9709/21 May/June 2022
2 (a) Express the equation 7 tan 4 cot sec 0 in terms of sin only. [3] 1 + 1 −13 1 = 1 … … … … … … … … … … … … (b) Hence solve the equation 7 tan 4 cot sec 0 for [3] 1 + 1 −13 1 = 0Å < 1 < 360Å. … … … … … … … … … … … …
6 marks
Mark scheme: 2(a) M1 with at least two of the terms correct and no missing s Condone use of x instead of Obtain 2 2 7sin 4cos 13sin 0 A1 SOI, OE Obtain 2 3sin 13sin 4 0 A1 Allow if missing s are recovered SC Allow full marks for 4 3sin 13 0 sin Must be in terms of for final A mark 3 Question Answer Marks Guidance 2(b) Attempt solution of 3-term quadratic equation for sin M1 Obtain 1 3 sin and hence 19.5 A1 or greater accuracy Obtain second value 160.5 A1 or greater accuracy; and no other values within the given range FT on o o 180 19.5 their 3
This question in 9709/21 May/June 2022
Q72 · The diagram shows the curve with equation y 3 sin x sin 2x for 0 The curve meets the… 9709/21 May/June 2022
3 The diagram shows the curve with equation y 3 sin x sin 2x for 0 The curve meets the x-axis at the origin and at the points with x-coordinates= −3a and ≤x ≤π. π. (a) Find the exact value of a. [3] … … … … … … … (b) Find the area of the shaded region. [4] … … … … … … … … …
7 marks
Mark scheme: 3(a) Attempt to find x-value from 3sin 3sin 2 0 x x using identity for sin 2x M1 Obtain at least 1 2 cos x A1 Obtain 1 3 A1 SC B3 can be spotted from sin sin 2 x x 3 3(b) Integrate to obtain form 1 2 cos cos2 k x k x *M1 non-zero constants 1 2 , k k M0 for 3cos 6cos2 x x Obtain correct 3 2 3cos cos2 x x A1 Attempt value of integral using their lower limit (in radians) and correctly DM1 Allow one sign error Obtain 27 4 A1 OE 4
This question in 9709/21 May/June 2022
Q73 · Solve the equation sec 1 = 5 cosec 1 for 0Å < 1 < 360Å 9709/22 Oct/Nov 2022
1 Solve the equation sec 1 = 5 cosec 1 for 0Å < 1 < 360Å. [4] … … … … … … … … … … … … … … … … … … … … … … … … …
4 marks
Mark scheme: Question Answer Marks Guidance 1 1 1 B1 Must be using sec 2 = 1 + tan 2 and Use sec= and cosec= or other appropriate identities cos sin cosec 2= 1 + cot 2 . Obtain tan= k using correct identities M1 OE For any non-zero constant k, if using other identities, must come from a 3-term quadratic equation. Obtain tan= 5 and hence 78.7 A1 AWRT Obtain 258.7 and no other solutions in the range A1 AWRT 4
This question in 9709/22 Oct/Nov 2022
Q74 · Y P x O The diagram shows the curve with parametric equations x k tan t, y 3 sin 2t sin… 9709/22 Feb/March 2023
7 y P x O The diagram shows the curve with parametric equations x k tan t, y 3 sin 2t sin t, = = −4 for 0 t 1 It is given that k is a positive constant. The curve crosses the x-axis at the point P. < < 2π. (a) Find the value of cos t at P, giving your answer as an exact fraction. [3] … … … … … … … … … … … … … … … … … dy (b) Express in terms of k and cos t. [4] dx … … … … … … … … … … … … (c) Given that the normal to the curve at P has gradient 10,9 find the value of k, giving your answer as an exact fraction. [3] … … … … … … … … … … …
10 marks
Mark scheme: 7(a) Use identity sin2t = 2sin t cos t B1 Attempt solution of y = 0 for cost M1 Obtain cost = 23 A1 Or exact equivalent. 3 7(b) d x d y *M1 Differentiate to obtain at least one of and correct d t d t dy 6cos2t − 4cos t A1 Obtain = dx k sec 2 t d y DM1 Using correct identities. Attempt to express in terms of cost d x dy 6(2cos 2 t − 1)cos 2 t − 4cos 3 t A1 OE Obtain = dx k 4 7(c) d y *M1 Substitute value from part (a) in expression for involving k d x and cost Equate to − 109 and solve for k DM1 Obtain k = 43 A1 3
This question in 9709/22 Feb/March 2023
Q75 · Express 7 cos 24 sin in the form R cos , where R 0 and Give the value of correct… 9709/21 May/June 2023
7 (a) Express 7 cos 24 sin in the form R cos , where R 0 and Give the value of correct to12+decimal1 places. 1 −! > 0Å < ! < 90Å. [3] ! … … … … … … … … … … (b) Solve the equation 7 cos 24 sin 18 for [4] 1 + 1 = 0Å < 1 < 360Å. … … … … … … … … … … … … … (c) As varies, the greatest possible value of " 150 7 cos 1 24 sin 1 50 2" + 2" + is denoted by V. Find the value of V and determine the smallest positive value of (in degrees) for which the value of V occurs. " [4] … … … … … … … … … … … … … … … … … … … … …
11 marks
Mark scheme: 7(a) State 25 R Use appropriate trigonometry to find M1 Allow if found in radians . Obtain 73.74 A1 or greater accuracy. 3 Question Answer Marks Guidance 7(b) Use correct method to find one value of M1 Obtain 29.8 (or 117.7) A1 or greater accuracy. Use correct method to find second value of between 0 and 360 M1 Obtain 117.7 (or 29.8) A1 or greater accuracy; and no others between 0 and 360. 4 7(c) State or imply expression is 1 2 150 25cos( 73.74) 50 B1 FT following their R and . Obtain 6 V B1 Attempt complete method to find positive value from 1 2 cos( 73.74) 1 M1 for their . Obtain 507.5 A1 or greater accuracy. 4
This question in 9709/21 May/June 2023
Q76 · Solve the equation 5 9 17 sec sec21 + tan21 = + 1 for [5] 0Å < 1 < 360Å 9709/22 May/June 2023
1 Solve the equation 5 9 17 sec sec21 + tan21 = + 1 for [5] 0Å < 1 < 360Å. … … … … … … … … … … … … … … … … … … … … … … … …
5 marks
Mark scheme: 1 identity, allow if ‘5’ omitted. Obtain 2 6sec 17sec 14 0 A1 or 2 14cos 17cos 6 0 . Attempt solution of 3-term quadratic equation to find one value of , from cos ... M1 Obtain 73.4 A1 or greater accuracy. Obtain 286.6 A1 or greater accuracy; and no others between 0 and 360. 5
This question in 9709/22 May/June 2023
Q77 · Show that 4 sin 1 cos 3 2 sin [4] 1 + 3π 1 −13π + 21 9709/22 May/June 2023
6 (a) Show that 4 sin 1 cos 3 2 sin [4] 1 + 3π 1 −13π + 21. … … … … … … … … … … … … … … … … (b) Find the exact value of 4 sin 17 cos 1 [2] 24π 24π. … … … … … … … 18π 1(c) Find the exact value of 4 sin 2x cos 2x dx. [4] Ó 0 + 3π −13π … … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 6(a) Obtain at least either 1 1 2 2 ( sin 3cos ) or 1 1 2 2 ( cos 3sin ) Expand and simplify with correct use of 2 2 sin cos 1 M1 Use 1 2 sin cos sin 2 M1 Confirm given result 3 2sin2 A1 AG necessary detail required. 4 6(b) Identify value of is 3 8 π *B1 OE Obtain 3 4 3 2sin π and conclude 3 2 DB1 or exact equivalent. 2 6(c) Identify integrand as 3 2sin4 x B1 Integrate to obtain form 1 2 cos4 k x k x M1 where 1 2 0 k k . Obtain correct 1 2 3 cos4 x x A1 Obtain 1 1 8 2 π 3 A1 or exact equivalent. 4
This question in 9709/22 May/June 2023
Q78 · Solve the equation 5 9 17 sec sec21 + tan21 = + 1 for [5] 0Å < 1 < 360Å 9709/23 May/June 2023
1 Solve the equation 5 9 17 sec sec21 + tan21 = + 1 for [5] 0Å < 1 < 360Å. … … … … … … … … … … … … … … … … … … … … … … … …
5 marks
Mark scheme: 1 identity, allow if ‘5’ omitted. Obtain 2 6sec 17sec 14 0 A1 or 2 14cos 17 cos 6 0 . Attempt solution of 3-term quadratic equation to find one value of , from cos ... M1 Obtain 73.4 A1 or greater accuracy. Obtain 286.6 A1 or greater accuracy; and no others between 0 and 360. 5
This question in 9709/23 May/June 2023
Q79 · Show that 4 sin 1 cos 3 2 sin [4] 1 + 3π 1 −13π + 21 9709/23 May/June 2023
6 (a) Show that 4 sin 1 cos 3 2 sin [4] 1 + 3π 1 −13π + 21. … … … … … … … … … … … … … … … … (b) Find the exact value of 4 sin 17 cos 1 [2] 24π 24π. … … … … … … … 18π 1(c) Find the exact value of 4 sin 2x cos 2x dx. [4] Ó 0 + 3π −13π … … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 6(a) Obtain at least either 1 1 2 2 ( sin 3cos ) or 1 1 2 2 ( cos 3sin ) B1 Allow if implied by decimal values. Expand and simplify with correct use of 2 2 sin cos 1 M1 Use 1 2 sin cos sin 2 M1 Confirm given result 3 2 sin 2 A1 AG necessary detail required. 4 6(b) Identify value of is 3 8 π *B1 OE Obtain 3 4 3 2sin π and conclude 3 2 DB1 or exact equivalent. 2 6(c) Identify integrand as 3 2 sin 4 x B1 Integrate to obtain form 1 2 cos 4 k x k x M1 where 1 2 0 k k . Obtain correct 1 2 3 cos4 x x A1 Obtain 1 1 8 2 π 3 A1 or exact equivalent. 4
This question in 9709/23 May/June 2023
Q80 · It is given that is an acute angle in degrees such that sin 23 9709/23 Oct/Nov 2023
1 It is given that is an acute angle in degrees such that sin 23. Find the exact value of sin 60 . [3] + … … … … … … … … … … … … … … … … … … … … … … … …
3 marks
Mark scheme: Question Answer Marks Guidance 1 1 B1 or exact equivalent. State or imply that cos= 5 3 Substitute appropriate values into sincos60+ cossin60 M1 1 1 A1 or exact equivalent. Obtain + 15 3 6 3
This question in 9709/23 Oct/Nov 2023
Q81 · Show that cosec 3 sin 2 4 sin3 4 6 cos 4 cos2 9709/23 Oct/Nov 2023
6 (a) Show that cosec 3 sin 2 4 sin3 4 6 cos 4 cos2 . [3] … … … … … … … … … … … … … … … … (b) Solve the equation cosec 3 sin 2 4 sin3 3 0 + + = for 0. [3] −π < < … … … … … … … … … … … … … … … (c) Find cosec 3 sin 2 4 sin3 d . [3] + … … … … … … … … … … … … … …
9 marks
Mark scheme: 6(a) 1 B1 Use cosec= sin Express in terms of sin and cos only M1 Dependent on B1. Obtain given result 4 + 6cos− 4cos 2 with sufficient detail A1 AG 3 6(b) Attempt use of formula to solve 3-term quadratic equation as far as M1 where −1 k1 .1 cos= k1 Solve 4cos 2 − 6cos− 7 = 0 to obtain at least cos= − 0.770... A1 6 − 148 or exact equivalent cos= . 8 Obtain −2.45 A1 or greater accuracy; and no others between − π and 0. 3 6(c) Express cos 2 term in the form k 2 + k3 cos2 M1 where k 2 k3 0 . Obtain integrand 6cos+ 2 − 2cos2 A1 Following the 3-term expression in cos from part (a). Integrate to obtain correct 6sin+ 2− sin2 A1 Condone absence of + c , but all in terms of . 3
This question in 9709/23 Oct/Nov 2023
Q82 · The polynomial p( )x is defined by p( x) = 6 x 3 + ax 2 + 3 x - 10 , where a is a constant 9709/22 Feb/March 2024
3 The polynomial p( )x is defined by p( x) = 6 x 3 + ax 2 + 3 x - 10 , where a is a constant. It is given that ( 2 x - 1) is a factor of p( )x . (a) Find the value of a and hence factorise p( )x completely. [5] … … … … … … … … … … … … … … … … (b) Solve the equation p( cosec i) = 0 for - 90° 1 i 1 90° . [2] … … … … … … …
7 marks
Mark scheme: 3(a) Substitute x = 1 , equate to zero and attempt solution M1 2 Obtain a = 31 A1 Divide by 2 x − 1 at least as far as 3x 2 + mx M1 Or equivalent (for example by inspection, …). Obtain 3 x 2 + 17 x + 10 A1 Obtain (2 x − 1)(3 x + 2)( x + 5) A1 5 3(b) Attempt solution of sin= k where k is valid constant from answer to part (a) M1 Obtain −11.5 A1 Or greater accuracy. 2
This question in 9709/22 Feb/March 2024
Q83 · Prove that sin 2i ( a cot i + b tan i) / a + b + ( a - b) cos 2i, where a and b are… 9709/22 Feb/March 2024
7 (a) Prove that sin 2i ( a cot i + b tan i) / a + b + ( a - b) cos 2i, where a and b are constants. [4] … … … … … … … … … … … … … … … … … … … … … … … … … 1 6 (b) Find the exact value of sin 2 i ( 5 cot i + 3 tan i)d i . [3] ry1 r 12 … … … … … … … … … … … … a = 11 for - r 1 a 1 r . [3](c) Solve the equation sin 23 a 2 cot 13 a + 7 tan 13 ` j … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 7(a) Use correct identity for sin2 or for cot B1 Express left-hand side in terms of sin and cos M1 Attempt to express k1 cos 2 + k 2 sin 2 in terms of cos2 only M1 Confirm a + b + ( a − b)cos2 A1 Answer given – necessary detail needed. 4 7(b) Obtain integrand 8 + 2cos2 B1 Integrate to obtain 8+ sin2 B1 Apply limits correctly to obtain 2 π + 1 3 − 1 B1 Or exact equivalent. 3 2 2 3 7(c) Use identity to obtain 5cos 2 = −2 B1 3 Obtain = 2.97 B1 Or greater accuracy. Obtain = −2.97 B1 Or greater accuracy; and no others between − π and π . 3
This question in 9709/22 Feb/March 2024
Q84 · Show that 3 tan 2i + tan ( i + 45 °) / i 2 i 9709/21 May/June 2024
4 (a) Show that 3 tan 2i + tan ( i + 45 °) / i 2 i . [4] 1 - tan i … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence solve the equation 3 tan 2i + tan ( i + 45°) = 4 for 0° 1 i 1 180 ° . [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …
7 marks
Mark scheme: 4(a) State 2 6tan 3tan 2 1 tan B1 Allow for 2 2tan 3tan 2 3 1 tan . State tan 1 tan( 45 ) 1 tan B1 Attempt to express left-hand side in terms of tan as a single fraction M1 Condone sign errors in identities. Allow if o tan 45 is not evaluated. Allow if 2 separate terms with a common denominator. Confirm 2 2 tan 8tan 1 1 tan A1 Answer given – necessary detail needed. 4 4(b) Solve quadratic equation 2 5tan 8tan 3 0 to obtain at least one value of M1 Must be using a correct method. Obtain 17.4 A1 Or greater accuracy. Obtain 117.6 A1 Or greater accuracy; and no others between 0 and 180. 3
This question in 9709/21 May/June 2024
Q85 · Prove that 2 sin i cosec 2 i / sec i 9709/22 May/June 2024
7 (a) Prove that 2 sin i cosec 2 i / sec i. [2] … … … … … … … … … … … … … … … … … (b) Solve the equation tan 2i + 7 sin i cosec 2 i = 8 for - r 1 i 1 r . [5] … … … … … … … … … … … … … … … … … … (c) Find 8 sin 2 12 x cosec 2 x d x . [3] y … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 7(a) 1 cosec2 sin 2 M1 Obtain 1 cos and confirm sec A1 Answer given – necessary detail needed. 2 7(b) Attempt to obtain quadratic equation in sec or cos only *M1 Obtain 2 7 2 sec 1 sec 8 involving one trigonometric ratio A1 Or equivalent, may be unsimplified, but reduce to 2 2sec 7sec 18 0 2 18cos 7cos 2 0 . Attempt to solve 3-term quadratic equation for sec, using a correct method, to find at least one value of DM1 Or equivalent using cos. Obtain any two of the four correct solutions 0.952, 1.76 A1 Or greater accuracy. Obtain remaining two correct solutions A1 Or greater accuracy; and no others between π and π . 5 7(c) Identify integrand as 2 1 2 2sec x B1 Integrate 2 1 2 sec k x to obtain 1 2 2 tan k x M1 Obtain correct 1 2 4tan x A1 Condone omission of ...c . 3
This question in 9709/22 May/June 2024
Q86 · Prove that 2 sin i cosec 2 i / sec i 9709/23 May/June 2024
7 (a) Prove that 2 sin i cosec 2 i / sec i. [2] … … … … … … … … … … … … … … … … … (b) Solve the equation tan 2i + 7 sin i cosec 2 i = 8 for - r 1 i 1 r . [5] … … … … … … … … … … … … … … … … … … (c) Find 8 sin 2 1 x cosec 2 x d x . [3] y 2 … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 7(a) 1 cosec2 sin 2 M1 Obtain 1 cos and confirm sec A1 Answer given – necessary detail needed. 2 7(b) Attempt to obtain quadratic equation in sec or cos only *M1 Obtain 2 7 2 sec 1 sec 8 involving one trigonometric ratio A1 Or equivalent, may be unsimplified, but reduce to 2 2sec 7sec 18 0 2 18cos 7cos 2 0 . Attempt to solve 3-term quadratic equation for sec, using a correct method, to find at least one value of *DM1 Or equivalent using cos. Obtain any two of the four correct solutions 0.952, 1.76 A1 Or greater accuracy. Obtain remaining two correct solutions A1 Or greater accuracy; and no others between π and π . 5 7(c) Identify integrand as 2 1 2 2sec x B1 Integrate 2 1 2 sec k x to obtain 1 2 2 tan k x M1 Obtain correct 1 2 4tan x A1 Condone omission of ...c . 3
This question in 9709/23 May/June 2024
Q87 · The polynomial p ( x) is defined by p ( x) = ax 3 - ax 2 - 15 x + 18 , where a is a… 9709/21 Oct/Nov 2024
4 The polynomial p ( x) is defined by p ( x) = ax 3 - ax 2 - 15 x + 18 , where a is a constant. It is given that ( x + 2) is a factor of p ( x) . (a) Find the value of a. [2] … … … … … … … … … … … … (b) Hence factorise p ( x) completely. [3] … … … … … … … … … … … (c) Solve the equation p ( cosec 2i) = 0 for - 90° 1 i 1 90° . [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …
8 marks
Mark scheme: 4(a) Substitute x = −2, equate to zero and attempt solution M1 Obtain a = 4 A1 2 4(b) Divide by x + 2 at least as far as k1 x 2 + k 2 x M1 Obtain 4 x 2 − 12 x + 9 A1 Obtain ( x + 2)(2 x − 3) 2 or equivalent with integer coefficients only A1 3 4(c) Equate sin 2 to appropriate value from factorised form and attempt solution M1 2 Using their . 3 Obtain 54.7 A1 Or greater accuracy. Obtain –54.7 A1 Or greater accuracy. No others in −90 90. 3
This question in 9709/21 Oct/Nov 2024
Q88 · The polynomial p ( )x is defined by p ( )x = ax 3 + bx 2 - ax + 8 , where a and b are… 9709/22 Oct/Nov 2024
5 The polynomial p ( )x is defined by p ( )x = ax 3 + bx 2 - ax + 8 , where a and b are constants. It is given that ( x + 2) is a factor of p ( )x , and that the remainder is 24 when p ( )x is divided by ( x - 2) . (a) Find the values of a and b. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Factorise p ( )x and hence show that the equation p ( )x = 0 has exactly one real root. [3] … … … … … … … … … … … … (c) Solve the equation p b 1 cosec il = 0 for - 90° 1 i 1 90° . [3] 2 … … … … … … … … … … … … … …
10 marks
Mark scheme: 5(a) Substitute x = −2 and equate to zero M1 −8a + 4b + 2a + 8 = 0 Substitute x = 2 and equate to 24 M1 8a + 4b − 2a + 8 = 24 Obtain −6a + 4b + 8 = 0 and 6a + 4b − 16 = 0 A1 OE Obtain a = 2 and b = 1 A1 4 5(b) Divide by x + 2 at least as far as the x term M1 OE Obtain ( x + 2)(2 x 2 − 3 x + 4) A1 SOI Conclude with reference to root –2, discriminant is –23 and no further root A1 Or complete equivalent. 3 5(c) State cosec= −4 B1 1 May be implied by sin= − 4 1 M1 Allow for 14.5, 14.4. Attempt to find at least one value of from sin= 4 Obtain –14.5 only and no others in the range A1 Or greater accuracy (14.4775…). 3
This question in 9709/22 Oct/Nov 2024
Q89 · Express 4 sin i sin ( i+ 60 °) in the form a + R sin ( 2i - a) , where a and R are… 9709/22 Oct/Nov 2024
7 (a) Express 4 sin i sin ( i+ 60 °) in the form a + R sin ( 2i - a) , where a and R are positive integers and 0° 1 a 1 90° . [6] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the smallest positive value of i satisfying the equation 1 + 4 sin i sin ( i + 60°) = 0 . [3] 5 … … … … … … … … … … … … … … … … … … … … … … … … … …
9 marks
Mark scheme: 7(a) Expand to obtain 2sin 2 + 2 3sincos B1 Needs to be simplified but may be implied by later work. Attempt to express in terms of sin2 and cos2 *M1 Needs to be in the form k + p cos2+ q sin2. Obtain 1 − cos2+ 3sin 2 A1 Obtain R = 2 A1 FT Following their p sin2+ q cos2 form. Use appropriate trigonometry to find DM1 For their p sin2+ q cos2 form. Allow for 60 o . Obtain 1 + 2sin(2− 30) A1 Alternative Method for Question 7(a): Expand to obtain 2sin 2 + 2 3sincos B1 Needs to be simplified. For expansion of a + R sin ( 2− ) and comparison with *M1 Need comparison of at least 2 like terms: 2 R sin= 1 , R cos= 3 , a − R sin= 0. 2sin + 2 3sincos R sin= 1 and R cos= 3 leading to values of R and DM1 Obtain R = 2 A1 FT Obtain = 30 o and a = 1 A1 Obtain 1 + 2sin(2− 30) A1 6 7(b) State sin(2− 30) = − 106 *B1 FT Following their answer to part (a). 1 Must be using sin(2− ) = − − their (a). Must have a 5 value for a. Attempt complete method to find smallest positive value DM1 Obtain 123.4 only A1 Or greater accuracy. 3
This question in 9709/22 Oct/Nov 2024
Q90 · The polynomial p ( x) is defined by p ( x) = ax 3 - ax 2 - 15 x + 18 , where a is a… 9709/23 Oct/Nov 2024
4 The polynomial p ( x) is defined by p ( x) = ax 3 - ax 2 - 15 x + 18 , where a is a constant. It is given that ( x + 2) is a factor of p ( x) . (a) Find the value of a. [2] … … … … … … … … … … … … (b) Hence factorise p ( x) completely. [3] … … … … … … … … … … … (c) Solve the equation p ( cosec 2i) = 0 for - 90° 1 i 1 90° . [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …
8 marks
Mark scheme: 4(a) Substitute x = −2, equate to zero and attempt solution M1 Obtain a = 4 A1 2 4(b) Divide by x + 2 at least as far as k1 x 2 + k 2 x M1 Obtain 4 x 2 − 12 x + 9 A1 Obtain ( x + 2)(2 x − 3) 2 or equivalent with integer coefficients only A1 3 4(c) Equate sin 2 to appropriate value from factorised form and attempt solution M1 2 Using their . 3 Obtain 54.7 A1 Or greater accuracy. Obtain –54.7 A1 Or greater accuracy. No others in −90 90. 3
This question in 9709/23 Oct/Nov 2024
Q91 · Express 6 sin i - 4 cos i in the form R sin ( i - a) , where R 2 0 and 0° 1 a 1 90° 9709/22 Feb/March 2025
7 (a) Express 6 sin i - 4 cos i in the form R sin ( i - a) , where R 2 0 and 0° 1 a 1 90° . Give the exact value of R and the value of a correct to 2 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence solve the equation 6 sin i - 4 cos i + 5 = 0 for 0° 1 i 1 360° . [4] … … … … … … … … … … … (c) As the value of b varies, find the greatest possible value of ( 3 sin 4b - 2 cos 4 b) 2 + 15 and determine the smallest positive value of b, in degrees, for which this greatest value occurs. [3] … … … … … … … … … … … … …
10 marks
Mark scheme: 7(a) State R = 52 or 2 13 B1 Use appropriate trigonometry to find M1 Obtain = 33.69 A1 Or greater accuracy. 3 7(b) Use correct method to find one value of M1 Obtain 257.6 (or 349.8) A1 Or greater accuracy 257.587… (or 349.792…). Use correct method to find second value of between 0 and 360 M1 Obtain 349.8 (or 257.6) and no others between 0 and 360 A1 Or greater accuracy. 4 7(c) Obtain greatest value 28 B1 Attempt complete method to find positive value of using M1 sin(4− ) = 1 Obtain 14 (90 + 33.69) and hence 30.9 A1 Or greater accuracy (30.922…). 3
This question in 9709/22 Feb/March 2025
Q92 · The polynomial p( )x is defined by p( )x = ax 3 + bx 2 - ax - 24 , where a and b are… 9709/21 May/June 2025
5 The polynomial p( )x is defined by p( )x = ax 3 + bx 2 - ax - 24 , where a and b are constants. It is given that ( 2x - 3 ) is a factor of p( )x and that the remainder is -15 when p( )x is divided by ( x + 1 ) . (a) Find the values of a and b. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence factorise p( )x completely. [3] … … … … … … … … … … … … … (c) Hence solve the equation p(3 cosec i) = 0 for 90° 1 i 1 270° . [2] … … … … … … … … … … … … …
9 marks
Mark scheme: 5(a) Substitute x = 32 equating to 0, and x = −1 equating to −15 M1 Allow algebraic long division, but must result in two expressions correctly equated to 0 and −15. SC B1 for b = 9 if M0 scored. Obtain 278 a + 94 b − 32 a − 24 = 0 A1 OE Powers of 32 must be evaluated. Obtain −+a b + a − 24 = −15 A1 Solve to obtain a = 2 and b = 9 A1 4 5(b) Divide p( x ) by 2 x − 3 M1 OE method, such as inspection. For algebraic long division, must go as far as the term in x. Allow attempt at synthetic division. 3 2 9 –2 –24 2 3 18 24 2 x 2 + 12 x + 16 Allow one sign error. 2 A1 2 Obtain quotient x + 6 x + 8 Allow 2 x + 12 x + 16 from synthetic division. Conclude (2 x − 3)( x + 2)( x + 4) A1 3 5(c) Use factorised form to determine value of sin, with −1 sin 1 M1 Obtain finally sin= − 34 only and hence angle 228.6 A1 Or greater accuracy 228.590… Ignore solutions outside the range. 2
This question in 9709/21 May/June 2025
Q93 · Prove that sin 2 2x + 4 cos 2 x cos 2x / 4 cos 4 x 9709/21 May/June 2025
7 (a) Prove that sin 2 2x + 4 cos 2 x cos 2x / 4 cos 4 x . [3] … … … … … … … … … … … … … … … (b) Find the set of possible values of the constant k for which the equation sin 2 2x + 4 cos 2 x cos 2x + 5 = k has no real solutions. [2] … … … … … … … … … 1 r 3 (c) Find the exact value of y 1 sin 2 t + 4 cos 2 b 12 tl cos t td . [4] - 3 r … … … … … … … … … … … … … … … … … … … … … … … … … …
9 marks
Mark scheme: 7(a) Use correct identities to express in terms of sin x and cos x M1 Allow for 2sin 2 x cos 2 x + 4cos 2 x cos 2 x − sin 2 x , OE. ( ) Obtain 4sin 2 x cos 2 x + 4cos 2 x (cos 2 x − sin 2 x ) A1 OE 4 A1 Confirm 4cos x 3 7(b) Use identity from part (a) to obtain k 5 or k 9 B1 Obtain k ,5 k 9 B1 2 7(c) 2 1 4 t M1* State or imply integrand is 2cos 2 t or 4cos 2 Use double-angle identity to express in terms of cost M1 Obtain 1 + cost and integrate to obtain t + sin t A1 Use limits to obtain 23 π + 3 A1 4
This question in 9709/21 May/June 2025
Q94 · The polynomial p ( )x is defined by p ( )x = ax 4 + bx 3 + 13 x 2 - 35 x + 15 , where a… 9709/22 May/June 2025
5 The polynomial p ( )x is defined by p ( )x = ax 4 + bx 3 + 13 x 2 - 35 x + 15 , where a and b are constants. It is given that ( 2x - 1 ) and ( x - 3 ) are factors of p ( )x . (a) Find the values of a and b. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence factorise p ( )x . [3] … … … … … … … … … … … … … (c) Find the least positive value of i in radians such that p ( cot2 i) = 0 . [2] … … … … … … … … … … … … …
9 marks
Mark scheme: 5(a) Substitute x = 1 and x = 3, and equate each to zero to produce two equations M1 SC B1 for a correct equation, if M0 otherwise. 2 Obtain 1 a + 1 b = − 3 A1 OE 16 8 4 a + 2b + 12 = 0 Obtain 81a + 27b = −27 A1 OE 3a + b + 1 = 0 Solve simultaneous equations to obtain a = 2 and b = −7 A1 4 5(b) Divide by 2 x 2 − 7 x + 3 or successively by 2 x − 1 and x − 3 M1 OE method, such as inspection. from synthetic division. Obtain quotient x 2 + 5 A1 Condone 2 ( x 2 + 5 ) State fully factorised form (2 x − 1)( x − 3)( x 2 + 5) A1 3 5(c) Attempt solution of at least cot2= 3 M1 Obtain tan 2= 1 and hence = 0.161 A1 Or greater accuracy 0.16087… 3 2
This question in 9709/22 May/June 2025
Q95 · Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1… 9709/22 May/June 2025
7 (a) Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1 90° and k is a constant. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence solve the equation 12 cos 2 z sin ( 2 z+ 30° ) = 5 for 0° 1 z 1 90 ° . [5] … … … … … … … … … … … … … … … … … … … … … … … … …
11 marks
Mark scheme: 7(a) Expand to obtain the form k1 cossin+ k 2 cos 2 M1 2 A1 Obtain 2 3cossin+ 2cos with exact coefficients Obtain 3sin2+ cos2+ 1 A1 FT Following their expression in sin and cos. State R = 2 B1 FT Following their expression in sin2 and cos2. Use appropriate trigonometry to find value of M1 Must be of the form a sin2+ b cos2. Conclude 2cos(2− 60) + 1 A1 6 7(b) Attempt to express equation in form cos(4− ) = k *M1 Use correct process to find one value of DM1 Obtain cos(4− 60) = 1 and hence 32.6 A1 Or greater accuracy 32.6321… 3 Use correct process to find second value of DM1 Obtain 87.4 and no others between 0 and 90 A1 Or greater accuracy 87.3678… 5
This question in 9709/22 May/June 2025
Q96 · The polynomial p ( )x is defined by p ( )x = ax 4 + bx 3 + 13 x 2 - 35 x + 15 , where a… 9709/23 May/June 2025
5 The polynomial p ( )x is defined by p ( )x = ax 4 + bx 3 + 13 x 2 - 35 x + 15 , where a and b are constants. It is given that ( 2x - 1 ) and ( x - 3 ) are factors of p ( )x . (a) Find the values of a and b. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence factorise p ( )x . [3] … … … … … … … … … … … … … (c) Find the least positive value of i in radians such that p ( cot2 i) = 0 . [2] … … … … … … … … … … … … …
9 marks
Mark scheme: 5(a) Substitute x = 1 and x = 3, and equate each to zero to produce two equations M1 SC B1 for a correct equation, if M0 otherwise. 2 Obtain 1 a + 1 b = − 3 A1 OE 16 8 4 a + 2b + 12 = 0 Obtain 81a + 27b = −27 A1 OE 3a + b + 1 = 0 Solve simultaneous equations to obtain a = 2 and b = −7 A1 4 5(b) Divide by 2 x 2 − 7 x + 3 or successively by 2 x − 1 and x − 3 M1 OE method, such as inspection. from synthetic division. Obtain quotient x 2 + 5 A1 Condone 2 ( x 2 + 5 ) State fully factorised form (2 x − 1)( x − 3)( x 2 + 5) A1 3 5(c) Attempt solution of at least cot2= 3 M1 Obtain tan 2= 1 and hence = 0.161 A1 Or greater accuracy 0.16087… 3 2
This question in 9709/23 May/June 2025
Q97 · Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1… 9709/23 May/June 2025
7 (a) Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1 90° and k is a constant. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence solve the equation 12 cos 2 z sin ( 2 z+ 30° ) = 5 for 0° 1 z 1 90 ° . [5] … … … … … … … … … … … … … … … … … … … … … … … … …
11 marks
Mark scheme: 7(a) Expand to obtain the form k1 cossin+ k 2 cos 2 M1 2 A1 Obtain 2 3cossin+ 2cos with exact coefficients Obtain 3sin2+ cos2+ 1 A1 FT Following their expression in sin and cos. State R = 2 B1 FT Following their expression in sin2 and cos2. Use appropriate trigonometry to find value of M1 Must be of the form a sin2+ b cos2. Conclude 2cos(2− 60) + 1 A1 6 7(b) Attempt to express equation in form cos(4− ) = k *M1 Use correct process to find one value of DM1 Obtain cos(4− 60) = 1 and hence 32.6 A1 Or greater accuracy 32.6321… 3 Use correct process to find second value of DM1 Obtain 87.4 and no others between 0 and 90 A1 Or greater accuracy 87.3678… 5
This question in 9709/23 May/June 2025
Q98 · Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1… 9709/25 May/June 2025
7 (a) Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1 90° and k is a constant. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence solve the equation 12 cos 2 z sin ( 2 z+ 30° ) = 5 for 0° 1 z 1 90 ° . [5] … … … … … … … … … … … … … … … … … … … … … … … … …
11 marks
Mark scheme: 7(a) Expand to obtain the form k1 cossin+ k 2 cos 2 M1 2 A1 Obtain 2 3cossin+ 2cos with exact coefficients Obtain 3sin2+ cos2+ 1 A1 FT Following their expression in sin and cos. State R = 2 B1 FT Following their expression in sin2 and cos2. Use appropriate trigonometry to find value of M1 Must be of the form a sin2+ b cos2. Conclude 2cos(2− 60) + 1 A1 6 7(b) Attempt to express equation in form cos(4− ) = k *M1 Use correct process to find one value of DM1 Obtain cos(4− 60) = 1 and hence 32.6 A1 Or greater accuracy 32.6321… 3 Use correct process to find second value of DM1 Obtain 87.4 and no others between 0 and 90 A1 Or greater accuracy 87.3678… 5
This question in 9709/25 May/June 2025
Q99 · Solve the equation cot i tan ( i+ 45 °) = 7 for 0° 1 i 1 90° 9709/21 Oct/Nov 2025
4 Solve the equation cot i tan ( i+ 45 °) = 7 for 0° 1 i 1 90° . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …
5 marks
Mark scheme: 4 Attempt to express equation in terms of tan only M1* 1 tan+ tan45 A1 OE Obtain = 7 tan 1 − tantan45 Simplify to obtain equation 7tan 2 − 6tan+ 1 = 0 A1 Attempt solution of three-term quadratic equation in tan to obtain at least one DM1 value of Obtain tan= 17 (3 2) or equivalent and hence 12.8 and 32.2, and no others in the A1 Or greater accuracy 12.764…, 32.235… given range 5
This question in 9709/21 Oct/Nov 2025
Q100 · Solve the equation 2 tan 2i + 3 sec i = 18 for –180° 1 i 1 180° 9709/22 Oct/Nov 2025
2 Solve the equation 2 tan 2i + 3 sec i = 18 for –180° 1 i 1 180° . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …
5 marks
Mark scheme: 2 Attempt identity involving tan 2 and sec 2 to obtain quadratic equation in sec or *M1 cos Obtain 2sec 2 + 3sec− 20 = 0 or 20cos 2 − 3cos− 2 = 0 A1 Solve three-term quadratic equation to obtain at least one value of DM1 Obtain at least two values of 66.4 and 104.5 A1 Or greater accuracy 66.421…, 104.477… Obtain all four values and no others between –180 and 180 A1 Or greater accuracy. 5
This question in 9709/22 Oct/Nov 2025
Q101 · Solve the equation 2x - 3 = 5 x + 2 9709/23 Oct/Nov 2025
3 (a) Solve the equation 2x - 3 = 5 x + 2 . [3] … … … … … … … … … … … … (b) Hence solve the equation 2 sec i - 3 = 5 sec i + 2 for r 1 i 1 2 r . Give your answer correct to 3 significant figures. [3] … … … … … … … … … … … … … …
6 marks
Mark scheme: 3(a) Solve 2 x −=3 5 x + 2 to obtain − 53 B1 Or exact equivalent. Attempt solution of linear equation where 2x and 5x have different signs M1 Obtain 1 A1 OE 7 Alternative Method for Question 3(a) State or imply non-modulus equation (2 x − 3) 2 = (5 x + 2) 2 B1 Attempt complete solution of three-term quadratic equation M1 Obtain − 53 and 17 or equivalents A1 OE 3 3(b) State or imply cos= − 53 B1 FT Following an appropriate answer from part (a). Attempt correct process for finding third quadrant angle M1 Condone working in degrees for this mark. Obtain 4.07 A1 Or greater accuracy 4.0688… 3
This question in 9709/23 Oct/Nov 2025
![Question 1: (i) By expanding sin(2x + x) and using double-angle formulae, show that sin 3x = 3 sin x −4 sin3x. [5] (ii) Hence show that 13π sin3x dx = …](https://img.pastlit.com/crops/7a956b3c-7353-4f83-b00d-78c47c47f8ad/q7.webp)
![Question 3: (i) Express cos2x in terms of cos 2x. [1] (ii) Hence show that 13π cos2x dx = 1 + 1 √3. [4] 6π 8 0 (iii) By using an appropriate trigonomet…](https://img.pastlit.com/crops/40afc916-026f-4686-9a09-fffcd26bfa94/q6.webp)
1 / 87![Question 5: (i) Prove the identity (cos x + 3 sin x)2 ≡5 −4 cos 2x + 3 sin 2x. [4] (ii) Using the identity, or otherwise, find the exact value of 14π (c…](https://img.pastlit.com/crops/2a2a50d4-ddfb-4b31-9298-333fcdc5103e/q7.webp)
![Question 7: (i) Show that the equation sin(x + 30◦) = 2 cos(x + 60◦) can be written in the form (3 √3) sin x = cos x. [3] (ii) Hence solve the equation…](https://img.pastlit.com/crops/f4198a1f-0cee-4541-bbf6-e9a383d01bad/q4.webp)
2 / 87![Question 9: (i) (a) Prove the identity + sin x sec2x + sec x tan x ≡1 . cos2x (b) Hence prove that 1 sec2x + sec x tan x ≡ 1 −sin x. [3] 1 dy (ii) By d…](https://img.pastlit.com/crops/f4198a1f-0cee-4541-bbf6-e9a383d01bad/q8.webp)
![Question 10: Solve the equation sec x 4 tan2x, giving all solutions in the interval [6] = −2 0◦≤x ≤180◦.](https://img.pastlit.com/crops/4fe5e427-155b-4864-bc8e-84d07929a1fe/q5.webp)
![Question 11: (i) Show that the equation 2 sin x can be written in the form tan x k, where k is a sin(60◦−x) = = constant. [4] (ii) Hence solve the equat…](https://img.pastlit.com/crops/022fd03f-43a0-4b00-9b56-653d6b52dad5/q4.webp)
![Question 12: (i) Express cos2 2x in terms of cos 4x. [2] 18π (ii) Hence find the exact value of cos2 2x dx. [4] ã 0](https://img.pastlit.com/crops/022fd03f-43a0-4b00-9b56-653d6b52dad5/q5.webp)
![Question 14: (i) Show that the equation 6 tan x can be written in the form tan(x + 45◦) = 6 tan2x tan x 1 0. −5 + = [3] (ii) Hence solve the equation [3…](https://img.pastlit.com/crops/5f1a17a1-a695-4b99-96d7-9a9c5c93fb58/q3.webp)
3 / 87![Question 15: (i) Prove the identity sin x. sin(x −30◦) + cos(x −60◦) ≡(√3) [3] (ii) Hence solve the equation 1 2 sec x, sin(x −30◦) + cos(x −60◦) = for …](https://img.pastlit.com/crops/a41c4a4d-3d65-4642-a77e-cc2d1613f35a/q8.webp)
![Question 16: Solve the equation 8 cot θ 2 cosec2θ, giving all solutions in the interval [6] + = 0◦≤θ ≤360◦.](https://img.pastlit.com/crops/9f23fc01-0acc-43bc-b604-48e2a8de678c/q5.webp)
![Question 17: Solve the equation 8 cot θ 2 cosec2θ, giving all solutions in the interval [6] + = 0◦≤θ ≤360◦.](https://img.pastlit.com/crops/a55466cd-b8d6-4219-b467-70415da7e069/q5.webp)
![Question 18: (a) Find dx. [2] ã e1−2x (b) Express sin23x in terms of cos 6x and hence find sin23x dx. [4] ã](https://img.pastlit.com/crops/fcc42d41-5740-4929-b684-4c2a07f82c67/q4.webp)
4 / 87![Question 21: (i) Prove that cos 2θ. [3] sin22θ(cosec2θ −sec2θ) ≡4 (ii) Hence (a) solve for equation 3, [4] 0◦≤θ ≤180◦the sin22θ(cosec2θ −sec2θ) = (b) fin…](https://img.pastlit.com/crops/464644a3-9acd-4034-b173-c915ae63376d/q8.webp)
![Question 22: (i) Prove that cos 2θ. [3] sin22θ(cosec2θ −sec2θ) ≡4 (ii) Hence (a) solve for equation 3, [4] 0◦≤θ ≤180◦the sin22θ(cosec2θ −sec2θ) = (b) fin…](https://img.pastlit.com/crops/2672de66-a34a-4b21-9044-ba1bc2d34e4d/q8.webp)
![Question 23: (i) By first expanding show that cos(2x + x), cos 3x cos3x cos x. ≡4 −3 [5] (ii) Hence show that 16π cos3x dx 12.5 ã 0 (2 −cos x) = [5]](https://img.pastlit.com/crops/4cc97a69-9a61-42c3-91ee-3b299f1ace0d/q8.webp)
![Question 24: (i) By sketching a suitable pair of graphs, show that the equation 1 sin x, x = [2] where x is in radians, has only one root for 0 x < ≤12π…](https://img.pastlit.com/crops/d0387d23-e673-4ca5-8723-81912b69fb8c/q5.webp)
5 / 87![Question 26: (i) Express cos2x in terms of cos 2x. [1] (ii) Hence show that 16π 1 1 sin dx 1 √3 8 12π 4. + + [5] ã 0 (cos2x + 2x) =](https://img.pastlit.com/crops/2f40fd85-e003-4715-9aec-848937714507/q4.webp)
![Question 27: (i) Given that 35 sec2θ 12 tan θ, find the value of tan θ. [3] + = (ii) Hence, showing the use of an appropriate formula in each case, find t…](https://img.pastlit.com/crops/57e0a1ae-48ed-409c-83a9-5615a6e7d331/q4.webp)
![Question 28: (i) Show that sin x cos can be written in the form 5 2 sin 2x cos 2x. [5] 2 2 (2 + x)2 + −3 14π (ii) Hence find the exact value of sin x cos…](https://img.pastlit.com/crops/57e0a1ae-48ed-409c-83a9-5615a6e7d331/q7.webp)
![Question 29: α4 (i) Express 9 sin θ 0 and Give the value cos θ in the form R where R −12 sin(θ −α), > < 90◦. 0◦< of α correct to 2 decimal places. [3] H…](https://img.pastlit.com/crops/28e6bda5-823c-479d-ba2a-ebb4359da100/q4.webp)
6 / 87![Question 30: (i) Show that tan2x cos2x 1 cos 2x and hence find the exact value of + ≡sec2x + 2 −12 14π dx. ã 0 (tan2x + cos2x) [7] (ii) y 1 x O 4p 0 is s…](https://img.pastlit.com/crops/28e6bda5-823c-479d-ba2a-ebb4359da100/q7.webp)
![Question 31: (i) Given that 35 sec2θ 12 tan θ, find the value of tan θ. [3] + = (ii) Hence, showing the use of an appropriate formula in each case, find t…](https://img.pastlit.com/crops/6ccf1951-906b-4e45-b3d8-7a79988b53f5/q4.webp)
![Question 32: (i) Show that sin x cos can be written in the form 5 2 sin 2x cos 2x. [5] 2 2 (2 + x)2 + −3 14π (ii) Hence find the exact value of sin x cos…](https://img.pastlit.com/crops/6ccf1951-906b-4e45-b3d8-7a79988b53f5/q7.webp)
![Question 33: Solve the equation 2 cos 2θ 4 cos θ = −3, for [4] 0◦≤θ ≤180◦.](https://img.pastlit.com/crops/d881e80a-3799-4e14-b14b-ea95d9aea1ca/q3.webp)
![Question 34: (a) Given that tan A t and 4, find tan B in terms of t. [3] = tan(A + B) = (b) Solve the equation 2 3 tan x, tan(45◦−x) = giving all solutio…](https://img.pastlit.com/crops/d37bbec1-1777-407a-8af2-ca11cc2ca40a/q8.webp)
7 / 87![Question 35: Solve the equation 2 cos 2θ 4 cos θ = −3, for [4] 0◦≤θ ≤180◦.](https://img.pastlit.com/crops/128f0754-dc4a-46f9-a92f-f72c9d0a8687/q3.webp)
![Question 36: (i) Show that 12 sin2x cos2x [3] ≡32(1 −cos 4x). (ii) Hence show that 13π π 3√3 12 sin2x cos2x dx . ã 1 = 8 + 16 [3] 4π](https://img.pastlit.com/crops/8fff98ef-f907-4ca1-b8f4-5adfd430bc1b/q3.webp)
![Question 37: (i) By sketching a suitable pair of graphs, show that the equation cot x 4x = −2, where x is in radians, has only one root for 0 2π. [2] ≤x…](https://img.pastlit.com/crops/8fff98ef-f907-4ca1-b8f4-5adfd430bc1b/q6.webp)
![Question 39: (i) Show that 12 sin2x cos2x [3] ≡32(1 −cos 4x). (ii) Hence show that 13π π 3√3 12 sin2x cos2x dx . ã 1 = 8 + 16 [3] 4π](https://img.pastlit.com/crops/0376ca22-09a5-4e10-b2f4-446915b14e89/q3.webp)
8 / 87![Question 40: (i) By sketching a suitable pair of graphs, show that the equation cot x 4x = −2, where x is in radians, has only one root for 0 2π. [2] ≤x…](https://img.pastlit.com/crops/0376ca22-09a5-4e10-b2f4-446915b14e89/q6.webp)
![Question 42: Solve the equation 2 cosec 10, giving all solutions in the interval [6] cot21 −5 1 = 0Å ≤1 ≤360Å.](https://img.pastlit.com/crops/85ee4199-1adc-431b-ad49-a6ac407b6913/q3.webp)
![Question 43: 5 (i) Prove that tan cot [3] sin 1 + 1 21. (ii) Hence (a) find the exact value of tan 1 cot 1 [2] 80 + 80, 1 20 6 (b) evaluate [3] tan cot…](https://img.pastlit.com/crops/846eaef2-9a62-40d8-bf14-9a2f6d0114c7/q5.webp)
![Question 44: Solve the equation 3 sin tan 2 for [4] 21 1 = 0Å < 1 < 180Å.](https://img.pastlit.com/crops/d778ed31-64aa-473d-b13d-3e4860f35afc/q2.webp)
![Question 45: Solve the equation 3 sin tan 2 for [4] 21 1 = 0Å < 1 < 180Å.](https://img.pastlit.com/crops/23ca553e-eef3-47cc-8f0b-10190d0a3d31/q2.webp)
9 / 87![Question 46: (a) Find 4 cos2 1 [3] Ó 21 d1. 6 1 (b) Find the exact value of dx. [4] Ô 2x 3 −1 +](https://img.pastlit.com/crops/1f45d539-201c-4c61-b287-e5a3a3875ca3/q3.webp)
![Question 47: (i) Prove that 2 cosec tan sec2 [3] 21 1 1. (ii) Hence (a) solve the equation 2 cosec tan 5 for 0 [3] 21 1 = < 1 < 0, 160 (b) find the exa…](https://img.pastlit.com/crops/0701d7ce-784e-433d-aa35-8ba32c7d29a4/q6.webp)
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![Question 50: (i) Show that sin 2x cotx 2 cos2x. [2] (ii) Using the identity in part (i), (a) find the least possible value of 3 sin 2x cotx 5 cos 2x 8 …](https://img.pastlit.com/crops/5bc1803a-27ac-4e8c-a820-e34648f3562a/q8.webp)
![Question 51: Solve the equation 5 tan 4 cot for [5] 21 = 1 0Å < 1 < 180Å.](https://img.pastlit.com/crops/31668970-5b37-4357-808b-0d4019b02891/q2.webp)
![Question 52: (i) Solve the equation 3u 1 2u . [3] + = −5 (ii) Hence solve the equation 3 cotx 1 2 cotx for 0 x 1 giving your answer correct to 3 signific…](https://img.pastlit.com/crops/b27b0693-bda4-4366-8e65-8375d61cb82a/q3.webp)
11 / 87![Question 53: cos 2x 9 cosx 5 5 (i) Show that 2 cos x 1. [3] + + cosx 4 + + 0 cos 4x 9 cos 2x 5 (ii) Hence find the exact value of dx. [4] + + Ô cos 2x …](https://img.pastlit.com/crops/e9a82e13-6422-4f37-ad7a-3dd180eeb165/q5.webp)
12 / 87![Question 54: (i) Given that tan cot 8, show that 34. [3] 21 1 = tan21 = ................................................................................…](https://img.pastlit.com/crops/22064034-148b-41fa-bb16-9ac3eedf46a9/q2.webp)
13 / 87![Question 55: Solve the equation 3 cosec for [5] sec21 = 1 0Å < 1 < 180Å. ...............................................................................…](https://img.pastlit.com/crops/462484bd-3fbd-4e23-a054-8e366851bd81/q3.webp)
14 / 87![Question 56: Solve the equation 3 cosec for [5] sec21 = 1 0Å < 1 < 180Å. ...............................................................................…](https://img.pastlit.com/crops/c5d77967-98d5-45b7-a3d5-06ae66fa25cc/q3.webp)
15 / 87![Question 57: Solve the equation sec2 tan2 5 tan 4 for Show all necessary working. [4] 1 + 1 = 1 + 0Å < 1 < 180Å. .......................................…](https://img.pastlit.com/crops/22562e63-cbb0-4b37-acd0-ebc392e55970/q1.webp)
16 / 87![Question 58: (a) Find tan2 3x dx. [3] Ó ................................................................................................................…](https://img.pastlit.com/crops/4e7a1150-e900-4261-b428-38edfacec7d8/q4.webp)
17 / 87![Question 59: (a) (i) Express 4 sin 4 cos in the form R sin , where R 0 and [3] 1 + 1 1 + ! > 0Å < ! < 90Å. .............................................…](https://img.pastlit.com/crops/5c6c1c8f-3fcd-45cb-a9e4-ba8bed272eb8/q7.webp)
![Question 60: (a) (i) Express 4 sin 4 cos in the form R sin , where R 0 and [3] 1 + 1 1 + ! > 0Å < ! < 90Å. .............................................…](https://img.pastlit.com/crops/a72e277c-2a6b-4727-b8e9-cef1075cd717/q7.webp)
![Question 61: (a) Show that 3 sin cot 6 [2] 21 1 cos21. ...............................................................................................…](https://img.pastlit.com/crops/24c4e74e-6ca8-4ae3-8e34-1a734d8e7a2f/q8.webp)
![Question 62: (a) Show that 3 sin cot 6 [2] 21 1 cos21. ...............................................................................................…](https://img.pastlit.com/crops/ccb7c0b2-0aa6-4725-8aea-c6b029b66087/q8.webp)
![Question 63: It is given that 3 sin cos where is an angle such that 21 = 1 1 0Å < 1 < 90Å. (a) Find the exact value of sin [2] 1. ......................…](https://img.pastlit.com/crops/544ff53d-81ac-49c6-b4b8-e004692f0604/q6.webp)
26 / 87![Question 64: Solve the equation 7 cot 3 cosec for [3] 1 = 1 0Å < 1 < 90Å. ..............................................................................…](https://img.pastlit.com/crops/57a73ab9-f9ee-4b14-827e-a02af0d684a2/q1.webp)
27 / 87![Question 65: It is given that 3 sin cos where is an angle such that 21 = 1 1 0Å < 1 < 90Å. (a) Find the exact value of sin [2] 1. ......................…](https://img.pastlit.com/crops/b921a5be-d042-416c-8f54-b2ca5392efca/q6.webp)
28 / 87![Question 66: Solve the equation cot 8 for 0 [5] sec21 1 = < 1 < π. .....................................................................................…](https://img.pastlit.com/crops/7395370d-46b0-4a0d-8306-981f2d400ed7/q2.webp)
29 / 87![Question 67: (a) Express 5 3 cos x 5 sin x in the form R cos x , where R 0 and 0 1 [3] + −! > < ! < 2π. ................................................…](https://img.pastlit.com/crops/7395370d-46b0-4a0d-8306-981f2d400ed7/q7.webp)
![Question 70: (a) Show that sin cot 1. [3] 21 1 −cos 21 ...............................................................................................…](https://img.pastlit.com/crops/78e0ff9a-4191-46fc-b902-529d83907108/q4.webp)
36 / 87![Question 71: (a) Express the equation 7 tan 4 cot sec 0 in terms of sin only. [3] 1 + 1 −13 1 = 1 ......................................................…](https://img.pastlit.com/crops/92123ff4-bddc-4cb2-9dee-929a5c54ffc4/q2.webp)
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38 / 87![Question 73: Solve the equation sec 1 = 5 cosec 1 for 0Å < 1 < 360Å. [4] ...............................................................................…](https://img.pastlit.com/crops/50af6ad3-d3e8-4fa4-b9e6-3d24bf0e9834/q1.webp)
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![Question 75: (a) Express 7 cos 24 sin in the form R cos , where R 0 and Give the value of correct to12+decimal1 places. 1 −! > 0Å < ! < 90Å. [3] ! .....…](https://img.pastlit.com/crops/b65f6d56-c748-46d8-b545-48f4caab6779/q7.webp)
![Question 76: Solve the equation 5 9 17 sec sec21 + tan21 = + 1 for [5] 0Å < 1 < 360Å. ..................................................................…](https://img.pastlit.com/crops/991f44d7-ca99-4c8d-b0b8-16fa4e0a9c68/q1.webp)
44 / 87![Question 77: (a) Show that 4 sin 1 cos 3 2 sin [4] 1 + 3π 1 −13π + 21. ...............................................................................…](https://img.pastlit.com/crops/991f44d7-ca99-4c8d-b0b8-16fa4e0a9c68/q6.webp)
![Question 78: Solve the equation 5 9 17 sec sec21 + tan21 = + 1 for [5] 0Å < 1 < 360Å. ..................................................................…](https://img.pastlit.com/crops/f634b703-55aa-4335-bb33-a453d78b85be/q1.webp)
47 / 87![Question 79: (a) Show that 4 sin 1 cos 3 2 sin [4] 1 + 3π 1 −13π + 21. ...............................................................................…](https://img.pastlit.com/crops/f634b703-55aa-4335-bb33-a453d78b85be/q6.webp)
![Question 80: It is given that is an acute angle in degrees such that sin 23. Find the exact value of sin 60 . [3] + ....................................…](https://img.pastlit.com/crops/ce64d611-8ee5-4767-a5d0-855ff9999d9a/q1.webp)
50 / 87![Question 81: (a) Show that cosec 3 sin 2 4 sin3 4 6 cos 4 cos2 . [3] ...................................................................................…](https://img.pastlit.com/crops/ce64d611-8ee5-4767-a5d0-855ff9999d9a/q6.webp)
53 / 87![Question 83: (a) Prove that sin 2i ( a cot i + b tan i) / a + b + ( a - b) cos 2i, where a and b are constants. [4] ....................................…](https://img.pastlit.com/crops/d6d070e0-1f94-4d0e-a74c-fedc8385abe8/q7.webp)
![Question 84: (a) Show that 3 tan 2i + tan ( i + 45 °) / i 2 i . [4] 1 - tan i ..........................................................................…](https://img.pastlit.com/crops/186dcd81-ff6a-4f9c-a118-7635815fad32/q4.webp)
![Question 85: (a) Prove that 2 sin i cosec 2 i / sec i. [2] .............................................................................................…](https://img.pastlit.com/crops/86cc7f75-06a5-4c35-921c-848bc20ba024/q7.webp)
![Question 86: (a) Prove that 2 sin i cosec 2 i / sec i. [2] .............................................................................................…](https://img.pastlit.com/crops/996c2aaa-eb6e-457e-89c4-427f68d500be/q7.webp)
![Question 89: (a) Express 4 sin i sin ( i+ 60 °) in the form a + R sin ( 2i - a) , where a and R are positive integers and 0° 1 a 1 90° . [6] ...........…](https://img.pastlit.com/crops/a9cdb0c5-a86e-4fc4-ae14-8f5be763c246/q7.webp)
![Question 93: (a) Prove that sin 2 2x + 4 cos 2 x cos 2x / 4 cos 4 x . [3] ..............................................................................…](https://img.pastlit.com/crops/475e4c1c-252b-426b-a7a8-783031f129b9/q7.webp)
![Question 95: (a) Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1 90° and k is a constant. [6] ...................…](https://img.pastlit.com/crops/28b52077-78a4-4ee1-9e5c-e7ffbd38b216/q7.webp)
![Question 97: (a) Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1 90° and k is a constant. [6] ...................…](https://img.pastlit.com/crops/5f38ec11-22c9-4ea4-9e10-51d96dd30b0a/q7.webp)
![Question 98: (a) Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1 90° and k is a constant. [6] ...................…](https://img.pastlit.com/crops/e9bffbf7-007d-47a6-97ad-51b6ca69f755/q7.webp)
![Question 99: Solve the equation cot i tan ( i+ 45 °) = 7 for 0° 1 i 1 90° . [5] ........................................................................…](https://img.pastlit.com/crops/b3391461-fd40-4d24-8c11-cb80687bb14d/q4.webp)
85 / 87![Question 100: Solve the equation 2 tan 2i + 3 sec i = 18 for –180° 1 i 1 180° . [5] .....................................................................…](https://img.pastlit.com/crops/d398e996-9105-41cd-b602-6c67d9fe52eb/q2.webp)
86 / 87![Question 101: (a) Solve the equation 2x - 3 = 5 x + 2 . [3] .............................................................................................…](https://img.pastlit.com/crops/9ff6723d-ee18-4484-99fa-4e3de655c115/q3.webp)
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