Trigonometry — Paper 3 · A Level Mathematics 9709

5 By expressing 8 sin θ −6 cos θ in the form R sin(θ −α), solve the equation 8 sin θ −6 cos θ = 7, for 0◦≤θ ≤360◦. [7]

7 marks

Mark scheme: 5 State or imply that R = 10 or R = −10 B1 Use trig formula to find α M1 Obtain α = 36.9° if R = 10 or α = 216.9° if R = −10, with no errors seen A1 Carry out evaluation of sin −1 ( 7 ) (≈ 44.427...°) M1 10 Obtain answer 81.3° A1 Carry out correct method for second answer M1 Obtain answer 172.4° and no others in the range A1 [7] [Ignore answers outside the given range.] GCE A/AS LEVEL – November 2005 9709, 8719 3 dx

This question in 9709/31 Oct/Nov 2005

3 (i) Prove the identity cosec 2θ cot 2θ θ. [3] + ≡cot (ii) Hence solve the equation cosec 2θ cot 2θ 2, for [2] + = 0◦≤θ ≤360◦.

5 marks

Mark scheme: 3 (i) Use cot A = 1/tan A or cos A/sin A and/or cosec A = 1/sin A on at least two terms M1 Use a correct double angle formula or the sin(A – B) formula at least once M1 Obtain given result A1 3 (ii) Solve cot θ = 2 for θ and obtain answer 26.6° B1 Obtain answer 206.6° and no others in the given range B1 √ 2 [Ignore answers outside the given range. Treat answers given in radians as a misread] 3

This question in 9709/31 May/June 2009

9 (i) Prove the identity cos 4θ 4 cos 2θ cos4θ [4] + ≡8 −3. (ii) Hence (a) solve the equation cos 4θ 4 cos 2θ 1 for 2π 2π, [3] + = −1 ≤θ ≤1 14π (b) find the exact value of cos4θ dθ. [3] ã 0 [Question 10 is printed on the next page.]

10 marks

Mark scheme: 9 (i) Express cos 4θ as 2 cos2 2θ – 1 or cos2 2θ – sin2 2θ or 1 – 2 sin2 2θ B1 Express cos 4θ in terms of cosθ M1 Obtain 8 cos4θ – 8 cos2θ + 1 A1 Use cos2θ = 2 cos2θ – 1 to obtain given answer 8 cos4θ – 3 AG A1 [4] (ii) (a) State or imply cos4θ = 12 B1 Obtain 0.572 B1 Obtain –0.572 B1 [3] (b) Integrate and obtain form k1θ + k2 sin 4θ + k3 sin 2θ M1 Obtain 83 θ + 321 sin 4θ + 14 sin 2θ A1 Obtain 323 π + 14 following completely correct work A1 [3]

This question in 9709/31 May/June 2011

α 0 and giving the exact6 (i) Express cos x 3 sin x in the form R where R + cos(x −α), > < 90◦, 0◦< value of R and the value of α correct to 2 decimal places. [3] θ 3 sin 2θ 2, for [5] (ii) Hence solve the equation cos 2θ + = < 90◦. 0◦<

8 marks

Mark scheme: 6 (i) State or imply R = 10 B1 Use trig formulae to find α M1 Obtain α = 71.57° with no errors seen A1 [3] [Do not allow radians in this part. If the only trig error is a sign error in cos(x – α) give M1A0] (ii) Evaluate cos–1 ( 2 / 10 ) correctly to at least 1 d.p. (50.7684…°) (Allow 50.7° here) B1√ Carry out an appropriate method to find a value of 2θ in 0° < 2θ < 180° M1 Obtain an answer for θ in the given range, e.g. θ = 61.2° A1 Use an appropriate method to find another value of 2θ in the above range M1 Obtain second angle, e.g. θ = 10.4°, and no others in the given range A1 [5] [Ignore answers outside the given range.] [Treat answers in radians as a misread and deduct A1 from the answers for the angles.] [SR: The use of correct trig formulae to obtain a 3-term quadratic in tan θ, sin 2θ, cos 2θ,or tan 2θ earns M1; then A1 for a correct quadratic, M1 for obtaining a value of θ in the given range, and A1 + A1 for the two correct answers (candidates who square must reject the spurious roots to get the final A1).] GCE AS/A LEVEL – October/November 2011 9709 31

This question in 9709/31 Oct/Nov 2011

α 0 and giving the exact6 (i) Express cos x 3 sin x in the form R where R + cos(x −α), > < 90◦, 0◦< value of R and the value of α correct to 2 decimal places. [3] θ 3 sin 2θ 2, for [5] (ii) Hence solve the equation cos 2θ + = < 90◦. 0◦<

8 marks

Mark scheme: 6 (i) State or imply R = 10 B1 Use trig formulae to find α M1 Obtain α = 71.57° with no errors seen A1 [3] [Do not allow radians in this part. If the only trig error is a sign error in cos(x – α) give M1A0] (ii) Evaluate cos–1 ( 2 / 10 ) correctly to at least 1 d.p. (50.7684…°) (Allow 50.7° here) B1√ Carry out an appropriate method to find a value of 2θ in 0° < 2θ < 180° M1 Obtain an answer for θ in the given range, e.g. θ = 61.2° A1 Use an appropriate method to find another value of 2θ in the above range M1 Obtain second angle, e.g. θ = 10.4°, and no others in the given range A1 [5] [Ignore answers outside the given range.] [Treat answers in radians as a misread and deduct A1 from the answers for the angles.] [SR: The use of correct trig formulae to obtain a 3-term quadratic in tan θ, sin 2θ, cos 2θ,or tan 2θ earns M1; then A1 for a correct quadratic, M1 for obtaining a value of θ in the given range, and A1 + A1 for the two correct answers (candidates who square must reject the spurious roots to get the final A1).] GCE AS/A LEVEL – October/November 2011 9709 32

This question in 9709/32 Oct/Nov 2011

4 Solve the equation cosec 2θ sec θ cot θ, = + θ giving all solutions in the interval [6] 0◦< < 360◦.

6 marks

Mark scheme: 4 Use trig formulae to express equation in terms of cos θ and sin θ M1 Use Pythagoras to obtain an equation in sin θ M1 Obtain 3-term quadratic 2 sin 2 θ − 2 sin θ − 1 = 0 , or equivalent A1 Solve a 3-term quadratic and obtain a value of θ M1 Obtain answer, e.g. 201.5° A1 Obtain second answer, e.g. 338.5°, and no others in the given interval A1 [6] [Ignore answers outside the given interval. Treat answers in radians (3.52, 5.91) as a misread and deduct A1 from the marks for the angles.]

This question in 9709/32 May/June 2012

5 (i) Prove that cot + tan 2 cosec 2 . [3] 1 3 1 (ii) Hence show that cosec 2 d = ln 3. [4] 1 2 6

7 marks

Mark scheme: 5 (i) Use Pythagoras M1 Use the sin2A formula M1 Obtain the given result A1 [3] (ii) Integrate and obtain a k ln sin θ or m ln cosθ term, or obtain integral of the form p ln tan θ M1* 1 1 1 Obtain indefinite integral ln sin θ − ln cos θ , or equivalent, or ln tan θ A1 2 2 2 Substitute limits correctly M1(dep)* Obtain the given answer correctly having shown appropriate working A1 [4] 2 2 2 ( )

This question in 9709/32 Oct/Nov 2013

5 (i) Prove the identity cos 8 [4] cos41 −4 21  sin41 −3. (ii) Hence solve the equation cos 4 cos 3, 41 = 21 + for [4] 0Å ≤1 ≤360Å.

8 marks

Mark scheme: 5 (i) EITHER: Express cos 4θ in terms of cos 2θ and/or sin 2θ B1 Use correct double angle formulae to express LHS in terms of sin θ and/or cos θ M1 Obtain a correct expression in terms of sin θ alone A1 Reduce correctly to the given form A1 OR: Use correct double angle formula to express RHS in terms of cos 2θ M1 Express cos 2 2θ in terms of cos 4θ B1 Obtain a correct expression in terms of cos 4θ and cos 2θ A1 Reduce correctly to the given form A1 [4] (ii) Use the identity and carry out a method for finding a root M1 Obtain answer 68.5° A1 Obtain a second answer, e.g. 291.5° A1 Obtain the remaining answers, e.g. 111.5° and 248.5°, and no others in the given interval A1 [4] [Ignore answers outside the given interval. Treat answers in radians as a misread.]

This question in 9709/32 May/June 2016

6 (i) By sketching a suitable pair of graphs, show that the equation 1 cosec 2x1 = 13x + has one root in the interval 0 x [2] < ≤0. (ii) Show by calculation that this root lies between 1.4 and 1.6. [2] (iii) Show that, if a sequence of values in the interval 0 x given by the iterative formula < ≤0 @ A 3 2 sin−1 xn+1 = xn 3 + converges, then it converges to the root of the equation in part (i). [2] (iv) Use this iterative formula to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

9 marks

Mark scheme: 6 (i) Make recognizable sketch of a relevant graph B1 Sketch the other relevant graph and justify the given statement B1 [2] (ii) Use calculations to consider the value of a relevant expression at x = 1.4 and x = 1.6, or the values of relevant expressions at x = 1.4 and x = 1.6 M1 Complete the argument correctly with correct calculated values A1 [2] −1 3  (iii) State x = 2sin   B1  x + 3  Rearrange this in the form cosec 12 x = 13 x + 1 B1 [2] x  3  If working in reverse, need sin = for first B1   2  x + 3  (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.471 A1 Show sufficient iterations to 5 d.p. to justify 1.471 to 3 d.p., or show there is a sign change in the interval (1.4705, 1.4715) A1 [3]

This question in 9709/31 Oct/Nov 2016

3 Express the equation cot 1 tan as a quadratic equation in tan Hence solve this equation for 21 = + 1 1. [6] 0Å < 1 < 180Å.

6 marks

Mark scheme: 3 Use the tan 2A formula to obtain an equation in tan θ only M1 Obtain a correct horizontal equation A1 Rearrange equation as a quadratic in tan θ, e.g. 3tan 2 θ+ 2tanθ−=1 0 A1 Solve for θ (usual requirements for solution of quadratic) M1 Obtain answer, e.g. 18.4° A1 Obtain second answer, e.g. 135° , and no others in the given interval A1 [6] 2

This question in 9709/33 Oct/Nov 2016

cos 3x sin 3x 5 (a) Show that 2 cot 2x. [4] sin x + cos x = … … … … … … … … … … … … … … … … … … … … … … … … … cos 3x sin 3x (b) Hence solve the equation 4, for 0 x [3] sin x + cosx = < < π. … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) Express LHS correctly as a single fraction B1 Use ( ) cos ± A B formula to simplify the numerator M1 Use sin 2A formula to simplify the denominator M1 Obtain the given result. A1 4 Question Answer Marks Guidance 5(b) Obtain an equation in tan2x and use correct method to solve for x M1 Obtain answer, e.g. 0.232 A1 Obtain second answer, e.g. 1.80 A1 Ignore answers outside the given interval. 3

This question in 9709/32 Feb/March 2020

6 (a) Show that the equation cot21 + 2 cos 21 = 4 can be written in the form 4 sin41 + 3 sin21 −1 = 0. [3] … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence solve the equation cot21 + 2 cos 21 = 4, for 0Å < 1 < 360Å. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 6(a) Use correct Pythagoras cot2 = cosec2– 1 or cot2 = 1/sin2– 1 M1 If consistent omission of brackets, e.g. (sinθ)2 written as or cot²= cos²/sin² and then cos²= 1 – sin², sinθ2 then SC B1 in place of M1A1. together with double angle formula cos2 = 1 – 2sin2, to obtain an equation in sin 𝜃 or sin 𝜃 and cosec2 Obtain a correct equation in sin 𝜃 in any form A1 e.g. 1/sin2 − 1 + 2(1 – 2sin2) = 4 1– sin² 2 or + 2 1– 2sin = 4 . ( ) sin² cos² 2 If + 2 1– 2sin = 4 then ( ) sin² e.g. 1 − sin² + 2 1– 2sin 2 sin² = 4 . ( ) (missing sin 2 on right) allow M1A1A0. Reduce to the given answer of 4sin 4 + 3sin 2 −=1 0 correctly A1 AG Must follow from a horizontal equation (no denominators). If s = sin 𝜃 used and defined, allow all marks. If not defined, award M1A1A0. 3 6(b) Solve the given quadratic to obtain a value for 𝜃 M1 (4sin2 − 1)(sin2 + 1) = 0 and solve for 𝜃. Incorrect sign in solution of quadratic seen, e.g. (4sin2 − 1)(sin2 – 1) = 0 then M0 A0 A0 but if only see (4sin2− 1) = 0 and nothing incorrect seen allow 3/3. Obtain answer, e.g. 𝜃 = 30° A1 /6 award A0 Obtain three further answers, e.g. 𝜃 = 150°, 210° and 330° and no others A1 Ignore any answers outside interval. in the interval 5/6 7/6 11/6 award A1. 3

This question in 9709/33 Oct/Nov 2023

4 By first expressing the equation tan ( x - 60°) = 2 cot x as a quadratic equation in tanx, solve the equation for 0° G x G 180° . [6] … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4 tan x − 3 B1 OE State Allow decimals throughout. 1 + 3 tan x 2 B1 SOI 2 cot x replaced by tan x B1 May be implied by further work. Reduce the equation to tan² x – 3 3 tan x – 2 = 0, or three-term equivalent Solve a three-term quadratic in tan x, for x M1 FT their 3-term quadratic. Allow tan–1(..). Obtain answer, e.g. 79.8° A1 AWRT 79.8. Obtain the second answer, e.g. 160.2° and no other in the interval A1 Allow 160, or AWRT 160.2. Treat answers in radians as a misread. Ignore answers outside the given interval. 6

This question in 9709/32 Feb/March 2025

5 (a) Show that cos 4x + 2 sin 2 x - 1 / 8 sin 4 x - 6 sin 2 x . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence solve the equation cos 4x + 2 sin 2 x - 1 = 0 for - 180° G x G 180° . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Use correct double angle formula to express cos 4x in terms of sin2 2x B1 Need to see 1 – 2 sin2 2x or (1 – sin2 2x) – sin2 2x. May be implied by further work. Use correct double angle formula to express cos 4x in terms of single angles M1 Allow with cos x but not cos 2x. E.g. 1 – 2(2sin x cos x)2 or 2(1 – 2sin2 x)2 – 1. Obtain a correct expression in sin x in any form A1 E.g. 1 – 2[4 sin² x(1 – sin² x)] + 2 sin² x – 1, or 2(1 – 2 sin2 x)2 – 1 + 2 sin² x – 1. Obtain the given answer 8 sin4 x – 6 sin² x A1 AG Must show at least one intermediate line of working including sin4 x between first A1 and AG. Allow, e.g., A for x except in the final answer. Allow recovery on the next line after a slip. Allow recovery on the next line after missing x. Alternative Method for Question 5(a) Use correct double angle formula to express whole expression in terms of co s2x M1 E.g. (1 – cos 2x)(2 – 2 cos 2x – 3). Use correct double angle formula to express 2 cos2 2x – 1 as cos 4x B1 Use correct double angle formula to express whole expression as cos 4x – cos2x A1 Obtain the given answer cos 4x + 2sin2 x – 1 A1 AG Allow, e.g., A for x except in the final answer. Allow recovery on the next line after a slip. Allow recovery on the next line after missing x. 4 5(b) Obtain answers 0°, 180° and –180° B1 Carry out a correct method to find a value of x in the given interval for M1 3 8 sin2 x – 6 = 0 Condone a wrong value of x if sin x = OE 4 seen. Allow M1A1A1 if dividing by sin2 x, but B1 is not scored. Obtain answer, e.g. 60° A1 In radians, would be 13 π. Obtain remaining answers, e.g. –60°, 120° and –120° and no other in the interval A1 Ignore answers outside the given interval. 4

This question in 9709/33 Oct/Nov 2025