Trigonometry — Paper 2 · A Level Mathematics 9709

7 y M 1 x O 2p The diagram shows the curve y x2 cos x, for 0 2π, and its maximum point M. = ≤x ≤1 (i) Show by differentiation that the x-coordinate of M satisfies the equation 2 tan x x. = [4] (ii) Verify by calculation that this equation has a root (in radians) between 1 and 1.2. [2] 2 (iii) Use the iterative formula tan−1 to determine this root correct to 2 decimal places. xn+1 = xn Give the result of each iteration to 4 decimal places. [3]

9 marks

Mark scheme: 7 (i) Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and express tan x in terms of x M1 Obtain given answer A1 [4] 2 (ii) Consider sign of tan x – at x = 1 and x = 1.2, or equivalent M1 x Complete the argument with correct calcuations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.08 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.075, 1.085) A1 [3] 1

This question in 9709/22 Oct/Nov 2009

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 9709/22 May/June 2010

5 Solve the equation 5 sec22θ tan 2θ 9, giving all solutions in the interval [6] = + 0◦≤θ ≤180◦.

6 marks

Mark scheme: 5 Use trig identity correctly to obtain a quadratic in tan 2θ M1 Solve the quadratic correctly M1 4 Obtain tan 2θ = 1 or – A1 5 Obtain one correct answer A1 Carry out correct method for second answer from either root M1 Obtain remaining 3 answers from 22.5°, 112.5°, 70.7°, 160.7° and no others in the range A1 [Ignore answers outside the given range] [6] GCE AS/A LEVEL – October/November 2011 9709 23

This question in 9709/23 Oct/Nov 2011

5 y B (q, cos q ) C R x O A 12p The diagram shows the curve y cos x, for 0 2π. A rectangle OABC is drawn, where B is the = ≤x ≤1 point on the curve with x-coordinate θ, and A and C are on the axes, as shown. The shaded region R is bounded by the curve and by the lines x θ and y 0. = = (i) Find the area of R in terms of θ. [2] (ii) The area of the rectangle OABC is equal to the area of R. Show that 1 θ θ . −sinθ cos = [1] 1 θn (iii) Use the iterative formula −sin , with initial value θ1 0.5, to determine the value θn+1 = cos θn = of θ correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

6 marks

Mark scheme: 5 (i) Attempt to integrate and use limits θ and π M1 Obtain 1– sin θ A1 [2] (ii) State that area of rectangle = θcos θ, equate area of rectangle to area of R and rearrange to given equation B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.56 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.555, 0.565) B1 [3]

This question in 9709/23 Oct/Nov 2012

3 (i) Solve the equation 3u 1 2u . [3] + = −5 (ii) Hence solve the equation 3 cotx 1 2 cot x 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 9709/23 May/June 2016

7 (i) Show that 2 cosec cot [3] 21 1  cosec21. … … … … … … … … … … … … … (ii) Hence show that tan 4. [2] cosec215Å 15Å = … … … … … … … … … … … (iii) Solve the equation 2 cosec cot 1 cosec 1 12 for Show all necessary working. & 2& + 2& = −360Å < & < 360Å. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(i) State or imply 1 cosec2 2sin cos θ θ θ = B1 Attempt to express left-hand side in terms of sinθ and cosθ only M1 Simplify to confirm 2 cosec θ AG A1 3 7(ii) Use identity to express left-hand side in terms of sin30 or cosec30 M1 Obtain 2 sin30 or 2cosec30 and confirm 4 AG A1 2 7(iii) Solve quadratic equation of the form 2 cosec cosec 12 0 2 2 k φ φ + − = or *M1 Allow sign errors 2 12sin sin 0 2 2 k φ φ − − = correctly for 1 2 cosec φ or 1 2 sin φ to find two values of 1 2 sin φ or 1 2 cosec φ Obtain 1 1 1 2 4 3 sin , φ = − A1 Use correct process to find at least one correct value of φ from 1 1 1 2 4 3 sin , φ = ± ± DM1 Allow for any rounded or truncated value Obtain any two of –331.0, –29.0, 38.9, 321.1 A1 Allow greater accuracy Obtain all four values and no others between –360 and 360 A1 Allow greater accuracy 5

This question in 9709/21 May/June 2019

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 9709/23 May/June 2021

7 (a) By first expanding cos 2 , show that cos 3 4 cos3 3 cos . [3] … … … … … … … … … … … … … (b) Find the exact value of 2 cos3 5 cos 5 . [2] 18π −32 18π … … … … … … … … … … (c) Find 12 cos3x cos3 3x dx. [4] −4 … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) θ θ θ θ − B1 Attempt correct relevant identities to express in terms of cosθ only M1 M0 if moving terms from side to side. Confirm 3 4cos 3cos θ θ − with sufficient detail A1 AG 3 7(b) Use identity with 5 18 π θ = M1 Obtain 5 1 2 6 cos π and hence 1 4 3 − A1 2 7(c) Express integrand in form 1 2 (cos3 3cos ) (cos9 3cos3 ) + + + k x x k x x M1 Obtain correct integrand 9cos cos9 − x x A1 OE (allow unsimplified). Integrate to obtain form 3 4 sin sin9 + k x k x M1 Obtain correct 1 9 9sin sin9 − x x A1 Now simplified; condone missing ... + c . 4

This question in 9709/23 Oct/Nov 2021

7 (a) Prove that cos ( i + 30°) cos ( i + 60°) / 1 3 - 1 sin 2 i . [4] 4 2 … … … … … … … … … … … … … … … (b) Solve the equation 5 cos ( 2 a + 30°) cos ( 2 a + 60 °) = 1 for 0° 1 a 1 90° . [4] … … … … … … … … … … … … … … … … … … … … … (c) Show that the exact value of cos 20° cos 50° + cos 40° cos 70° is 1 3 . [3] 2 … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) State (coscos30 − sinsin30)(coscos60 − sinsin60) B1 Expand and use correct exact values M1 Obtain 1 4 3(cos 2 + sin 2 ) − sincos or similarly simplified equivalent A1 Conclude 1 4 3 − 12 sin2 A1 AG – necessary detail needed. 4 7(b) Use identity to obtain value for sin4 *M1 Obtain sin4= 12 3 − 52 or 0.466… A1 Show correct process to obtain one value of  DM1 Obtain 6.9 and 38.1 A1 Or greater accuracy; and no others between 0 and 90. 4 7(c) Substitute = −10 to obtain cos20cos50 = 14 3 − 12 sin(−20) B1 3 1 B1 Substitute = 10 to obtain cos40cos70 = − sin20 4 2 Add and confirm 1 2 3 with clear indication that sin( −20) = − sin20 B1 AG – necessary detail needed. Alternative solution for Question 7(c) Rewrite as sin70cos50 + sin50cos70 or cos20sin40 + cos40sin20 or B1 sin70sin40 + cos70cos40 Obtain sin120 or sin60 or cos30 B1 Confirm 1 2 3 B1 AG – necessary detail needed. 3

This question in 9709/21 Oct/Nov 2024

7 (a) Prove that cos ( i + 30°) cos ( i + 60°) / 1 3 - 1 sin 2 i . [4] 4 2 … … … … … … … … … … … … … … … (b) Solve the equation 5 cos ( 2 a + 30°) cos ( 2 a + 60 °) = 1 for 0° 1 a 1 90° . [4] … … … … … … … … … … … … … … … … … … … … … (c) Show that the exact value of cos 20° cos 50° + cos 40° cos 70° is 1 3 . [3] 2 … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) State (coscos30 − sinsin30)(coscos60 − sinsin60) B1 Expand and use correct exact values M1 Obtain 1 4 3(cos 2 + sin 2 ) − sincos or similarly simplified equivalent A1 Conclude 1 4 3 − 12 sin2 A1 AG – necessary detail needed. 4 7(b) Use identity to obtain value for sin4 *M1 Obtain sin4= 12 3 − 52 or 0.466… A1 Show correct process to obtain one value of  DM1 Obtain 6.9 and 38.1 A1 Or greater accuracy; and no others between 0 and 90. 4 7(c) Substitute = −10 to obtain cos20cos50 = 14 3 − 12 sin(−20) B1 3 1 B1 Substitute = 10 to obtain cos40cos70 = − sin20 4 2 Add and confirm 1 2 3 with clear indication that sin( −20) = − sin20 B1 AG – necessary detail needed. Alternative solution for Question 7(c) Rewrite as sin70cos50 + sin50cos70 or cos20sin40 + cos40sin20 or B1 sin70sin40 + cos70cos40 Obtain sin120 or sin60 or cos30 B1 Confirm 1 2 3 B1 AG – necessary detail needed. 3

This question in 9709/23 Oct/Nov 2024

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 9709/25 May/June 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 9709/21 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 − tantan45 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 9709/23 Oct/Nov 2025

3 (a) Solve the equation 2x - 3 = 5x + 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 9709/25 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 − tantan45 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 9709/25 Oct/Nov 2025