# IB Math Analysis & Approaches Questionbank-Topic: SL 3.6 Double angle identities for sine and cosine SL Paper 1

## Question

Given that $$\cos A = \frac{1}{3}$$ and $$0 \le A \le \frac{\pi }{2}$$ , find $$\cos 2A$$ .


a.

Given that $$\sin B = \frac{2}{3}$$  and $$\frac{\pi }{2} \le B \le \pi$$ , find $$\cos B$$ .


b.

## Markscheme

evidence of choosing the formula $$\cos 2A = 2{\cos ^2}A – 1$$     (M1)

Note: If they choose another correct formula, do not award the M1 unless there is evidence of finding $${\sin ^2}A = 1 – \frac{1}{9}$$

correct substitution     A1

e.g.$$\cos 2A = {\left( {\frac{1}{3}} \right)^2} – \frac{8}{9}$$ , $$\cos 2A = 2 \times {\left( {\frac{1}{3}} \right)^2} – 1$$

$$\cos 2A = – \frac{7}{9}$$     A1     N2

[3 marks]

a.

METHOD 1

evidence of using $${\sin ^2}B + {\cos ^2}B = 1$$     (M1)

e.g. $${\left( {\frac{2}{3}} \right)^2} + {\cos ^2}B = 1$$ , $$\sqrt {\frac{5}{9}}$$ (seen anywhere),

$$\cos B = \pm \sqrt {\frac{5}{9}}$$ $$\left( { = \pm \frac{{\sqrt 5 }}{3}} \right)$$     (A1)

$$\cos B = – \sqrt {\frac{5}{9}}$$ $$\left( { = – \frac{{\sqrt 5 }}{3}} \right)$$     A1     N2

METHOD 2

diagram     M1 for finding third side equals $$\sqrt 5$$     (A1)

$$\cos B = – \frac{{\sqrt 5 }}{3}$$     A1     N2

[3 marks]

b.

## Question

Let  $$f(x) = {\sin ^3}x + {\cos ^3}x\tan x,\frac{\pi }{2} < x < \pi$$ .

Show that $$f(x) = \sin x$$ .


a.

Let $$\sin x = \frac{2}{3}$$ . Show that $$f(2x) = – \frac{{4\sqrt 5 }}{9}$$ .


b.

## Markscheme

changing $$\tan x$$ into $$\frac{{\sin x}}{{\cos x}}$$     A1

e.g. $${\sin ^3}x + {\cos ^3}x\frac{{\sin x}}{{\cos x}}$$

simplifying     A1

e.g $$\sin x({\sin ^2}x + {\cos ^2}x)$$ , $${\sin ^3}x + \sin x – {\sin ^3}x$$

$$f(x) = \sin x$$    AG     N0

[2 marks]

a.

recognizing $$f(2x) = \sin 2x$$ , seen anywhere     (A1)

evidence of using double angle identity $$\sin (2x) = 2\sin x\cos x$$ , seen anywhere     (M1)

evidence of using Pythagoras with $$\sin x = \frac{2}{3}$$     M1

e.g. sketch of right triangle, $${\sin ^2}x + {\cos ^2}x = 1$$

$$\cos x = – \frac{{\sqrt 5 }}{3}$$ (accept $$\frac{{\sqrt 5 }}{3}$$ )     (A1)

$$f(2x) = 2\left( {\frac{2}{3}} \right)\left( { – \frac{{\sqrt 5 }}{3}} \right)$$     A1

$$f(2x) = – \frac{{4\sqrt 5 }}{9}$$     AG     N0

[5 marks]

b.

## Question

A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below. The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is $$\theta$$ radians, where $$0 \le \theta \le \frac{\pi }{2}$$ .

Write down an expression in terms of $$\theta$$ for

(i)     $$x$$ ;

(ii)    $$y$$ .


a.

Let the area of the rectangle be A.

Show that $$A = 18\sin 2\theta$$ .


b.

(i)     Find $$\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}$$ .

(ii)    Hence, find the exact value of $$\theta$$ which maximizes the area of the rectangle.

(iii)   Use the second derivative to justify that this value of $$\theta$$ does give a maximum.


c.

## Markscheme

(i) $$x = 3\cos \theta$$     A1     N1

(ii) $$y = 3\sin \theta$$     A1     N1

[2 marks]

a.

finding area     (M1)

e.g. $$A = 2x \times 2y$$ , $$A = 8 \times \frac{1}{2}bh$$

substituting     A1

e.g. $$A = 4 \times 3\sin \theta \times 3\cos \theta$$ , $$8 \times \frac{1}{2} \times 3\cos \theta \times 3\sin \theta$$

$$A = 18(2\sin \theta \cos \theta )$$    A1

$$A = 18\sin 2\theta$$     AG     N0

[3 marks]

b.

(i) $$\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta$$     A2     N2

(ii) for setting derivative equal to 0     (M1)

e.g. $$36\cos 2\theta = 0$$ , $$\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0$$

$$2\theta = \frac{\pi }{2}$$     (A1)

$$\theta = \frac{\pi }{4}$$     A1     N2

(iii) valid reason (seen anywhere)     R1

e.g. at $$\frac{\pi }{4}$$, $$\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0$$ ; maximum when $$f”(x) < 0$$

finding second derivative $$\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} = – 72\sin 2\theta$$     A1

evidence of substituting $$\frac{\pi }{4}$$     M1

e.g. $$– 72\sin \left( {2 \times \frac{\pi }{4}} \right)$$ , $$– 72\sin \left( {\frac{\pi }{2}} \right)$$ , $$– 72$$

$$\theta = \frac{\pi }{4}$$ produces the maximum area     AG     N0

[8 marks]

c.

## Question

Solve $$\cos 2x – 3\cos x – 3 – {\cos ^2}x = {\sin ^2}x$$ , for $$0 \le x \le 2\pi$$ .

## Markscheme

evidence of substituting for $$\cos 2x$$     (M1)

evidence of substituting into $${\sin ^2}x + {\cos ^2}x = 1$$     (M1)

correct equation in terms of $$\cos x$$ (seen anywhere)     A1

e.g. $$2{\cos ^2}x – 1 – 3\cos x – 3 = 1$$ , $$2{\cos ^2}x – 3\cos x – 5 = 0$$

evidence of appropriate approach to solve     (M1)

appropriate working     A1

e.g. $$(2\cos x – 5)(\cos x + 1) = 0$$ , $$(2x – 5)(x + 1)$$ , $$\cos x = \frac{{3 \pm \sqrt {49}}}{4}$$

correct solutions to the equation

e.g. $$\cos x = \frac{5}{2}$$ , $$\cos x = – 1$$ , $$x = \frac{5}{2}$$ , $$x = – 1$$     (A1)

$$x = \pi$$     A1     N4

[7 marks]

## Question

The straight line with equation $$y = \frac{3}{4}x$$ makes an acute angle $$\theta$$ with the x-axis.

Write down the value of $$\tan \theta$$ .


a.

Find the value of

(i)     $$\sin 2\theta$$ ;

(ii)    $$\cos 2\theta$$ .


b(i) and (ii).

## Markscheme

$$\tan \theta = \frac{3}{4}$$ (do not accept $$\frac{3}{4}x$$ )     A1     N1

[1 mark]

a.

(i) $$\sin \theta = \frac{3}{5}$$ , $$\cos \theta = \frac{4}{5}$$     (A1)(A1)

correct substitution     A1

e.g. $$\sin 2\theta = 2\left( {\frac{3}{5}} \right)\left( {\frac{4}{5}} \right)$$

$$\sin 2\theta = \frac{{24}}{{25}}$$     A1     N3

(ii) correct substitution     A1

e.g. $$\cos 2\theta = 1 – 2{\left( {\frac{3}{5}} \right)^2}$$ ,  $${\left( {\frac{4}{5}} \right)^2} – {\left( {\frac{3}{5}} \right)^2}$$

$$\cos 2\theta = \frac{7}{{25}}$$     A1     N1

[6 marks]

b(i) and (ii).

## Question

Let $$f(x) = \cos 2x$$ and $$g(x) = 2{x^2} – 1$$ .

Find $$f\left( {\frac{\pi }{2}} \right)$$ .


a.

Find $$(g \circ f)\left( {\frac{\pi }{2}} \right)$$ .


b.

Given that $$(g \circ f)(x)$$ can be written as $$\cos (kx)$$ , find the value of k, $$k \in \mathbb{Z}$$ .


c.

## Markscheme

$$f\left( {\frac{\pi }{2}} \right) = \cos \pi$$     (A1)

$$= – 1$$     A1     N2

[2 marks]

a.

$$(g \circ f)\left( {\frac{\pi }{2}} \right) = g( – 1)$$ $$( = 2{( – 1)^2} – 1)$$    (A1)

$$= 1$$     A1     N2

[2 marks]

b.

$$(g \circ f)(x) = 2{(\cos (2x))^2} – 1$$ $$( = 2{\cos ^2}(2x) – 1)$$     A1

evidence of $$2{\cos ^2}\theta – 1 = \cos 2\theta$$ (seen anywhere)     (M1)

$$(g \circ f)(x) = \cos 4x$$

$$k = 4$$     A1     N2

[3 marks]

c.

## Question

Show that $$4 – \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin \theta + 3$$ .


a.

Hence, solve the equation $$4 – \cos 2\theta + 5\sin \theta = 0$$ for $$0 \le \theta \le 2\pi$$ .


b.

## Markscheme

attempt to substitute $$1 – 2{\sin ^2}\theta$$ for $$\cos 2\theta$$     (M1)

correct substitution     A1

e.g. $$4 – (1 – 2{\sin ^2}\theta ) + 5\sin\theta$$

$$4 – \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin\theta + 3$$     AG     N0

[2 marks]

a.

evidence of appropriate approach to solve     (M1)

correct working     A1

e.g. $$(2\sin \theta + 3)(\sin \theta + 1)$$ , $$(2x + 3)(x + 1) = 0$$ , $$\sin x = \frac{{ – 5 \pm \sqrt 1 }}{4}$$

correct solution $$\sin \theta = – 1$$ (do not penalise for including $$\sin \theta = – \frac{3}{2}$$     (A1)

$$\theta = \frac{{3\pi }}{2}$$     A2     N3

[5 marks]

b.

## Question

Let $$\sin \theta = \frac{2}{{\sqrt {13} }}$$ , where $$\frac{\pi }{2} < \theta < \pi$$ .

Find $$\cos \theta$$ .


a.

Find $$\tan 2\theta$$ .


b.

## Markscheme

METHOD 1

evidence of choosing $${\sin ^2}\theta + {\cos ^2}\theta = 1$$     (M1)

correct working     (A1)

e.g. $${\cos ^2}\theta = \frac{9}{{13}}$$ , $$\cos \theta = \pm \frac{3}{{\sqrt {13} }}$$ , $$\cos \theta = \sqrt {\frac{9}{{13}}}$$

$$\cos \theta = – \frac{3}{{\sqrt {13} }}$$    A1     N2

Note: If no working shown, award N1 for $$\frac{3}{{\sqrt {13} }}$$ .

METHOD 2

approach involving Pythagoras’ theorem     (M1)

e.g. $${2^2} + {x^2} = 13$$ , finding third side equals 3     (A1)

$$\cos \theta = – \frac{3}{{\sqrt {13} }}$$     A1     N2

Note: If no working shown, award N1 for $$\frac{3}{{\sqrt {13} }}$$ .

[3 marks]

a.

correct substitution into $$\sin 2\theta$$ (seen anywhere)     (A1)

e.g. $$2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { – \frac{3}{{\sqrt {13} }}} \right)$$

correct substitution into $$\cos 2\theta$$ (seen anywhere)     (A1)

e.g. $${\left( { – \frac{3}{{\sqrt {13} }}} \right)^2} – {\left( {\frac{2}{{\sqrt {13} }}} \right)^2}$$ , $$2{\left( { – \frac{3}{{\sqrt {13} }}} \right)^2} – 1$$ , $$1 – 2{\left( {\frac{2}{{\sqrt {13} }}} \right)^2}$$

valid attempt to find $$\tan 2\theta$$     (M1)

e.g. $$\frac{{2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { – \frac{3}{{\sqrt {13} }}} \right)}}{{{{\left( { – \frac{3}{{\sqrt {13} }}} \right)}^2} – {{\left( {\frac{2}{{\sqrt {13} }}} \right)}^2}}}$$ , $$\frac{{2\left( { – \frac{2}{3}} \right)}}{{1 – {{\left( { – \frac{2}{3}} \right)}^2}}}$$

correct working     A1

e.g. $$\frac{{\frac{{(2)(2)( – 3)}}{{13}}}}{{\frac{9}{{13}} – \frac{4}{{13}}}}$$ , $$\frac{{ – \frac{{12}}{{{{\left( {\sqrt {13} } \right)}^2}}}}}{{\frac{{18}}{{13}} – 1}}$$ , $$\frac{{ – \frac{{12}}{{13}}}}{{\frac{5}{{13}}}}$$

$$\tan 2\theta = – \frac{{12}}{5}$$     A1     N4

Note: If students find answers for $$\cos \theta$$ which are not in the range $$[ – 1{\text{, }}1]$$, award full FT in (b) for correct FT working shown.

[5 marks]

b.

## Examiners report

While the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle in the second quadrant will have a negative value.

a.

In part (b), many candidates incorrectly tried to calculate $$\tan 2\theta$$ as $$2 \times \tan \theta$$ , rather than using the double-angle identities.

b.

## Question

The following diagram shows a right-angled triangle, $$\rm{ABC}$$, where $$\sin \rm{A} = \frac{5}{{13}}$$. Show that $$\cos A = \frac{{12}}{{13}}$$.


a.

Find $$\cos 2A$$.


b.

## Markscheme

METHOD 1

approach involving Pythagoras’ theorem     (M1)

eg     $${5^2} + {x^2} = {13^2}$$, labelling correct sides on triangle

finding third side is 12 (may be seen on diagram)     A1

$$\cos A = \frac{{12}}{{13}}$$     AG     N0

METHOD 2

approach involving $${\sin ^2}\theta + {\cos ^2}\theta = 1$$     (M1)

eg     $${\left( {\frac{5}{{13}}} \right)^2} + {\cos ^2}\theta = 1,{\text{ }}{x^2} + \frac{{25}}{{169}} = 1$$

correct working     A1

eg     $${\cos ^2}\theta = \frac{{144}}{{169}}$$

$$\cos A = \frac{{12}}{{13}}$$     AG     N0

[2 marks]

a.

correct substitution into $$\cos 2\theta$$     (A1)

eg     $$1 – 2{\left( {\frac{5}{{13}}} \right)^2},{\text{ }}2{\left( {\frac{{12}}{{13}}} \right)^2} – 1,{\text{ }}{\left( {\frac{{12}}{{13}}} \right)^2} – {\left( {\frac{5}{{13}}} \right)^2}$$

correct working     (A1)

eg     $$1 – \frac{{50}}{{169}},{\text{ }}\frac{{288}}{{169}} – 1,{\text{ }}\frac{{144}}{{169}} – \frac{{25}}{{169}}$$

$$\cos 2A = \frac{{119}}{{169}}$$     A1     N2

[3 marks]

b.

## Question

Given that $$\sin x = \frac{3}{4}$$, where $$x$$ is an obtuse angle,

find the value of $$\cos x;$$


a.

find the value of $$\cos 2x.$$


b.

## Markscheme

valid approach     (M1)

eg$$\;\;\;$$ , $${\sin ^2}x + {\cos ^2}x = 1$$

correct working     (A1)

eg$$\;\;\;{4^2} – {3^2},{\text{ }}{\cos ^2}x = 1 – {\left( {\frac{3}{4}} \right)^2}$$

correct calculation     (A1)

eg$$\;\;\;\frac{{\sqrt 7 }}{4},{\text{ }}{\cos ^2}x = \frac{7}{{16}}$$

$$\cos x = – \frac{{\sqrt 7 }}{4}$$     A1     N3

[4 marks]

a.

correct substitution (accept missing minus with cos)     (A1)

eg$$\;\;\;1 – 2{\left( {\frac{3}{4}} \right)^2},{\text{ }}2{\left( { – \frac{{\sqrt 7 }}{4}} \right)^2} – 1,{\text{ }}{\left( {\frac{{\sqrt 7 }}{4}} \right)^2} – {\left( {\frac{3}{4}} \right)^2}$$

correct working     A1

eg$$\;\;\;2\left( {\frac{7}{{16}}} \right) – 1,{\text{ }}1 – \frac{{18}}{{16}},{\text{ }}\frac{7}{{16}} – \frac{9}{{16}}$$

$$\cos 2x = – \frac{2}{{16}}\;\;\;\left( { = – \frac{1}{8}} \right)$$     A1     N2

[3 marks]

Total [7 marks]

b.

## Question

Let $$f(x) = 6x\sqrt {1 – {x^2}}$$, for $$– 1 \leqslant x \leqslant 1$$, and $$g(x) = \cos (x)$$, for $$0 \leqslant x \leqslant \pi$$.

Let $$h(x) = (f \circ g)(x)$$.

Write $$h(x)$$ in the form $$a\sin (bx)$$, where $$a,{\text{ }}b \in \mathbb{Z}$$.


a.

Hence find the range of $$h$$.


b.

## Markscheme

attempt to form composite in any order     (M1)

eg$$\,\,\,\,\,$$$$f\left( {g(x)} \right),{\text{ }}\cos \left( {6x\sqrt {1 – {x^2}} } \right)$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$6\cos x\sqrt {1 – {{\cos }^2}x}$$

correct application of Pythagorean identity (do not accept $${\sin ^2}x + {\cos ^2}x = 1$$)     (A1)

eg$$\,\,\,\,\,$$$${\sin ^2}x = 1 – {\cos ^2}x,{\text{ }}6\cos x\sin x,{\text{ }}6\cos x \left| \sin x\right|$$

valid approach (do not accept $$2\sin x\cos x = \sin 2x$$)     (M1)

eg$$\,\,\,\,\,$$$$3(2\cos x\sin x)$$

$$h(x) = 3\sin 2x$$    A1     N3

[5 marks]

a.

valid approach     (M1)

eg$$\,\,\,\,\,$$amplitude $$= 3$$, sketch with max and min $$y$$-values labelled, $$– 3 < y < 3$$

correct range     A1     N2

eg$$\,\,\,\,\,$$$$– 3 \leqslant y \leqslant 3$$, $$[ – 3,{\text{ }}3]$$ from $$– 3$$ to 3

Note:     Do not award A1 for $$– 3 < y < 3$$ or for “between $$– 3$$ and 3”.

[2 marks]

b.

## Question

Let $$\sin \theta = \frac{{\sqrt 5 }}{3}$$, where $$\theta$$ is acute.

Find $$\cos \theta$$.


a.

Find $$\cos 2\theta$$.


b.

## Markscheme

evidence of valid approach     (M1)

eg$$\,\,\,\,\,$$right triangle, $${\cos ^2}\theta = 1 – {\sin ^2}\theta$$

correct working     (A1)

eg$$\,\,\,\,\,$$missing side is 2, $$\sqrt {1 – {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}}$$

$$\cos \theta = \frac{2}{3}$$     A1     N2

[3 marks]

a.

correct substitution into formula for $$\cos 2\theta$$     (A1)

eg$$\,\,\,\,\,$$$$2 \times {\left( {\frac{2}{3}} \right)^2} – 1,{\text{ }}1 – 2{\left( {\frac{{\sqrt 5 }}{3}} \right)^2},{\text{ }}{\left( {\frac{2}{3}} \right)^2} – {\left( {\frac{{\sqrt 5 }}{3}} \right)^2}$$

$$\cos 2\theta = – \frac{1}{9}$$    A1     N2

[2 marks]

b.

## Question

Solve $${\log _2}(2\sin x) + {\log _2}(\cos x) = – 1$$, for $$2\pi < x < \frac{{5\pi }}{2}$$.

## Markscheme

correct application of $$\log a + \log b = \log ab$$     (A1)

eg$$\,\,\,\,\,$$$${\log _2}(2\sin x\cos x),{\text{ }}\log 2 + \log (\sin x) + \log (\cos x)$$

correct equation without logs     A1

eg$$\,\,\,\,\,$$$$2\sin x\cos x = {2^{ – 1}},{\text{ }}\sin x\cos x = \frac{1}{4},{\text{ }}\sin 2x = \frac{1}{2}$$

recognizing double-angle identity (seen anywhere)     A1

eg$$\,\,\,\,\,$$$$\log (\sin 2x),{\text{ }}2\sin x\cos x = \sin 2x,{\text{ }}\sin 2x = \frac{1}{2}$$

evaluating $${\sin ^{ – 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}{\text{ }}(30^\circ )$$     (A1)

correct working     A1

eg$$\,\,\,\,\,$$$$x = \frac{\pi }{{12}} + 2\pi ,{\text{ }}2x = \frac{{25\pi }}{6},{\text{ }}\frac{{29\pi }}{6},{\text{ }}750^\circ ,{\text{ }}870^\circ ,{\text{ }}x = \frac{\pi }{{12}}$$and $$x = \frac{{5\pi }}{{12}}$$, one correct final answer

$$x = \frac{{25\pi }}{{12}},{\text{ }}\frac{{29\pi }}{{12}}$$ (do not accept additional values)     A2     N0

[7 marks]

## Question

Let $$f(x) = 15 – {x^2}$$, for $$x \in \mathbb{R}$$. The following diagram shows part of the graph of $$f$$ and the rectangle OABC, where A is on the negative $$x$$-axis, B is on the graph of $$f$$, and C is on the $$y$$-axis. Find the $$x$$-coordinate of A that gives the maximum area of OABC.

## Markscheme

attempt to find the area of OABC     (M1)

eg$$\,\,\,\,\,$$$${\text{OA}} \times {\text{OC, }}x \times f(x),{\text{ }}f(x) \times ( – x)$$

correct expression for area in one variable     (A1)

eg$$\,\,\,\,\,$$$${\text{area}} = x(15 – {x^2}),{\text{ }}15x – {x^3},{\text{ }}{x^3} – 15x$$

valid approach to find maximum area (seen anywhere)     (M1)

eg$$\,\,\,\,\,$$$$A’(x) = 0$$

correct derivative     A1

eg$$\,\,\,\,\,$$$$15 – 3{x^2},{\text{ }}(15 – {x^2}) + x( – 2x) = 0,{\text{ }} – 15 + 3{x^2}$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$15 = 3{x^2},{\text{ }}{x^2} = 5,{\text{ }}x = \sqrt 5$$

$$x = – \sqrt 5 {\text{ }}\left( {{\text{accept A}}\left( { – \sqrt 5 ,{\text{ }}0} \right)} \right)$$     A2     N3

[7 marks]

## Question

The first two terms of an infinite geometric sequence are u1 = 18 and u2 = 12sin2 θ , where 0 < θ < 2$$\pi$$ , and θ ≠ $$\pi$$.

Find an expression for r in terms of θ.


a.i.

Find the possible values of r.


a.ii.

Show that the sum of the infinite sequence is $$\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}$$.


b.

Find the values of θ which give the greatest value of the sum.


c.

## Markscheme

valid approach     (M1)

eg   $$\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}$$

$$r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)$$      A1 N2

[2 marks]

a.i.

recognizing that sinθ is bounded      (M1)

eg    0 sin2 θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1

0 < r ≤ $$\frac{2}{3}$$      A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).
If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

a.ii.

correct substitution into formula for infinite sum       A1

eg  $$\frac{{18}}{{1 – \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}$$

evidence of choosing an appropriate rule for cos 2θ (seen anywhere)         (M1)

eg   cos 2θ = 1 − 2 sin2 θ

correct substitution of identity/working (seen anywhere)      (A1)

eg   $$\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}$$

eg  $$\frac{{18 \times 3}}{{2 + \left( {1 – 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 – \left( {1 – {\text{cos}}\,2\theta } \right)}}$$

$$\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}$$    AG N0

[4 marks]

b.

METHOD 1 (using differentiation)

recognizing $$\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0$$ (seen anywhere)       (M1)

finding any correct expression for $$\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}$$       (A1)

eg  $$\frac{{0 – 54 \times \left( { – 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, – 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ – 2}}\,\left( { – 2\,{\text{sin}}\,2\,\theta } \right)$$

correct working       (A1)

eg  sin 2θ = 0

any correct value for sin−1(0) (seen anywhere)       (A1)

eg  0, $$\pi$$, … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values)      (A1)

2θ = $$\pi$$, 3$$\pi$$ (accept values in degrees)

both correct answers $$\theta = \frac{\pi }{2},\,\frac{{3\pi }}{2}$$      A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

METHOD 2 (using denominator)

recognizing when S is greatest      (M1)

eg 2 + cos 2θ is a minimum, 1−r is smallest
correct working      (A1)

eg  minimum value of 2 + cos 2θ is 1, minimum r = $$\frac{2}{3}$$

correct working      (A1)

eg  $${\text{cos}}\,2\,\theta = – 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\theta = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\theta = 1$$

EITHER (using cos 2θ)

any correct value for cos−1(−1) (seen anywhere)      (A1)

eg  $$\pi$$, 3$$\pi$$, … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values for 2θ  (ignore additional values)      (A1)

2θ = $$\pi$$, 3$$\pi$$ (accept values in degrees)

OR (using sinθ)

sinθ = ±1     (A1)

sin−1(1) = $$\frac{\pi }{2}$$ (accept values in degrees) (seen anywhere)      A1

THEN

both correct answers $$\theta = \frac{\pi }{2},\,\frac{{3\pi }}{2}$$       A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

[6 marks]

c.

## Question

The expression $$6\sin x\cos x$$ can be expressed in the form $$a\sin bx$$ .

Find the value of a and of b .


a.

Hence or otherwise, solve the equation $$6\sin x\cos x = \frac{3}{2}$$ , for $$\frac{\pi }{4} \le x \le \frac{\pi }{2}$$ .


b.

## Markscheme

recognizing double angle     M1

e.g. $$3 \times 2\sin x\cos x$$ , $$3\sin 2x$$

$$a = 3$$ , $$b = 2$$     A1A1     N3

[3 marks]

a.

substitution $$3\sin 2x = \frac{3}{2}$$     M1

$$\sin 2x = \frac{1}{2}$$     A1

finding the angle     A1

e.g. $$\frac{\pi }{6}$$ , $$2x = \frac{{5\pi }}{6}$$

$$x = \frac{{5\pi }}{{12}}$$     A1     N2

Note: Award A0 if other values are included.

[4 marks]

b.