C3 Past Trig Questions

1. (a) Show that

tan2cos1

2sin

(2)

(b) Hence find, for -180° θ < 180°, all the solutions of

12cos1

2sin2

Give your answers to 1 decimal place.(3)

(Total 5 marks)

2. (a) Express 2 sin θ – 1.5 cos θ in the form R sin (θ – α), where R > 0 and 0 < α < .2

Give the value of α to 4 decimal places.(3)

(b) (i) Find the maximum value of 2 sin θ – 1.5 cos θ.

(ii) Find the value of θ, for 0 θ < , at which this maximum occurs.(3)

Tom models the height of sea water, H metres, on a particular day by the equation

,120,25

4cos5.125

4sin26

tttH

where t hours is the number of hours after midday.

(c) Calculate the maximum value of H predicted by this model and the value of t, to 2 decimal places, when this maximum occurs.

(3)

(d) Calculate, to the nearest minute, the times when the height of sea water is predicted, by this model, to be 7 metres.

(6)(Total 15 marks)

3. (a) Express 5 cos x – 3 sin x in the form R cos(x + α), where R > 0 and 0 < α < .21

(4)

(b) Hence, or otherwise, solve the equation

5 cos x – 3 sin x = 4

for 0 x < 2 ,giving your answers to 2 decimal places.(5)

(Total 9 marks)

4. Solve

cosec2 2x – cot 2x = 1

for 0 x 180°.(Total 7 marks)

5. (a) Use the identity cos2 θ + sin2 θ = 1 to prove that tan2 θ = sec2 θ – 1.(2)

(b) Solve, for 0 ≤ θ < 360°, the equation

2 tan2 θ + 4 sec θ + sec2 θ = 2(6)

(Total 8 marks)

6. (a) Use the identity cos(A + B) = cosA cosB – sinA sinB, to show that

cos 2A = 1 – 2sin2A(2)

The curves C1 and C2 have equations

C1: y = 3sin 2x

C2: y = 4 sin2x – 2cos 2x

(b) Show that the x-coordinates of the points where C1 and C2 intersect satisfy the equation

4cos 2x + 3sin 2x = 2(3)

(c) Express 4cos2x + 3sin 2x in the form R cos(2x – α), where R > 0 and 0 < α < 90°, giving the value of α to 2 decimal places.

(3)

(d) Hence find, for 0 ≤ x < 180°, all the solutions of

4cos 2x + 3sin 2x = 2

giving your answers to 1 decimal place.(4)

(Total 12 marks)

7. (a) Write down sin 2x in terms of sin x and cos x.(1)

(b) Find, for 0 < x < π, all the solutions of the equation

cosec x – 8 cos x = 0

giving your answers to 2 decimal places.(5)

(Total 6 marks)

8. (a) (i) By writing 3θ = (2θ + θ), show that

sin 3θ = 3 sinθ – 4 sin3θ.(4)

(ii) Hence, or otherwise, for ,3

0 solve

8 sin3θ – 6 sin θ + 1 = 0.

Give your answers in terms of π.(5)

(b) Using ,sincoscossin)sin( or otherwise, show that

).26(4115sin

(4)(Total 13 marks)

9. (a) Express 3 cos θ + 4 sin θ in the form R cos(θ – α), where R and α are constants, R > 0 and 0 < α < 90°.

(4)

(b) Hence find the maximum value of 3 cos θ + 4 sin θ and the smallest positive value of θ for which this maximum occurs.

(3)

The temperature, f (t), of a warehouse is modelled using the equation

f(t) = 10 + 3 cos(15t)° + 4 sin(15t)°,

where t is the time in hours from midday and 0 t < 24.

(c) Calculate the minimum temperature of the warehouse as given by this model.(2)

(d) Find the value of t when this minimum temperature occurs.(3)

(Total 12 marks)

10. f(x) = 5cosx + 12sinx

Given that f(x) = Rcos(x – α), where R > 0 and 2

0 ,

(a) find the value of R and the value of α to 3 decimal places.(4)

(b) Hence solve the equation

5 cosx + 12sinx = 6

for 0 ≤ x < 2π.(5)

(c) (i) Write down the maximum value of 5cosx + 12sinx.(1)

(ii) Find the smallest positive value of x for which this maximum value occurs.(2)

(Total 12 marks)

11. (a) Given that sin2θ + cos2θ ≡ 1, show that 1 + cot2θ ≡ cosec2θ.(2)

(b) Solve, for 0 ≤ θ < 180°, the equation

2cot2θ – 9cosecθ = 3,

giving your answers to 1 decimal place.(6)

(Total 8 marks)

12. (a) Use the double angle formulae and the identity

cos(A + B) ≡ cosA cosB – sin A sinB

to obtain an expression for cos 3x in terms of powers of cos x only.(4)

(b) (i) Prove that

2)12(,sec2

cossin1

sin1cos

nxxx

xx

x.

(4)

(ii) Hence find, for 0 < x < 2π, all the solutions of

4cos

sin1sin1

cos

xx

xx

.

(3)(Total 11 marks)

13. (a) Express 3 sin x + 2 cos x in the form R sin(x + α) where R > 0 and 0 < α < .2

(4)

(b) Hence find the greatest value of (3 sin x + 2 cos x)4.(2)

(c) Solve, for 0 < x < 2π, the equation

3 sin x + 2 cos x = 1,

giving your answers to 3 decimal places.(5)

(Total 11 marks)

14. (a) Prove that

.90,2cosec2sincos

cossin

n

(4)

(b) On the axes below, sketch the graph of y = 2 cosec 2θ for 0° < θ < 360°.

y

O 9 0 1 8 0 2 7 0 3 6 0

(2)

(c) Solve, for 0° < θ < 360°, the equation

,3sincos

cossin

giving your answers to 1 decimal place.(6)

(Total 12 marks)

15. (a) By writing sin 3θ as sin (2θ + θ), show that

sin 3θ = 3sinθ – 4sin3θ.(5)

(b) Given that sinθ = 43 , find the exact value of sin 3θ.

(2)(Total 7 marks)

16.

The diagram above shows an oscilloscope screen.

The curve shown on the screen satisfies the equation

.sincos3 xxy

(a) Express the equation of the curve in the form y = Rsin(x + α), where R and α are constants, R > 0

and 0 < α < .2

(4)

(b) Find the values of x, 0 ≤ x < 2, for which y = 1.(4)

(Total 8 marks)

17. (i) Prove that

sec2x – cosec2 x ≡ tan2x – cot2x.(3)

(ii) Given that

y = arccos x, –1 ≤ x ≤ 1, and 0 ≤ y ≤ π

(a) express arcsin x in terms of y.(2)

(b) Hence evaluate arccos x + arcsin x. Give your answer in terms of π.(1)

(Total 6 marks)

18. (a) Using sin2θ + cos2θ ≡ 1, show that cosec2θ – cot2 θ ≡ 1.(2)

(b) Hence, or otherwise, prove that

cosec4θ – cot4θ ≡ cosec2θ + cot2θ.(2)

(c) Solve, for 90° < θ < 180°,

cosec4θ – cot4θ = 2 – cot θ.(6)

(Total 10 marks)

19. (a) Given that cos 43

A , where 270° < A < 360°, find the exact value of sin 2A.

(5)

(b) (i) Show that xxx 2cos3

2cos3

2cos

(3)

Given that

32cos

32cossin3 2 xxxy ,

(ii) show that xxy 2sin

dd

(4)(Total 12 marks)

20. (a) Show that

(i) ,)(,sincossincos

2cos41

nxxx

xxx

n

(2)

(ii) 212

21 sincoscos)2sin2(cos xxxxx

(3)

(b) Hence, or otherwise, show that the equation

21

sincos2coscos

can be written as

sin 2θ = cos 2θ.(3)

(c) Solve, for 0 θ 2π,

sin 2θ = cos 2θ,

giving your answers in terms of π.(4)

(Total 12 marks)

21. (a) Differentiate with respect to x

(i) x2e3x+2,(4)

(ii) .3

)2cos( 3

xx

(4)

(b) Given that x = 4 sin(2y + 6), find xy

dd

in terms of x.

(5)(Total 13 marks)

22. f(x) = 12 cos x – 4 sin x.

Given that f(x) = R cos(x + α), where R 0 and 0 α 90°,

(a) find the value of R and the value of α.(4)

(b) Hence solve the equation

12 cos x – 4 sin x = 7

for 0 x 360°, giving your answers to one decimal place.(5)

(c) (i) Write down the minimum value of 12 cos x – 4 sin x.(1)

(ii) Find, to 2 decimal places, the smallest positive value of x for which this minimum value occurs.

(2)(Total 12 marks)

23. (a) Given that 2 sin( + 30) = cos( + 60), find the exact value of tan .(5)

(b) (i) Using the identity cos (A + B) cos A cos B – sin A sin B, prove that

cos 2A 1 – 2 sin2 A.(2)

(ii) Hence solve, for 0 x < 2,

cos 2x = sin x,

giving your answers in terms of .(5)

(iii) Show that sin 2y tan y + cos 2y 1, for 0 y < 21 .

(3)(Total 15 marks)

24. (a) Given that sin2 + cos2 1, show that 1 + tan2 sec2.(2)

(b) Solve, for 0 < 360, the equation

2 tan2 + sec = 1,

giving your answers to 1 decimal place.(6)

(Total 8 marks)

25. (a) Using the identity cos(A + B) cosA cosB – sinA sinB, prove that

cos 2A 1 – 2 sin2 A.(2)

(b) Show that

2 sin 2 – 3 cos 2 – 3 sin + 3 sin (4 cos + 6 sin – 3).(4)

(c) Express 4 cos + 6 sin in the form R sin( + ), where R > 0 and 0 < < 21 .

(4)

(d) Hence, for 0 < , solve

2 sin 2 = 3(cos 2 + sin – 1),

giving your answers in radians to 3 significant figures, where appropriate.(5)

(Total 15 marks)

26.

B G F C

ED

A

2 2 c m

2 cm2 c m

2 c m

This diagram shows an isosceles triangle ABC with AB = AC = 4 cm and BAC = 2 .

The mid-points of AB and AC are D and E respectively. Rectangle DEFG is drawn, with F and G on BC. The perimeter of rectangle DEFG is P cm.

(a) Show that DE = 4 sin .(2)

(b) Show that P = 8 sin + 4 cos.(2)

(c) Express P in the form R sin( + ), where R > 0 and 0 < < 2

.

(4)

Given that P = 8.5,

(d) find, to 3 significant figures, the possible values of .(5)

(Total 13 marks)

27. (a) Sketch, on the same axes, in the interval 0 x 180, the graphs of

y = tan x° and y = 2 cos x°,

showing clearly the coordinates of the points at which the graphs meet the axes.(4)

(b) Show that tan x° = 2 cos x° can be written as

2 sin2 x° + sin x° – 2 = 0.(3)

(c) Hence find the values of x, in the interval 0 x 180, for which tan x° = 2 cos x°.(4)

(Total 11 marks)

28. (i) Given that sin x = 53

, use an appropriate double angle formula to find the exact value of sec 2x.

(4)

(ii) Prove that

cot 2x + cosec 2x cot x, (x 2n

, n ).

(4)(Total 8 marks)

29. (i) (a) Express (12 cos – 5 sin ) in the form R cos ( + ), where R > 0 and0 < < 90°.

(4)

(b) Hence solve the equation

12 cos – 5 sin = 4,

for 0 < < 90°, giving your answer to 1 decimal place.(3)

(ii) Solve

8 cot – 3 tan = 2,

for 0 < < 90°, giving your answer to 1 decimal place.(5)

(Total 12 marks)

30. (a) Prove that

.2costan1tan1

2

2

(4)

(b) Hence, or otherwise, prove

tan2 8

= 3 – 22.

(5)(Total 9 marks)

31. (i) Given that cos(x + 30) = 3 cos(x – 30), prove that tan x = 23 .

(5)

(ii) (a) Prove that

2sin2cos1

tan .

(3)

(b) Verify that = 180 is a solution of the equation sin 2 = 2 – 2 cos 2. (1)

(c) Using the result in part (a), or otherwise, find the other two solutions, 0 < < 360, of the equation sin 2 = 2 – 2 cos 2.

(4)(Total 13 marks)

32. (a) Express sin x + 3 cos x in the form R sin (x + ), where R > 0 and 0 < < 90.(4)

(b) Show that the equation sec x + 3 cosec x = 4 can be written in the form

sin x + 3 cos x = 2 sin 2x.(3)

(c) Deduce from parts (a) and (b) that sec x + 3 cosec x = 4 can be written in the form

sin 2x – sin (x + 60) = 0.(1)

(d) Hence, using the identity sin X – sin Y = 2 cos 2

sin2

YXYX , or otherwise, find the values

of x in the interval 0 x 180, for which sec x + 3 cosec x = 4.(5)

(Total 13 marks)

33. On separate diagrams, sketch the curves with equations

(a) y = arcsin x, 1 x 1,

(b) y = sec x, 3

x 3

, stating the coordinates of the end points

of your curves in each case.(4)

Use the trapezium rule with five equally spaced ordinates to estimate the area of the region

bounded by the curve with equation y = sec x, the x-axis and the lines x = 3

and x = 3

, giving

your answer to two decimal places.(4)

(Total 8 marks)

34. (a) Prove that for all values of x,

sin x + sin (60 x) sin (60 + x).(4)

(b) Given that sin 84 sin 36 = sin , deduce the exact value of the acute angle .(2)

(c) Solve the equation

4 sin 2x + sin (60 2x) = sin (60° + 2x) – 1

for values of x in the interval 0 x < 360, giving your answers to one decimal place.(5)

(Total 11 marks)

35. Find, giving your answers to two decimal places, the values of w, x, y and z for which

(a) ew = 4,(2)

(b) arctan x = 1,(2)

(c) ln (y + 1) – ln y = 0.85(4)

(d) cos z + sin z = 31 , < z < .

(5)(Total 13 marks)

36. (a) Using the formulae

sin (A B) = sin A cos B cos A sin B,cos (A B) = cos A cos B sin A sin B,

show that

(i) sin (A + B) – sin (A – B) = 2 cos A sin B,(2)

(ii) cos (A – B) – cos (A + B) = 2 sin A sin B.(2)

(b) Use the above results to show that

)cos()cos()sin()sin(

BABABABA

= cot A.

(3)

Using the result of part (b) and the exact values of sin 60 and cos 60,

(c) find an exact value for cot 75 in its simplest form.(4)

(Total 11 marks)

37. In a particular circuit the current, I amperes, is given by

I = 4 sin – 3 cos , > 0,

where is an angle related to the voltage.

Given that I = R sin ( ), where R > 0 and 0 < 360,

(a) find the value of R, and the value of to 1 decimal place.(4)

(b) Hence solve the equation 4 sin – 3 cos = 3 to find the values of between 0 and 360.

(5)

(c) Write down the greatest value for I. (1)

(d) Find the value of between 0 and 360 at which the greatest value of I occurs.(2)

(Total 12 marks)

38. (a) 2

2sin cos1 2 cos 1

M1

2 sin cos 2 cos cos

tan (as required) AG A1 cso 2

Note

M1: Uses both a correct identity for sin 2θ and a correct identityfor cos2θ .Also allow a candidate writing 1+cos2θ =2cos2 θ on the denominator.Also note that angles must be consistent in when candidates applythese identities.A1: Correct proof. No errors seen.

(b)12 tan 1 tan2

M1

1 awrt 26.6 A1

2 awrt 153.4 A1 3

Note

1st M1 for either 2tanθ = 1 or tanθ = 21

, seen or implied.

A1: awrt 26.6A1ft: awrt – 153.4° or θ2 = – 180° + θ1

Special Case: For candidate solving, tanθ = k, where, k 21 , to give

θ1 and θ2 = – 180° + θ1, then award M0A0B1 in part (b).Special Case: Note that those candidates who writes tanθ = 1, andgives ONLY two answers of 45° and –135° that are inside the rangewill be awarded SC M0A0B1.

[5]

39. (a) 6.25 or 2.5R B1

1.5 32 4tan awrt 0.6435 M1 A1 3

Note

B1: R = 2.5 or R = 6.25 . For R = ± 2.5, award B0.M1: tan α = ± 2

5.1 or tan α = ± 5.12

A1: α = awrt 0.6435

(b) (i) Max Value 2.5 B1ft

(ii) sin 0.6435 1 or 2their ;awrt 2.21 M1; A1ft 3

Note

B1 : 2.5 or follow through the value of R in part (a).M1: For sin (θ – their α) = 1A1 : awrt 2.21 or 2

+ their α rounding correctly to 3 sf.

(c) Max 8.5 (m)H B1ft4sin 0.6435 125

t

or 4 their (b) answer25

t ;

awrt 4.41t M1; A1 3

Note

B1 : 8.5 or 6 + their R found in part (a) as long as the answer isgreater than 6.

M1: sin

theirt

254

= 1 or 254 t

= their (b) answer

A1: For sin–1 (0.4) This can be implied by awrt 4.41 or awrt 4.40.

(d)46 2.5sin 0.6435 725

t

;

4 1sin 0.6435 0.425 2.5

t

M1; M1

14 0.6435 sin (0.4)25

t

or awrt 0.41 A1

Either awrt 2.1 or awrt 6.7t A1

So, 4 0.6435 0.411517... or 2.730076...25

ct

ddM1

Times 14 : 06, 18 : 43 A1 6

Note

M1: 6 + (their R) sin

theirt

254

= 7, M1:

theirt

254

= Rtheir1

A1: For sin–1 (0.4). This can be implied by awrt 0.41 or awrt 2.73 or othervalues for different α ‘s. Note this mark can be implied by seeing 1.055.A1: Either t = awrt 2.1 or t = awrt 6.7ddM1: either π – their PVc. Note that this mark is dependent upon thetwo M marks.This mark will usually be awarded for seeing either 2.730… or 3.373…A1: Both t = 14 : 06 and t = 18 : 43 or both 126 (min) and 403 (min) orboth 2 hr 6 min and 6 hr 43 min.

[15]

40. (a) 5cosx – 3sin x = R cos(x + α), R > 0, 0 < x< 2

5cosx – 3sinx = Rcos x cos α – R sin x sin α

Equate cos x: 5 = R cos α

Equate sin x: 3 = R sin α

...83095.534;35 22 R R2 = 52 + 32 M1;

34 or awrt 5.8 A1

tan α = c...5404195003.053 tan α = ortanor 3

553

sin α = ortheir 3

R

cos α = ortheir 5

R M1

α = awrt 0.54 or α = awrt 0.17π or

or 5.8awrt A1 4

Hence, 5cos x – 3sin x = 5404.0cos34 x

(b) 5cos x – 3sin x = 4

45405.0cos34 x

cos(x + 0.5404) =

...68599.0344

cos(xtheir α) = Rtheir4

M1

(x + 0.5404) = 0.814826916...c For applying cos–1

Rtheir

4M1

x = 0.2744...c awrt 0.27c A1

(x + 0.5404) = 2π – 0.814826916...c

{ = 5.468358...c} 2π – their 0.8148 ddM1

x = 4.9279...c awrt 4.93c A1 5

Hence, x = {0.27, 4.93}

Note

If there are any EXTRA solutions inside the range 0 ≤ x < 2π,then withhold the final accuracy mark if the candidate wouldotherwise score all 5 marks. Also ignore EXTRA solutionsoutside the range 0 ≤ x < 2π .

[9]

41. cosec2 2x – cot 2x = 1, (eqn *)0 ≤ x ≤180°

Using cosec2 2x = 1 + cot22x gives Writing down or usingcosec22x = ±1 ± cot22x M1

1 + cot2 2x – cot 2x = 1 or cosec2θ = ±1 ± cot2θ

cot 2 2 x – cot 2 x = 0 orcot2 2x = cot 2x For either cot 2 2 x – cot 2 x {=0}

or cot2 2x = cot 2x A1

cot 2x(cot 2x–1)= 0 orcot 2x = 1 Attempt to factorise or solve a

quadratic (See rules forfactorising quadratics) or

cancelling out cot 2x from bothsides. dM1

cot 2x = 0 or cot 2x = 1 Both cot 2x = 0 and cot 2x = 1. A1

cot 2x = 0 (tan 2x→ ∞) 2x = 90, 270

Candidate attempts to divide at leastone of their principal anglesby 2.

This will be usually implied byseeingx = 22.5 resulting from

cot 2x = 1.

ddM1

x = 45, 135

cot 2x = 1 tan 2x = 1 2x = 45,225 x = 22.5, 112.5

Overall, x = {22.5, 45, 112.5, 135} Both x = 22.5 and x = 112.5 A1

Both x = 45 and x = 135 B1

If there are any EXTRA solutions inside the range 0 ≤ x ≤180°and the candidate would otherwise score FULL MARKS thenwithhold the final accuracy mark (the sixth mark in this question).

Also ignore EXTRA solutions outside the range 0 ≤ x ≤180° .[7]

42. (a) cos2θ +sin2θ = 1 (÷ cos2θ)

22

2

2

2

cos1

cossin

coscos

Dividing cos2θ +sin2θ = 1 by M1

cos2θ to give underlined equation.

1 + tan2θ = sec2θ

tan2θ = sec2θ – 1 (as required) AG Complete proof. A1 cso 2No errors seen.

(b) 2tan2θ + 4secθ + sec2θ = 2, (eqn *) 0 ≤θ< 360°

Substituting tan2θ = sec2θ – 12(sec2θ – 1) + 4secθ + sec2θ = 2 into eqn * to get a M1

quadratic in secθ only

2sec2θ – 2 + 4secθ + sec2θ = 2

3sec 2 θ + 4sec θ – 4 = 0 Forming a three term “one sided” M1quadratic expression in secθ .

(secθ + 2)(3secθ – 2) = 0 Attempt to factorise M1or solve a quadratic.

secθ = –2 or secθ = 32

2–cos

1

or

32

cos1

;–cos 21 or 2

3cos ;–cos 21 A1;

α = 120° or α = no solutions

θ1 = 120° 120° A1

θ2 = 240° 240° or θ2 =360° – θ1 when B1ft 6solving using cosθ = ...

θ = {120°, 240°} Note the final A1 mark has beenchanged to a B1 mark.

[8]

43. (a) A = B cos (A + A) = cos2A = cos A cos A – sin A sin A AppliesA = B to cos(A + B) to give the

underlined equation or M1cos 2A = cos 2 A – sin 2 A

cos 2A = cos2A – sin2A and cos2A + sin2A = 1gives

cos2 A = 1 – sin2A – sin2A = 1 – 2sin 2 A (as required)Complete proof, with a link between LHS and RHS. A1 AG 2No errors seen.

(b) C1 = C2 3sin 2x = 4sin2x – 2cos 2x Eliminating y correctly. M1

Using result in part (a) tosubstitute for sin2 x as

22cos–142sin3 xx –2cos2x

22cos1 x

or ksin2x as

22cos1 xk to produce an

equation in only double angles. M1

3sin 2x = 2(1 – cos2x) – 2cos2x

3sin 2x = 2 – 2cos2x – 2cos2x

3sin 2x + 4cos2x = 2 Rearranges to give correct result A1 AG 3

(c) 3sin 2x + 4cos 2x = R cos(2x – α)

3sin2x + 4cos2x = Rcos2xcosα + Rsin2xsinα

Equate sin 2x: 3 = R sin α

Equate cos 2x: 4 = R cos α

525;43 22 R R = 5 B1

5...º36.8698976tan 43 3

443 tanorsin or M1

RR their4

their3 cosorsin

awrt 36.87 A1 3

Hence, 3sin 2x + 4cos2x = 5cos(2x – 36.87)

(d) 3sin 2x + 4cos2x = 2

5cos(2x – 36.87) = 2

cos(2x – 36.87) = 52

cos(2x ± their α) = Rtheir2

M1

(2x – 36.87) = 66.42182...° awrt 66 A1

(2x – 36.87) = 360 – 66.42182...°

Hence, x = 51.64591...o, 165.22409...° One of either awrt51.6 or awrt 51.7 or awrt A1

165.2 or awrt 165.3 Both awrt 51.6AND awrt 165.2 A1 4

If there are any EXTRA solutionsinside the range 0 ≤ x < 180° thenwithhold the final accuracy mark.

Also ignore EXTRA solutionsoutside the range 0≤ x <180°.

[12]

44. (a) sin 2x = 2 sin x cos x B1 aef 1

(b) cosec x – 8 cos x = 0, 0 < x < π

0cos8–sin

1x

xUsing cosec x =

xsin1

M1

xx

cos8sin

1

1 = 8 sin x cos x

1 = 4(2 sin x cos x)

1 = 4 sin 2x

412sin x sin 2x = k, where –1 < k < 1 and M1

k ≠ 0

412sin x A1

Radians 2x = {0.25268..., 2.88891...}

Degrees 2x = {14.4775..., 165.5225...}

Either arwt 7.24 or 82.76 or 0.13Radians x = {0.12634..., 1.44445...} or 1.44 or 1.45

or awrt 0.04π or A1Degrees x = {7.23875..., 82.76124...} awrt 0.46π.

Both 0.13 and 1.44 A1 cao 5Solutions for the final two A

marks must be given in x only.If there are any EXTRA solutions

inside the range 0 < x < π thenwithhold the final accuracy mark.

Also ignore EXTRA solutionsoutside the range 0 < x < π.

[6]

45. (a) (i) sin 3θ = sin(2θ + θ)

= sin 2θ cosθ + cos 2θ sin θ

= 2sinθ cosθ.cosθ + (1 – 2sin2 θ)sin θ M1 A1

= 2sinθ (1 – sin2θ) + sinθ – 2sin3θ M1

= 3sinθ – 4sin3θ cso A1 4

(ii) 8sin3θ – 6sin θ + 1 = 0

–2sin3θ + 1 = 0 M1 A1

sin 3θ = 21

M1

65,

63

65,

18 A1 A1 5

(b) sin15° = sin(60° – 45°) = sin 60°cos 45° – cos 60°sin 45° M1

21

21–

21

23

M1 A1

)2–6(412

41–6

41

cso A1 4

Alternatives

1 sin15° = sin (45° – 30°) = sin 45°cos30° – cos 45°sin 30° M1

21

21–

23

21

M1 A1

)2–6(412

41–6

41

cso A1 4

2 Using cos 2θ = 1 – 2sin2θ, cos30° = 1 – 2sin2 15°

2sin215° = 1 – cos30° = 1 – 23

sin215° = 4

3–2M1 A1

43–2

)122–26(161)2–6(

41 2

M1

Hence sin15° = )2–6(41

cso A1 4

[13]

46. (a) R2 = 32 + 42 M1

R = 5 A1

tan α = 34

M1

α = 53...° awrt 53° A1 4

(b) Maximum value is 5 ft their R B1 ft

At the maximum, cos(θ – α) = 1 or θ – α = 0 M1

θ = α = 53 ...° ft their α A1 ft 3

(c) f(t) = 10 + 5cos(15t – α)°

Minimum occurs when cos (15t – α)° = –1 M1

The minimum temperature is (10 – 5)° = 5° A1 ft 2

(d) 15t – α = 180 M1

t = 15.5 awrt 15.5 M1 A1 3[12]

47. (a) R2 = 52 + 122 M1R = 13 A1

tan α = 5

12M1

α ≈ 1.176 cao A1 4

(b) cos(x – α) = 136

M1

x – = arccos136

= 1.091 … A1

x = 1.091... + 1.176 … ≈ 2.267… awrt 2.3 A1x – α = –1.091 … accept … = 5.19 … for M M1x = –1.091 … + 1.176 …≈ 0.0849… awrt 0.084 or 0.085 A1 5

(c) (i) Rmax = 13 ft their R B1ft

(ii) At the maximum, cos (x – α) = 1 or x – α = 0 M1x = α = 1.176 … awrt 1.2, ft their α A1ft 3

[12]

48. (a) sin2θ + cos2θ = 1

÷ sin2

22

2

2

2

sin1

sincos

sinsin

M1

1 + cot2 = cosec2 * cso A1 2

Alternative

1 + cot2 = 1 +

22

22

2

2

sin1

sincossin

sincos

M1

cosec2 * cso A1

(b) 2 (cosec2 θ – 1) – 9 cosec θ = 3 M12cosec2 θ – 9 cosec θ – 5 = 0 or 5 sin2 θ + 9 sin θ – 2 = 0 M1(2 cosec θ + 1) (cosec θ– 5) = 0 or (5 sin θ – 1) (sin θ + 2) = 0 M1

cosec θ = 5 or sin = 51

A1

= 11.5°, 168.5° A1A1 6[8]

49. (a) cos (2x + x) = cos 2x cos x – sin 2x sin x M1= (2 cos2 x – 1) cos x – (2 sin x cos x) sin x M1= (2 cos2 x – 1) cos x – 2(1 – cos2 x) cos x any correct expression A1= 4cos3 x – 3cosx A1 4

(b) (i)xxxx

xx

xx

cos)sin1()sin1(cos

cossin1

sin1cos 22

M1

xxxxx

cos)sin1(sinsin21cos 22

A1

xxx

cos)sin1()sin1(2

M1

(*)sec2cos

2 xx cso A1 4

(c) sec x = 2 or cos x = 21

M1

x = 3

5,3

accept awrt 1.05, 5.24 A1, A1 3

[11]

50. (a) Complete method for R: e.g. R cos = 3, R sin = 2, R = )23( 22 M1

R = 13 or 3.61 (or more accurate) A1

Complete method for tan = 32

[Allow tan = 23

] M1

= 0.588 (Allow 33.7°) A1 4

1st M1 for correct method for R2nd M1 for correct method for tanNo working at all: M1A1 for 13, M1A1 for 0.588 or 33.7°.N.B. R cos = 2, R sin = 3 used, can still score M1A1 for R,

but loses the A mark for .cos = 3, sin = 2: apply the same marking.

(b) Greatest value = ( 13 )4 = 169 M1, A1 2

M1 for realising sin(x + ) = 1, so finding R4.

(c) sin(x + 0.588) = 131

(= 0.27735...) sin(x + their ) = Rtheir

1M1

(x + 0.588) = 0.281 (03...) or 16.1°) A1(x + 0.588) = – 0.28103... M1

Must be – their 0.281 or 180° – their 16.1°or (x + 0.588) = 2 + 0.28103... M1

Must be 2 + their 0.281 or 360° + their 16.1°x = 2.273 or x = 5.976 (awrt) Both (radian only) A1 5If 0.281 or 16.1° not seen, correct answers imply this A mark

Alt (i) Squaring to form quadratic in sin x or cos x M1[13 cos2x – 4cosx – 8 = 0, 13 sin2x – 6sinx – 3 = 0]Correct values for cos x = 0.953.... –0.646;or sinx = 0.767, 2.27 awrt A1For any one value of cos x or sin x, correct method fortwo values of x M1x = 2.273 or x = 5.976 (awrt) Both seen anywhere A1Checking other values (0.307, 4.011 or 0.869, 3.449)and discarding M1

Alt (ii) Squaring and forming equation of form a cos2x + bsin2x = c9sin2x + 4cos2x + 12sin2x = 1 12sin2x + 5cos2x = 11Setting up to solve using R formula e.g. 13cos(2x – 1.176) = 11 M1

...0(562.013

11cos)176.12( 1

x () A1

(2x – 1.176) = 2 – , 2 + ,................ M1x = 2.273 or x = 5.976 (awrt) Both seen anywhere A1Checking other values and discarding M1

[11]

51. (a)

sincoscossin

sincos

cossin 22

M1

M1 Use of common denominator to obtain single fraction

= sincos

1M1

M1 Use of appropriate trig identity (in this case sin2 + cos2 = 1)

= 2sin21

1Use of sin 2 = 2sin cos M1

= 2cosec2 (*) A1cso 4

Alt. (a)

tan1tan

tan1tan

sincos

cossin 2

M1

= tan

sec 2

M1

= 2sin21

1sincos

1

M1

= 2cosec2 (*) (cso) A1If show two expressions are equal, need conclusion such as QED, tick true.

(b)y

2

– 2

O 9 0º 1 8 0 º 2 7 0 º 3 6 0 º

Shape (May be translated but need to see 4 “sections”) B1

T.P.s at y = 2, asymptotic at correct x-values(dotted lines not required) B1dep 2

(c) 2cosec2 = 3

sin2 = 32

Allow 32sin

2

[M1 for equation in sin2] M1, A1

(2) = [41.810...°, 138.189...°; 401.810...°, 498.189...°] M1; M11st M1 for , 180 – ; 2nd M1 adding 360° to at least one of values M1; M1 = 20.9°, 69.1°, 200.9°, 249.1° (1 d.p.) awrt

1st A1for any two correct, 2nd A1 for other two A1, A1 6Extra solutions in range lose final A1 onlySC: Final 4 marks: = 20.9°, after M0M0 is B1; record as M0M0A1A0

Alt. (c) 3tan

1tan

and form quadratic, tan2 – 3tan + 1 = 0 M1, A1

(M1 for attempt to multiply through by tan,A1 for correct equation above).

Solving quadratic 2

53[tan = 2.618... or = 0.3819...] M1

= 69.1°, 249.1° = 20.9°, 200.9° (1 d.p.) M1, A1, A1(M1 is for one use of 180° + °, A1A1 as for main scheme)

[12]

52. (a) sin 3 = sin (2 + ) = sin 2 cos + cos 2 sin B1= 2 sin cos2 + (1 – 2 sin2 ) sin B1B1= 2sin – 2 sin3 + sin – 2 sin3 M1= 3 sin – 4 sin3 * cso A1 5

(b)16

3916

334

33434

4333sin

3

or exact M1A1 2

equivalent[7]

53. (a) R2 = (3)2 +12 R = 2 M1A1

tan = 3 = 3

accept awrt 1.05 M1 A1 4

(b) sin(x + their ) = 21

M1

x + their =

613,

65

6

A1

xi = 6

11,2

accept awrt 1.57, 5.76 M1 A1 4

The use of degrees loses only one mark in this question.Penalise the first time it occurs in an answer and then ignore.

[8]

54. (i) sec2 x – cosec2 x = (1 + tan2 x) – (1 + cot2 x) M1 A1= tan2 x – cot2 x * cso A1 3

Alternatives

sec2x – tan2x = 1 = cosec2x – cot2x M1A1Rearranging sec2x – cosec2x = tan2x – cot2x * cso A1 3

xxxx

xx 22

22

22 sincoscossin

sin1

cos1LHS

RHS = xx

xxxxxxxx

xx

xx

22

2222

22

44

2

2

2

2

sincos)cos)(sincos(sin

sincoscossin

sincos

cossin

M1

xxxx

22

22

sincoscossin

A1

= LHS * or equivalent A1 3

(ii) (a) y = arccos x x = cos y B1

yxyx

2arcsin

2sin

B1 2

Accept arcsin x = arcsin cos y

(b) arccos x + arcsin x = 22

yy B1 1

[6]

55. (a) Dividing sin2 θ + cos2 θ 1 by sin2 θ to give M1

22

2

2

2

sin1

sincos

sinsin

Completion: 1 + cot2 θ = cosec2 cosec2θ cot2 θ 1 AG A1* 2

(b) cosec4θ cot4 θ (cosec2θ cot2 θ)(cosec2θ + cot2 θ) M1

(cosec2θ + cot2θ) using (a) AG A1* 2

Using LHS = (1 + cot2 θ)2 cot4 θ, using (a) & elim. cot4θ M1, conclusion {using (a) again} A1*

Conversion to sines and cosines: needs

4

22

sin)cos1)(cos1(

for M1

(c) Using (b) to form cosec2θ + cot2 θ 2 cotθ M1

Forming quadratic in cotθ M1 1 + cot2 θ + cot2 θ 2 cotθ {using (a)} M1

2cot2θ + cotθ 1 = 0 A1

Solving: (2 cot θ 1)(cotθ + 1 = 0) to cot θ = M1

21cot or cot θ = –1 A1

= 135° (or correct value(s) for candidate dep. on 3Ms) A1ft 6

Ignore solutions outside rangeExtra “solutions” in range loses A1ft, but candidate may possibly have more than one “correct” solution.

[10]

56. (a) Method for finding sin A M1

47sin A A1 A1

First A1 for 47 , exact

Second A1 for sign (even if dec. answer given)

Use of sin 2A 2 sin A cos A M1

sin 2A = 8

73 or equivalent exact A1ft 5

± f.t. Requires exact value, dependent on 2nd M

(b) (i)

32cos

32cos xx

3sin2sin

3cos2cos

3sin2sin

3cos2cos xxxx M1

3coscos2 x A1

[This can be just written down (using factor formulae) for M1 A1]

= cos2x AG A1* 3

M1 A1 earned, if 3

cos2cos2 x just written down, using factor

theoremFinal A1* requires some working after first result.

(b) (ii) xxxxy 2sin2cossin6

dd

B1 B1

or

32sin2

32sin2cossin6 xxxx

= 3sin2x 2sin2x M1

= sin 2x AG A1 4

First B1 for 6 sin x cos x ; second B1 for remaining term(s)[12]

57. (a) (i)xx

xxxx

xsincossincos

sincos2cos 22

M1

xxxxxx

sincos)sin)(cossin(cos

= cos x – sin x AG A1 2

(ii) 21 (cos 2x – sin 2x) = 2

1 (2 cos2 x – 1 – 2 sin x cos x) M1, M1

= cos2 x – 21 – sin x cos x AG A1 3

(b) 21

sincos2coscos

cos (cos – sin) = 21 M1

cos2 – cos sin = 21

21 (cos 2 + 1) – 2

1 sin 2 = 21 M1

21 (cos 2 – sin 2)= 0

sin 2 = cos 2 AG A1 3

(c) sin 2 = cos 2

tan 2 = 1 M1

2 = 413,

49,

45,

4 A1 for 1

= 813,

89,

85,

8 M1 (4 solns)

A1 4[12]

58. (a) (i) xdd (e3x+2) = 3e3x+2 (or 3e2e3x) At any stage B1

xy

dd = 3x2 e3x+2 + 2xe3x+2 Or equivalent M1 A1+A1 4

(ii) xdd (cos(2x3)) = –6x2 sin(2x3) At any stage M1 A1

xy

dd =

2

333

9)2cos(3)2sin(18

xxxx M1 A1 4

Alternatively using the product rule for second M1 A1

y = (3x)–1 cos(2x3)

xy

dd = – 3(3x)–2 cos(2x3) – 6x2 (3x)–1 sin(2x3)

Accept equivalent unsimplified forms

(b) 1 = 8 cos(2y + 6)xy

dd or y

xdd = 8cos(2y + 6) M1

xy

dd = )62cos(8

1y M1 A1

xy

dd =

)16(2

1)(

4arcsincos8

12xx M1 A1 5

[13]

59. (a) R cos α = 12, R sin α = 4R = (122 + 42) = 160 Accept if just written down, awrt 12.6 M1 A1

tan α = 124 , α 18.43° awrt 18.4° M1, A1 4

(b) cos (x + their α) = Rtheir 7 ( 0.5534) M1

x + their α = 56.4° awrt 56° A1

= ... , 303.6° 360° – their principal value M1x = °38.0, 285.2° Ignore solutions out of range A1, A1 5

If answers given to more than 1 dp, penalise first time thenaccept awrt above.

(c) (i) minimum value is – 160 ft their R B1ft

(ii) cos (x + their α) = –1 M1

x 161.57° cao A1 3[12]

60. (a) 2sin(θ + 30)° = cos(θ + 60)°2sinθ°cos30° + 2cosθ°sin30° = cosθ°cos60° – sinθ° sin60° B1B1

232 sinθ° +

22

cosθ° = 21

cosθ° – 23 sinθ° M1

Finding tanθ°, tanθ° = –33

1 or equiva. Exact M1,A1 5

(b) (i) Setting A = B to give cos 2A = cos2A – sin2 A M1Correct completion:= (1 – sin2A) – sin2 A = 1 – 2sin2 A A1 2

Need to see intermediate step above for A1

(ii) Forming quadratic in sinx [2 sin2 x + sin x – 1 = 0] M1

Solving [(2 sin x – 1)(sinx + 1) = 0 or formula] M1[sin = ½ or sin = – 1]

= 6

, 6

5; A1,A1ft

A1ft for – “”

= 2

3A1 5

(iii) LHS = 2siny cosy yy

cossin

+ (1 – 2 sin2 y) B1M1

B1 use of tany = yy

cossin

, M1 forming expression in siny,

cosy only

Completion: = 2 sin2 y + (1 – 2 sin2 y) = 1 AG A1 3

[Alternative: LHS = yyyyy

coscos2cossin2sin

B1M1

= yyy

cos)2cos(

= 1 A1]

[15]

61. (a) Dividing by cos2 :

22

2

2

2

cos1

coscos

cossin

M1

Completion: 1 + tan2 sec2 A1 2(no errors seen)

(b) use of 1 + tan2 = sec2 : 2(sec2 – 1) + sec = 1 M1[2sec2 + sec – 3 = 0]

Factorising or solving: (2 sec + 3)(sec – 1) = 0

[sec = –23

or sec = 1]

= 0 B1

cos = –32

; 1 = 131.8° M1 A1

2 = 228.2° A1 ft 6[A1ft for 2 = 360° – 1]

[8]

62. (a) cos 2A = cos2 A – sin2 A (+ use of cos2 A + sin2 A 1) M1= (1 – sin2 A); –sin2 A = 1 – 2sin2 A (*) A1 2

(b) 2sin 2 – 3cos 2 – 3sin + 3 4sin; –3(1 – 2sin2 ) – 3sin + 3 B1; M1 4sin cos + 6sin2 – 3sin M1 sin(4cos + 6sin – 3) (*) A1 4

(c) 4cos + 6sin Rsincos + Rcos sinComplete method for R (may be implied by correct answer)[R2 = 42 + 62, Rsin = 4, Rcos = 6] M1R = 52 or 7.21 A1Complete method for ; = 0.588 M1 A1 4

(allow 33.7°)

(d) sin(4cos + 6sin – 3) = 0 M1 = 0 B1

sin( + 0.588) = 523

= 0.4160.. (24.6°) M1

+ 0.588 = (0.4291), 2.7125 [or + 33.7° = (24.6°), 155.4°] dM1 = 2.12 cao A1 5

[15]

63. (a) Complete method for DE [e.g. split triangle ADE and sin,or sine or cos rule] M1

DE = 4 sin (*) ( c.s.o.) A1(*) 2

(b) P = 2 DE + 2EF or equivalent. With attempt at EF M1

= 8sin + 4cos (*) ( c.s.o.) A1(*) 2

(c) 8sin + 4cos = R sin ( + )= R sin cos + R cos sin

Method for R, method for M1 M1

need to use tan for 2nd M

[R cos = 8, R sin = 4 tan = 0.5, R = 22 48( ]

R = 4 5 or 8.94, = 0.464 (allow 26.6), A1 A1 4

awrt 0.464

(d) Using candidate’s R sin ( + ) = 8.5 to give ( + ) = sin–1R5.8

M1

Solving to give = sin–1R5.8

– , = 0.791 (allow 45.3) M1 A1

Considering second angle: + = (or 180) – sin–1R5.8

; M1

= 1.42 (allow 81.6) A1 5[13]

64. (a) 4 5 9 0 1 3 5 1 8 0

54321

– 1– 2– 3– 4– 5

Tangent graph shape M1180 indicated A1Cosine graph shape M12 and 90 indicated A1 4

Allow separate sketches.

(b) Using tan x = xx

cossin

and multiplying both sides by cos x. (sinx = 2cos2x) M1

Using sin2 x + cos2 x = 1 M12 sin2 x + sin x – 2 = 0 (*) A1 3

(c) Solving quadratic: sin x = 4

171 (or equiv.) M1 A1

x = 51.3 (3 s.f. or better, 51.33…) A1x = 128.7 (accept 129) (3 s.f. or better) 180 – ( 90n) B1ft 4

[11]

65. (i) A correct form of cos 2x used M1

1 – 22

53

or

22

53

54

or 1

542

2

257

A1

xx

2cos12sec ; =

7

25 or

743 M1A1 4

(ii) (a) xx

x2sin

12sin2cos

or (b) xx 2sin

12tan

1 M1

Forming single fraction (or ** multiplying both sides by sin2x) M1Use of correct trig. formulae throughout and producing expressionin terms of sinx and cosx M1

Completion (cso) e.g. xx

xcossin2

cos2 2

= xxx cot

sincos

(*) A1 4

[8]

66. (a) (i) 12cos – 5sin = Rcoscos – Rsinsin.

R 5

1 2R2 = 52 + 122, R = 13 M1, A1

tan = 125

, = 22.6° (AWRT 22.6)

or 0.39C (AWRT 0.39C) M1, A1 4

M1 for correct expression for R or R2

M1 for correct trig expression for α

(b) (i) cos( + 22.6) = 134

M1

+ 22.6 = 72.1, M1 = 49.5 (only) A1 3

M1 cos( + ) = R4

M1 + = … ft their R

(ii)tan

8– 3tan = 2 M1

i.e. 0 = 3tan2 + 2tan – 8 M1 0 = (3tan – 4)(tan + 2) M1

tan = 34

or –2

tan = 34

= 53.1 A1

[ignore not in range e.g. = 116.6] A1 5

M1 Use of cot = tan

1

M1 3TQ in tan = 0M1 Attempt to solve 3TQ = 0A1 For Final A mark must deal with tan = – 2

[12]

67. (a)

equivalentor

seccossin1

or

cossin1

cossin1

tan1tan1

2

2

2

2

2

2

2

2

2

M1M1

12cos

sincossincos

22

22

= cos2 (*) M1 A1 4

cso

(b) = 8

, cos2 = 2

1M1

21

11

2

2

tt

M1

t2 = 1212

M1

= 1212.

1212

= 3 – 2 2 (*) M1 A1 5

cso[9]

Alternative to (b)

212

tt

= tan2 = 1 M1

t2 + 2t – 1 = 0 M1t = 2 – 1 M1t2 = ( 2 – 1)2 = 3 – 2 2 (*) M1 A1 5

cso

68. (i) cos x cos 30 – sin x sin 30 = 3(cos x cos 30 + sin x sin 30) M1Correct use of cos(x 30)

3 cos x – sin x = 33 cos x + 3 sin x M1, A1Sub. for sin 30 etcdecimals M1, surds A1

i.e. –4 sin x = 23 cos x tan x = – 23 (*) M1, A1cso 5

Collect terms and use tan x = xcosxsin

(ii) (a) LHS =

cossin2)sin21(1 2 M1; A1

Use of cos 2A or sin 2A; both correct

=

cossin

= tan (*) A1 cso 3

(b) Verifying: 0 = 2 – 2 (since sin 360 = 0, cos 360 = 1) B1 cso 1

(c) Equation 1 =

2sin)2cos1(2

M1

Rearrange to form

2sin2cos1

tan = 21 or cot = 2 A1

i.e. = (26.6° or 206.6°) AWRT 27°, 207°

1st solution M1

must be tan = 21

or 2

(both) A1 4[13]

Alt 1

(c) 2 sin cos = 2 – 2(1 – 2 sin2) M1Use of cos 2A and sin 2A

0 = 2 sin (2 sin – cos )

(sin = 0) tan = 21 etc, as in scheme A1

Alt 2

(c) 2 cos 2 + sin 2 = 2 cos (2 – ) = 5

2M1

= 22.6 (or 27) A1

2 = 2, 360, 360 + 2 = , 180 + i.e. = 27° or 207° (or 1 dp)

= or 180 + M1A1 both

69. (a) sin x + 3 cos x = R sin (x + )

= R (sin x cos + cos x sin ) M1

R cos = 1, R sin = 3 A1

Method for R or , e.g. R = (1 + 3) or tan = 3 M1

Both R = 2 and = 60 A1 4

(b) sec x + 3 cosec x = 4 xcos

1 +

xsin3 = 4 B1

sin x + 3 cos x = 4 sin x cos x M1

= 2 sin 2x (*) M1 3

(c) Clearly producing 2 sin 2x = 2 sin (x + 60) A1 1

(d) sin 2x – sin (x + 60) = 0 cos 2

603 x sin

260x

= 0 M1

cos 2

603 x = 0 x = 40, 160 M1 A1 A1 ft

sin 260x

= 0 x = 60 B1 5

[13]

70. (a)

– 1x

y

(1 , )

(1 , –

–

)

2

2

2

2

O

y = arcsin x

(a) Shape correctpassing through O: G1;end-points: G1 2

(b) x

y

O

y x = se c

( , 2 )3(– , 2 )3

Shape correct,symmetry in Oy: G1end-points: G1 2

(c) x3

6

06

3

sec x 2 1.155 1 1.155 2

Area estimate = 3

3

dsec

xx =

155.11155.12

226

M1 A1 A1

= 2.78 (2 d.p.) A1 4[8]

71. (a) LHS = sin x + sin 60 cos 60 sin x M1

= sin x + 23 cos x

21

sin x A1

RHS = sin 60 cos x + cos 60 sin x M1

= 23 cos x +

21

sin x = LHS A1 4

(b) From (a), sin (60 + x) – sin (60 x) = sin x

x = 24 sin 84 sin 36 = sin 24 M1

= 24 A1 2

(c) 3 sin 2x + sin 2x + sin (60 2x) = sin (60 + 2x) – 1 M1

Using (a), 3 sin 2x = 1 A1

2x = 199.47 or 340.53 M1

x = 99.7, 170.3 A1

or 279.7, 350.3 A1 ft 5[11]

72. (a) ew = 0.25 w = 1.39 M1 A1 2

(b) arctan x = 1 x = 0.79 M1 A1 2

(c) ln yy 1

= 0.85 yy 1

= e0.85 M1 A1

y1

= 2.340 – 1 y = 0.75 M1 A1 4

(d) Putting cos z + sin z in the form 2 cos

4z or equivalent M1 A1

cos

4z =

231

attempt for z M1

z = 2.12, 0.55 A1, A1 ft 5[13]

73. (a) (i) sin (A + B) – sin (A – B)

= sin A cos B + sin B cos A – sin A cos B + sin B cos A M1

= 2 sin B cos A (*) A1 cso 2

(ii) cos (A – B) – cos (A + B)

= cos A cos B + sin A sin B – cos A cos B + sin A sin B M1

= 2 sin A sin B (*) A1 cso 2

(b) BABA

BABA

sin cossinsin

= BAAB

sin sin 2cos sin 2

M1

= AA

sincos

A1

= cot A (*) A1 cso 3

(c) Let A = 75 and B = 15 B1

ooo

oo

75cot 90 cos 60 cos60sin 90sin

M1

cot 75 = 0

21

231

= 2 3 M1 A1 4

[11]

74. (a) 4 sin – 3 cos = R sin cos R cos sin

sin terms give 4 = R cos cos terms give 3 = R sin tan = 0.75 M1 = 36.9o A1R2 = 42 + 32 = 25 R = 5 M1 A1 4

(b) 5 sin ( – 36.9o) = 3sin ( – 36.9o) = 0.6 M1 – 36.9o = 36.9o, 143.1 A1 M1 = 73.7o, 180

awrt 74 A1 A15

(c) Max value 5 B1 1

(d) sin ( –36.9o) = 1 M1 36.9 = 90 = 90 + 36.9 = 126.9 A1 2

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