# IB Math Analysis & Approaches Questionbank-Topic: SL 3.5 Definition of trigonometric ratios and Exact values SL Paper 1

## Question

Let $$p = \sin 40^\circ$$ and $$q = \cos 110^\circ$$ . Give your answers to the following in terms of p and/or q .

Write down an expression for

(i)     $$\sin 140^\circ$$ ;

(ii)    $$\cos 70^\circ$$ .

[2]
a(i) and (ii).

Find an expression for $$\cos 140^\circ$$ .

[3]
b.

Find an expression for $$\tan 140^\circ$$ .

[1]
c.

## Markscheme

(i) $$\sin 140^\circ = p$$     A1     N1

(ii) $$\cos 70^\circ = – q$$     A1     N1

[2 marks]

a(i) and (ii).

METHOD 1

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

e.g. diagram, $$\sqrt {1 – {p^2}}$$ (seen anywhere)

$$\cos 140^\circ = \pm \sqrt {1 – {p^2}}$$     (A1)

$$\cos 140^\circ = – \sqrt {1 – {p^2}}$$     A1     N2

METHOD 2

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

$$\cos 140^\circ = 2{\cos ^2}70 – 1$$     (A1)

$$\cos 140^\circ = 2{( – q)^2} – 1$$ $$( = 2{q^2} – 1)$$     A1     N2

[3 marks]

b.

METHOD 1

$$\tan 140^\circ = \frac{{\sin 140^\circ }}{{\cos 140^\circ }} = – \frac{p}{{\sqrt {1 – {p^2}} }}$$     A1     N1

METHOD 2

$$\tan 140^\circ = \frac{p}{{2{q^2} – 1}}$$     A1     N1

[1 mark]

c.

## 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)    $$y$$ .

[2]
a.

Let the area of the rectangle be A.

Show that $$A = 18\sin 2\theta$$ .

[3]
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.

[8]
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

Six equilateral triangles, each with side length 3 cm, are arranged to form a hexagon.
This is shown in the following diagram.

The vectors p , q and r are shown on the diagram.

Find p•(p + q + r).

## Markscheme

METHOD 1 (using |p| |2q| cosθ)

finding p + q + r (A1)

eg 2q,

| p + q + r | = 2 × 3 (= 6) (seen anywhere) A1

correct angle between p and q (seen anywhere) (A1)

$$\frac{\pi }{3}$$ (accept 60°)

substitution of their values (M1)

eg 3 × 6 × cos$$\left( {\frac{\pi }{3}} \right)$$

correct value for cos$$\left( {\frac{\pi }{3}} \right)$$ (seen anywhere) (A1)

eg $$\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}$$

p•(p + q + r) = 9 A1 N3

METHOD 2 (scalar product using distributive law)

correct expression for scalar distribution (A1)

eg pp + pq + pr

three correct angles between the vector pairs (seen anywhere) (A2)

eg 0° between p and p, $$\frac{\pi }{3}$$ between p and q, $$\frac{{2\pi }}{3}$$ between p and r

Note: Award A1 for only two correct angles.

substitution of their values (M1)

eg 3.3.cos0 +3.3.cos$$\frac{\pi }{3}$$ + 3.3.cos120

one correct value for cos0, cos$$\left( {\frac{\pi }{3}} \right)$$ or cos$$\left( {\frac{2\pi }{3}} \right)$$ (seen anywhere) A1

eg $$\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}$$

p•(p + q + r) = 9 A1 N3

METHOD 3 (scalar product using relative position vectors)

valid attempt to find one component of p or r (M1)

eg sin 60 = $$\frac{x}{3}$$, cos 60 = $$\frac{x}{3}$$, one correct value $$\frac{3}{2},\,\,\frac{{3\sqrt 3 }}{2},\,\,\frac{{ – 3\sqrt 3 }}{2}$$

one correct vector (two or three dimensions) (seen anywhere) A1

eg $$p = \left( \begin{gathered} \,\,\,\frac{3}{2} \hfill \\ \frac{{3\sqrt 3 }}{2} \hfill \\ \end{gathered} \right),\,\,q = \left( \begin{gathered} 3 \hfill \\ 0 \hfill \\ \end{gathered} \right),\,\,r = \left( \begin{gathered} \,\,\,\,\frac{3}{2} \hfill \\ – \frac{{3\sqrt 3 }}{2} \hfill \\ \,\,\,\,0 \hfill \\ \end{gathered} \right)$$

three correct vectors p + q + r = 2q (A1)

p + q + r = $$\left( \begin{gathered} 6 \hfill \\ 0 \hfill \\ \end{gathered} \right)$$ or $$\left( \begin{gathered} 6 \hfill \\ 0 \hfill \\ 0 \hfill \\ \end{gathered} \right)$$ (seen anywhere, including scalar product) (A1)

correct working (A1)
eg $$\left( {\frac{3}{2} \times 6} \right) + \left( {\frac{{3\sqrt 3 }}{2} \times 0} \right),\,\,9 + 0 + 0$$

p•(p + q + r) = 9 A1 N3

[6 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.

[2]
a.i.

Find the possible values of r.

[3]
a.ii.

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

[4]
b.

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

[6]
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   $$\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}$$

correct working that clearly leads to the given answer       A1

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

Let $$p = \sin 40^\circ$$ and $$q = \cos 110^\circ$$ . Give your answers to the following in terms of p and/or q .

Write down an expression for

(i)     $$\sin 140^\circ$$ ;

(ii)    $$\cos 70^\circ$$ .

[2]
a(i) and (ii).

Find an expression for $$\cos 140^\circ$$ .

[3]
b.

Find an expression for $$\tan 140^\circ$$ .

[1]
c.

## Markscheme

(i) $$\sin 140^\circ = p$$     A1     N1

(ii) $$\cos 70^\circ = – q$$     A1     N1

[2 marks]

a(i) and (ii).

METHOD 1

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

e.g. diagram, $$\sqrt {1 – {p^2}}$$ (seen anywhere)

$$\cos 140^\circ = \pm \sqrt {1 – {p^2}}$$     (A1)

$$\cos 140^\circ = – \sqrt {1 – {p^2}}$$     A1     N2

METHOD 2

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

$$\cos 140^\circ = 2{\cos ^2}70 – 1$$     (A1)

$$\cos 140^\circ = 2{( – q)^2} – 1$$ $$( = 2{q^2} – 1)$$     A1     N2

[3 marks]

b.

METHOD 1

$$\tan 140^\circ = \frac{{\sin 140^\circ }}{{\cos 140^\circ }} = – \frac{p}{{\sqrt {1 – {p^2}} }}$$     A1     N1

METHOD 2

$$\tan 140^\circ = \frac{p}{{2{q^2} – 1}}$$     A1     N1

[1 mark]

c.

## Question

Let $$f(x) = \sqrt 3 {{\rm{e}}^{2x}}\sin x + {{\rm{e}}^{2x}}\cos x$$ , for $$0 \le x \le \pi$$ . Solve the equation $$f(x) = 0$$ .

## Markscheme

$${{\rm{e}}^{2x}}\left( {\sqrt 3 \sin x + \cos x} \right) = 0$$     (A1)

$${{\rm{e}}^{2x}} = 0$$ not possible (seen anywhere)     (A1)

simplifying

e.g. $$\sqrt 3 \sin x + \cos x = 0$$ , $$\sqrt 3 \sin x = – \cos x$$ , $$\frac{{\sin x}}{{ – \cos x}} = \frac{1}{{\sqrt 3 }}$$     A1

EITHER

$$\tan x = – \frac{1}{{\sqrt 3 }}$$     A1

$$x = \frac{{5\pi }}{6}$$     A2     N4

OR

sketch of $$30^\circ$$ , $$60^\circ$$ , $$90^\circ$$ triangle with sides $$1$$, $$2$$, $$\sqrt 3$$     A1

work leading to $$x = \frac{{5\pi }}{6}$$     A1

verifying $$\frac{{5\pi }}{6}$$ satisfies equation     A1     N4

[6 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.

[2]
a.i.

Find the possible values of r.

[3]
a.ii.

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

[4]
b.

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

[6]
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   $$\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}$$

correct working that clearly leads to the given answer       A1

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.

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

[N/A]

c.

## Question

Let $$f(x) = {{\rm{e}}^{ – 3x}}$$ and $$g(x) = \sin \left( {x – \frac{\pi }{3}} \right)$$ .

Write down

(i)     $$f'(x)$$ ;

(ii)    $$g'(x)$$ .

[2]
a.

Let $$h(x) = {{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right)$$ . Find the exact value of $$h’\left( {\frac{\pi }{3}} \right)$$ .

[4]
b.

## Markscheme

(i) $$– 3{{\rm{e}}^{ – 3x}}$$     A1     N1

(ii) $$\cos \left( {x – \frac{\pi }{3}} \right)$$     A1     N1

[4 marks]

a.

evidence of choosing product rule     (M1)

e.g. $$uv’ + vu’$$

correct expression     A1

e.g. $$– 3{{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3x}}\cos \left( {x – \frac{\pi }{3}} \right)$$

complete correct substitution of $$x = \frac{\pi }{3}$$     (A1)

e.g. $$– 3{{\rm{e}}^{ – 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right)$$        

$$h’\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ – \pi }}$$     A1     N3

[4 marks]

b.

## Question

Let $$f(x) = \frac{{\cos x}}{{\sin x}}$$ , for $$\sin x \ne 0$$ .

In the following table, $$f’\left( {\frac{\pi }{2}} \right) = p$$ and $$f”\left( {\frac{\pi }{2}} \right) = q$$ . The table also gives approximate values of $$f'(x)$$ and $$f”(x)$$ near $$x = \frac{\pi }{2}$$ .

Use the quotient rule to show that $$f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}$$ .

[5]
a.

Find $$f”(x)$$ .

[3]
b.

Find the value of p and of q.

[3]
c.

Use information from the table to explain why there is a point of inflexion on the graph of f where $$x = \frac{\pi }{2}$$ .

[2]
d.

## Markscheme

$$\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x$$ , $$\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = – \sin x$$ (seen anywhere)     (A1)(A1)

evidence of using the quotient rule     M1

correct substitution     A1

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

$$f'(x) = \frac{{ – ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}$$     A1

$$f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}$$     AG      N0

[5 marks]

a.

METHOD 1

appropriate approach     (M1)

e.g. $$f'(x) = – {(\sin x)^{ – 2}}$$

$$f”(x) = 2({\sin ^{ – 3}}x)(\cos x)$$ $$\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)$$     A1A1     N3

Note: Award A1 for $$2{\sin ^{ – 3}}x$$ , A1 for $$\cos x$$ .

METHOD 2

derivative of $${\sin ^2}x = 2\sin x\cos x$$ (seen anywhere)     A1

evidence of choosing quotient rule     (M1)

e.g. $$u = – 1$$ ,  $$v = {\sin ^2}x$$ , $$f” = \frac{{{{\sin }^2}x \times 0 – ( – 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}$$

$$f”(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}$$ $$\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)$$     A1     N3

[3 marks]

b.

evidence of substituting $$\frac{\pi }{2}$$     M1

e.g. $$\frac{{ – 1}}{{{{\sin }^2}\frac{\pi }{2}}}$$ , $$\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}$$

$$p = – 1$$ ,  $$q = 0$$    A1A1     N1N1

[3 marks]

c.

second derivative is zero, second derivative changes sign     R1R1     N2

[2 marks]

d.

## Question

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

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

[2]
a.

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

[2]
b.

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

[3]
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

Let $$h(x) = \frac{{6x}}{{\cos x}}$$ . Find $$h'(0)$$ .

## Markscheme

METHOD 1 (quotient)

derivative of numerator is 6     (A1)

derivative of denominator is $$– \sin x$$     (A1)

attempt to substitute into quotient rule     (M1)

correct substitution     A1

e.g. $$\frac{{(\cos x)(6) – (6x)( – \sin x)}}{{{{(\cos x)}^2}}}$$

substituting $$x = 0$$     (A1)

e.g. $$\frac{{(\cos 0)(6) – (6 \times 0)( – \sin 0)}}{{{{(\cos 0)}^2}}}$$

$$h'(0) = 6$$     A1     N2

METHOD 2 (product)

$$h(x) = 6x \times {(\cos x)^{ – 1}}$$

derivative of 6x is 6     (A1)

derivative of $${(\cos x)^{ – 1}}$$ is $$( – {(\cos x)^{ – 2}}( – \sin x))$$     (A1)

attempt to substitute into product rule     (M1)

correct substitution     A1

e.g. $$(6x)( – {(\cos x)^{ – 2}}( – \sin x)) + (6){(\cos x)^{ – 1}}$$

substituting $$x = 0$$    (A1)

e.g. $$(6 \times 0)( – {(\cos 0)^{ – 2}}( – \sin 0)) + (6){(\cos 0)^{ – 1}}$$

$$h'(0) = 6$$     A1     N2

[6 marks]

## Question

Let $$\int_\pi ^a {\cos 2x{\text{d}}x} = \frac{1}{2}{\text{, where }}\pi < a < 2\pi$$. Find the value of $$a$$.

## Markscheme

correct integration (ignore absence of limits and “$$+C$$”)     (A1)

eg     $$\frac{{\sin (2x)}}{2},{\text{ }}\int_\pi ^a {\cos 2x = \left[ {\frac{1}{2}\sin (2x)} \right]_\pi ^a}$$

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     $$\frac{1}{2}\sin (2a) – \frac{1}{2}\sin (2\pi ),{\text{ }}\sin (2\pi ) – \sin (2a)$$

$$\sin (2\pi ) = 0$$     (A1)

setting their result from an integrated function equal to $$\frac{1}{2}$$     M1

eg     $$\frac{1}{2}\sin 2a = \frac{1}{2},{\text{ }}\sin (2a) = 1$$

recognizing $${\sin ^{ – 1}}1 = \frac{\pi }{2}$$     (A1)

eg     $$2a = \frac{\pi }{2},{\text{ }}a = \frac{\pi }{4}$$

correct value     (A1)

eg     $$\frac{\pi }{2} + 2\pi ,{\text{ }}2a = \frac{{5\pi }}{2},{\text{ }}a = \frac{\pi }{4} + \pi$$

$$a = \frac{{5\pi }}{4}$$     A1     N3

[7 marks]

## Question

The following diagram shows a triangle ABC and a sector BDC of a circle with centre B and radius 6 cm. The points A , B and D are on the same line.

$${\text{AB}} = 2\sqrt 3 {\text{ cm, BC}} = 6{\text{ cm, area of triangle ABC}} = 3\sqrt 3{\text{ c}}{{\text{m}}^{\text{2}}}{\rm{, A\hat BC}}$$ is obtuse.

Find $${\rm{A\hat BC}}$$.

[5]
a.

Find the exact area of the sector BDC.

[3]
b.

## Markscheme

METHOD 1

correct substitution into formula for area of triangle     (A1)

eg$$\,\,\,\,\,$$$$\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B,{\text{ }}6\sqrt 3 \sin B,{\text{ }}\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B = 3\sqrt 3$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$6\sqrt 3 \sin B = 3\sqrt 3 ,{\text{ }}\sin B = \frac{{3\sqrt 3 }}{{\frac{1}{2}(6)2\sqrt 3 }}$$

$$\sin B = \frac{1}{2}$$    (A1)

$$\frac{\pi }{6}(30^\circ )$$    (A1)

$${\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )$$     A1     N3

METHOD 2

(using height of triangle ABC by drawing perpendicular segment from C to AD)

correct substitution into formula for area of triangle     (A1)

eg$$\,\,\,\,\,$$$$\frac{1}{2}\left( {2\sqrt 3 } \right)(h) = 3\sqrt 3 ,{\text{ }}h\sqrt 3$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$h\sqrt 3 = 3\sqrt 3$$

height of triangle is 3     A1

$${\rm{C\hat BD}} = \frac{\pi }{6}(30^\circ )$$    (A1)

$${\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )$$     A1     N3

[5 marks]

a.

recognizing supplementary angle     (M1)

eg$$\,\,\,\,\,$$$${\rm{C\hat BD}} = \frac{\pi }{6},{\text{ sector}} = \frac{1}{2}(180 – {\rm{A\hat BC)(}}{{\text{6}}^2})$$

correct substitution into formula for area of sector     (A1)

eg$$\,\,\,\,\,$$$$\frac{1}{2} \times \frac{\pi }{6} \times {6^2},{\text{ }}\pi ({6^2})\left( {\frac{{30}}{{360}}} \right)$$

$${\text{area}} = 3\pi {\text{ }}({\text{c}}{{\text{m}}^2})$$     A1     N2

[3 marks]

b.

## Question

The following diagram shows triangle PQR.

Find PR.

## Markscheme

METHOD 1

evidence of choosing the sine rule     (M1)

eg$$\,\,\,\,\,$$$$\frac{a}{{\sin A}} = \frac{b}{{\sin B}}$$

correct substitution     A1

eg$$\,\,\,\,\,$$$$\frac{x}{{\sin 30}} = \frac{{13}}{{\sin 45}},{\text{ }}\frac{{13\sin 30}}{{\sin 45}}$$

$$\sin 30 = \frac{1}{2},{\text{ }}\sin 45 = \frac{1}{{\sqrt 2 }}$$     (A1)(A1)

correct working     A1

eg$$\,\,\,\,\,$$$$\frac{1}{2} \times \frac{{13}}{{\frac{1}{{\sqrt 2 }}}},{\text{ }}\frac{1}{2} \times 13 \times \frac{2}{{\sqrt 2 }},{\text{ }}13 \times \frac{1}{2} \times \sqrt 2$$

correct answer     A1     N3

eg$$\,\,\,\,\,$$$${\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}$$

METHOD 2 (using height of ΔPQR)

valid approach to find height of ΔPQR     (M1)

eg$$\,\,\,\,\,$$$$\sin 30 = \frac{x}{{13}},{\text{ }}\cos 60 = \frac{x}{{13}}$$

$$\sin 30 = \frac{1}{2}{\text{ or }}\cos 60 = \frac{1}{2}$$     (A1)

$${\text{height}} = 6.5$$     A1

correct working     A1

eg$$\,\,\,\,\,$$$$\sin 45 = \frac{{6.5}}{{{\text{PR}}}},{\text{ }}\sqrt {{{6.5}^2} + {{6.5}^2}}$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$\sin 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\cos 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\sqrt {\frac{{169 \times 2}}{4}}$$

correct answer     A1     N3

eg$$\,\,\,\,\,$$$${\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}$$

[6 marks]

## Question

The following table shows the probability distribution of a discrete random variable $$A$$, in terms of an angle $$\theta$$.

Show that $$\cos \theta = \frac{3}{4}$$.

[6]
a.

Given that $$\tan \theta > 0$$, find $$\tan \theta$$.

[3]
b.

Let $$y = \frac{1}{{\cos x}}$$, for $$0 < x < \frac{\pi }{2}$$. The graph of $$y$$between $$x = \theta$$ and $$x = \frac{\pi }{4}$$ is rotated 360° about the $$x$$-axis. Find the volume of the solid formed.

[6]
c.

## Markscheme

evidence of summing to 1     (M1)

eg$$\,\,\,\,\,$$$$\sum {p = 1}$$

correct equation     A1

eg$$\,\,\,\,\,$$$$\cos \theta + 2\cos 2\theta = 1$$

correct equation in $$\cos \theta$$     A1

eg$$\,\,\,\,\,$$$$\cos \theta + 2(2{\cos ^2}\theta – 1) = 1,{\text{ }}4{\cos ^2}\theta + \cos \theta – 3 = 0$$

evidence of valid approach to solve quadratic     (M1)

eg$$\,\,\,\,\,$$factorizing equation set equal to $$0,{\text{ }}\frac{{ – 1 \pm \sqrt {1 – 4 \times 4 \times ( – 3)} }}{8}$$

correct working, clearly leading to required answer     A1

eg$$\,\,\,\,\,$$$$(4\cos \theta – 3)(\cos \theta + 1),{\text{ }}\frac{{ – 1 \pm 7}}{8}$$

correct reason for rejecting $$\cos \theta \ne – 1$$     R1

eg$$\,\,\,\,\,$$$$\cos \theta$$ is a probability (value must lie between 0 and 1), $$\cos \theta > 0$$

Note:     Award R0 for $$\cos \theta \ne – 1$$ without a reason.

$$\cos \theta = \frac{3}{4}$$    AG  N0

a.

valid approach     (M1)

eg$$\,\,\,\,\,$$sketch of right triangle with sides 3 and 4, $${\sin ^2}x + {\cos ^2}x = 1$$

correct working

(A1)

eg$$\,\,\,\,\,$$missing side $$= \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}$$

$$\tan \theta = \frac{{\sqrt 7 }}{3}$$     A1     N2

[3 marks]

b.

attempt to substitute either limits or the function into formula involving $${f^2}$$     (M1)

eg$$\,\,\,\,\,$$$$\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} }$$

correct substitution of both limits and function     (A1)

eg$$\,\,\,\,\,$$$$\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x}$$

correct integration     (A1)

eg$$\,\,\,\,\,$$$$\tan x$$

substituting their limits into their integrated function and subtracting     (M1)

eg$$\,\,\,\,\,$$$$\tan \frac{\pi }{4} – \tan \theta$$

Note:     Award M0 if they substitute into original or differentiated function.

$$\tan \frac{\pi }{4} = 1$$    (A1)

eg$$\,\,\,\,\,$$$$1 – \tan \theta$$

$$V = \pi – \frac{{\pi \sqrt 7 }}{3}$$     A1     N3

[6 marks]

c.

## 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.

[2]
a.i.

Find the possible values of r.

[3]
a.ii.

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

[4]
b.

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

[6]
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   $$\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}$$

correct working that clearly leads to the given answer       A1

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 .

[3]
a.

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

[4]
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.

## Question

Let $$f(x) = \cos x + \sqrt 3 \sin x$$ , $$0 \le x \le 2\pi$$ . The following diagram shows the graph of $$f$$ .

The $$y$$-intercept is at ($$0$$, $$1$$) , there is a minimum point at A ($$p$$, $$q$$) and a maximum point at B.

Find $$f'(x)$$ .

[2]
a.

Hence

(i)     show that $$q = – 2$$ ;

(ii)    verify that A is a minimum point.

[10]
b(i) and (ii).

Find the maximum value of $$f(x)$$ .

[3]
c.

The function $$f(x)$$ can be written in the form $$r\cos (x – a)$$ .

Write down the value of r and of a .

[2]
d.

## Markscheme

$$f'(x) = – \sin x + \sqrt 3 \cos x$$     A1A1     N2

[2 marks]

a.

(i) at A, $$f'(x) = 0$$     R1

correct working     A1

e.g. $$\sin x = \sqrt 3 \cos x$$

$$\tan x = \sqrt 3$$     A1

$$x = \frac{\pi }{3}$$ , $$\frac{{4\pi }}{3}$$     A1

attempt to substitute their x into $$f(x)$$     M1

e.g. $$\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)$$

correct substitution     A1

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

correct working that clearly leads to $$– 2$$     A1

e.g. $$– \frac{1}{2} – \frac{3}{2}$$

 $$q = – 2$$     AG     N0

(ii) correct calculations to find $$f'(x)$$ either side of $$x = \frac{{4\pi }}{3}$$     A1A1

e.g. $$f'(\pi ) = 0 – \sqrt 3$$ ,  $$f'(2\pi ) = 0 + \sqrt 3$$   

$$f'(x)$$ changes sign from negative to positive     R1

so A is a minimum     AG     N0

[10 marks]

b(i) and (ii).

max when $$x = \frac{\pi }{3}$$     R1

correctly substituting $$x = \frac{\pi }{3}$$ into $$f(x)$$     A1

e.g. $$\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)$$

max value is 2     A1     N1

[3 marks]

c.

$$r = 2$$ , $$a = \frac{\pi }{3}$$     A1A1     N2

[2 marks]

d.

MAA SL 3.5-3.6 TRIGONOMETRIC IDENTITIES AND EQUATIONS [concise]-manav

### Question

[Maximum mark: 7] [without GDC]
Given that sin x = p , where x is an acute angle
(a) Find the value of cos x in terms of p ; [2]
(b) Hence, express the following in terms of p :
(i) tan x   (ii) cos 2x   (iii) sin 2x   (iv) tan 2x   (v) sin 4x  [5]

Ans:
(a) $$\cos x=\sqrt{1-p^{2}}$$
(b) $$\tan x=\frac{p}{\sqrt{1-p^{2}}}, \quad \cos 2 x=1-2 p^{2}$$
$$\sin 2 x=2 p \sqrt{1-p^{2}}, \quad \tan 2 x=\frac{2 p \sqrt{1-p^{2}}}{1-2 p^{2}}$$
$$\sin 4 x=4 p \sqrt{1-p^{2}}\left(1-2 p^{2}\right)$$

### Question

[Maximum mark: 17] [without GDC]
Given that sin 20° = p , and cos 20° = q (so that $$\left.p^{2}+q^{2}=1\right)$$ [5]
(a)
(b) By observing the unit circle   [12]

express the following in terms of p and/or q

Ans:
(a) $$\tan 20^{\circ}=\frac{p}{q}, \quad \sin 40^{\circ}=2 p q, \quad \cos 40^{\circ}=q^{2}-p^{2}=1-2 p^{2}=2 q^{2}-1$$
(b)

### Question

[Maximum mark: 4] [without GDC]
Given that $$\sin \theta=\frac{1}{2}, \quad \cos \theta=-\frac{\sqrt{3}}{2} \text { and } 0^{\circ}<\theta<360^{\circ}$$
(a) find the value of θ ; [2]
(b) write down the exact value of tanθ . [2]

Ans:
(a) Acute angle $$30^{\circ} \Rightarrow \theta=150^{\circ}$$ (2nd quadrant since sine positive and cosine negative)
(b) $$\tan 150^{\circ}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}$$

### Question

[Maximum mark: 6] [without GDC]
Given that $$\sin x=\frac{1}{3}$$, where x is an acute angle, find the exact value of
(a) cos x ; [4]
(b) cos 2x . [2]

Ans:
(a) x is an acute angle => cosx is positive.
$$\cos ^{2} x+\sin ^{2} x=1=>\cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=\sqrt{\frac{8}{9}}\left(=\frac{2 \sqrt{2}}{3}\right)$$
(b) $$\cos 2 x=1-2 \sin ^{2} x=1-2\left(\frac{1}{3}\right)^{2}=\frac{7}{9}$$

### Question

[Maximum mark: 6] [without GDC]
The following diagram shows a triangle ABC, where ACB is 90°, AB = 3, AC = 2 and BAC is  θ.(a) Show that $$\sin \theta=\frac{\sqrt{5}}{3}$$  [1]
(b) Show that  $$\sin 2 \theta=\frac{4 \sqrt{5}}{9}$$ [2]
(c) Find the exact value of cos 2θ. [3]

Ans:
(a) $$\sin \theta=\frac{\mathrm{BC}}{\mathrm{AB}}, \mathrm{BC}=\sqrt{3^{2}-2^{2}}=\sqrt{5} \quad \sin \theta=\frac{\sqrt{5}}{3}$$
(b) $$\sin 2 \theta=2 \sin \theta \cos \theta=2\left(\frac{\sqrt{5}}{3}\right)\left(\frac{2}{3}\right)=\frac{4 \sqrt{5}}{9}$$
(c) $$\cos 2 \theta=\frac{4}{9}-\frac{5}{9}=-\frac{1}{9} \quad \text { OR } \quad \cos 2 \theta=1-2 \times \frac{5}{9}=-\frac{1}{9}$$

### Question

[Maximum mark: 4] [without GDC]
If A is an obtuse angle in a triangle and $$\sin A=\frac{5}{13}$$, calculate the exact value of sin 2A.

Ans:
$$\sin A=\frac{5}{13} \Rightarrow \cos A=\pm \frac{12}{13}$$  But A is obtuse $$\Rightarrow \cos A=-\frac{12}{13}$$
$$\sin 2 A=2 \sin A \cos A=2 \times \frac{5}{13} \times\left(-\frac{12}{13}\right)=-\frac{120}{169}$$

### Question

[Maximum mark: 6] [without GDC]
Let p = sin 40° , q = cos110°. Give your answers to the following in terms of p and/or q
(a) Write down an expression for (i) sin140° ; (ii) cos 70°. [2]
(b) Find an expression for cos140°. [3]
(c) Find an expression for tan140°. [1]

Ans:
(a) (i) sin 140° = p (ii) cos 70° = -q
(b) METHOD 1
using $$\sin ^{2} \theta+\cos ^{2} \theta=1$$
$$\cos 140^{\circ}=\pm \sqrt{1-p^{2}} \quad \cos 140^{\circ}=-\sqrt{1-p^{2}}$$
METHOD 2
$$\text { using } \cos ^{2} \theta=2 \cos ^{2} \theta-1$$
$$\cos 140^{\circ}=2 \cos ^{2} 70-1=2(-q)^{2}-1=2 q^{2}-1$$
(c) METHOD 1  $$\tan 140^{\circ}=\frac{\sin 140^{\circ}}{\cos 140^{\circ}}=-\frac{p}{\sqrt{1-p^{2}}}$$
METHOD 2 $$\tan 140^{\circ}=\frac{p}{2 q^{2}-1}$$

### Question

[Maximum mark: 6] [without GDC]
(a) Given that $$\cos A=\frac{1}{3} \text { and } 0 \leq A \leq \frac{\pi}{2}$$,  find cos 2A.  [3]
(b) Given that $$\sin B=\frac{2}{3} \text { and } \frac{\pi}{2} \leq B \leq \pi$$, find cos B.  [3]

Ans:
(a) $$\cos 2 A=2 \cos ^{2} A-1=2 \times\left(\frac{1}{3}\right)^{2}-1 !=-\frac{7}{9}$$
(b) METHOD 1
using $$\sin ^{2} B+\cos ^{2} B=1$$
$$\cos B=\pm \sqrt{\frac{5}{9}}\left(=\pm \frac{\sqrt{5}}{3}\right)$$    $$\cos B==-\frac{\sqrt{5}}{3}$$
METHOD 2

Diagram, eg
third side equals $$\sqrt{5}$$
$$\cos B=-\frac{\sqrt{5}}{3}$$

### Question

[Maximum mark: 7] [without GDC]
The straight line with equation $$y=\frac{3}{4}$$makes an acute angle θ with the x-axis.
(a) Write down the value of tanθ [1]
(b) Find the value of (i) sin 2θ ; (ii) cos 2θ. [6]

Ans:
(a) $$\tan \theta=\frac{3}{4}$$
(b) (i) by using a right-angles triangle with sides 3,4,5
$$\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$$
$$\sin 2 \theta=2 \sin x \cos x=\frac{24}{25}$$
(ii)  $$\cos 2 \theta=1-2\left(\frac{3}{5}\right)^{2}=\frac{7}{25} \quad \text { OR } \quad \cos 2 \theta=\left(\frac{3}{5}\right)^{2},\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}=\frac{7}{25}$$

### Question

[Maximum mark: 7] [without GDC]
Let $$f(x)=\sin ^{3} x+\cos ^{3} x \tan x, \quad \frac{\pi}{2}<x<\pi$$.
(a) Show that $$f(x)=\sin x$$ [2]
(b) Let $$\sin x=\frac{2}{3}$$. Show that $$f(2 x)=-\frac{4 \sqrt{5}}{9}$$. [5]

Ans:
(a) $$f(x)=\sin ^{3} x+\cos ^{3} x \frac{\sin x}{\cos x}=\sin x\left(\sin ^{2} x+\cos ^{2} x\right)=\sin x$$
(b) $$f(2 x)=\sin 2 x=2 \sin x \cos x$$
$$\sin ^{2} x+\cos ^{2} x=1 \Rightarrow \cos x=-\frac{\sqrt{5}}{3}$$
$$f(2 x)=2\left(\frac{2}{3}\right)\left(-\frac{\sqrt{5}}{3}\right)=-\frac{4 \sqrt{5}}{9} f(2 x)$$

### Question

[Maximum mark: 28] [without GDC]
Complete the solutions of the following equations, within the given domain.

Ans:

### Question

[Maximum mark: 6] [without GDC]
Solve the equation, $$\sin x=-\frac{1}{2}$$, (i) for $$0^{\circ} \leq x \leq 360^{\circ}(\mathrm{deg})$$ (ii) $$\text { for } 0 \leq x \leq 2 \pi(\mathrm{rad})$$

Ans:
(i) x = 210°, x = 300°
(ii) x = 7π/6 , x =11π/6

### Question

[Maximum mark: 32] [without GDC]
Write down the solutions of the following equations, within the given domain.

Ans:

### Question

[Maximum mark: 6] [without GDC]
Solve the equation $$\cos x=\frac{1}{2}$$ (i) for $$0^{\circ} \leq x \leq 360^{\circ}(\mathrm{deg})$$  (ii) for $$0 \leq x \leq 2 \pi(\mathrm{rad})$$

Ans:
(i) x = 60° ,x = 300°      (ii) x = π/3 , x = 5π/3

### Question

[Maximum mark: 6] [without GDC]
Solve the equation $$\cos x=-\frac{1}{2}$$  (i) for $$0^{\circ} \leq x \leq 360^{\circ}(\mathrm{deg})$$  (ii) for $$0 \leq x \leq 2 \pi(\mathrm{rad})$$

Ans:
(i) x = 120° ,x = 240°      (ii) x = 2π/3 , x = 4π/3

### Question

[Maximum mark: 24] [without GDC]
Write down the solutions of the following equations, within the given domain.

Ans:

### Question

[Maximum mark: 24] [without GDC]
Write down the solutions of the following equations, within the given domain.

Ans:

### Question

[Maximum mark: 7] [without GDC]
Solve the equation $$2 \cos x=\sin 2 x \text {, for } 0 \leq x \leq 3 \pi$$.

Ans:
METHOD 1
$$2 \cos x=\sin 2 x \Leftrightarrow 2 \cos x=2 \sin x \cos x \Leftrightarrow 2 \cos x(1-\sin x)=0 \Leftrightarrow \cos x=0 \text { OR } \sin x=1$$
$$x=\frac{\pi}{2}, x=\frac{3 \pi}{2}, x=\frac{5 \pi}{2}$$
intersection points at $$x=\frac{\pi}{2}, x=\frac{3 \pi}{2}, x \frac{5 \pi}{2}$$

### Question

[Maximum mark: 6] [with / without GDC]
Solve the equation $$2 \cos ^{2} x=\sin 2 x \text { for } 0 \leq x \leq \pi$$, giving your answers in terms of π.

Ans:
$$2 \cos ^{2} x=2 \sin x \cos x \Leftrightarrow 2 \cos ^{2} x-2 \sin x \cos x=0 \Leftrightarrow 2 \cos x(\cos x-\sin x)=0$$
cos x = 0, (cos x – sin x) = 0
cos x = 0 $$\Leftrightarrow x=\frac{\pi}{2}$$
$$\cos x-\sin x=0 \Leftrightarrow \cos x=\sin x \Leftrightarrow \tan x=1 \Leftrightarrow x=\frac{\pi}{4}$$

### Question

[Maximum mark: 6] [with GDC]
(a) Factorize the expression $$3 \sin ^{2} x-11 \sin x+6$$. [2]
(b) Consider the equation $$3 \sin ^{2} x-11 \sin x+6=0$$.
(i) Find the two values of sin x which satisfy this equation,
(ii) Solve the equation, for $$0^{\circ} \leq x \leq 180^{\circ}$$. [4]

Ans:
(a) (3 sin x – 2)(sin x – 3)
(b) (i) (3sinx – 2)(sinx – 3) = 0 $$\Rightarrow$$ sin x $$\frac{2}{3}$$      sin x = 3
(ii) x = 41.8°, 138°

### Question

[Maximum mark: 4] [with GDC]
(a) Write the expression $$3 \sin ^{2} x+4 \cos x$$ in the form $$a \cos ^{2} x+b \cos x+c$$.  [1]
(b) Hence or otherwise, solve the equation $$3 \sin ^{2} x+4 \cos x-4=0,0^{\circ} \leq x \leq 90^{\circ}$$ [3]

Ans:
(a) $$3 \sin ^{2} x+4 \cos x=3\left(1-\cos ^{2} x\right)+4 \cos x=3-3 \cos ^{2}+4 \cos x$$
(b) $$3 \sin ^{2} x+4 \cos x-4=0 \Rightarrow 3-3 \cos ^{2} x+4 \cos x-4=0$$
$$\Rightarrow 3 \cos ^{2} x-4 \cos x+1=0$$
$$\cos x=\frac{1}{3} \text { or } \cos x=1$$
x = 70.5° or x = 0

### Question

[Maximum mark: 6] [without GDC]
(a) Express 2cos2 + sin x  in terms of sin x only. [2]
(b) Solve the equation 2cos2x + sinx = 2 for x in the interval $$0 \leq x \leq \pi$$  [4]

Ans:
(a) $$2 \cos ^{2} x+\sin x=2\left(1-\sin ^{2} x\right)+\sin x=2-2 \sin ^{2} x+\sin x$$
(b) $$2 \cos ^{2} x+\sin x=2 \Rightarrow 2-2 \sin ^{2} x+\sin x=2$$
$$\sin x-2 \sin ^{2} x=0$$
sin x(1 – 2sin x) = 0
sin x = 0 or sin x = $$\frac{1}{2}$$
sin x = 0 ⇒ x = 0 or π
$$\sin x=\frac{1}{2} \Rightarrow x=\frac{\pi}{6} \text { or } \frac{5 \pi}{6}$$

### Question

[Maximum mark: 6] [without GDC]
Consider the trigonometric equation $$2 \sin ^{2} x=1+\cos x$$.
(a) Write this equation in the form f (x) = 0, where $$f(x)=a \sin ^{2} x+b \sin x+c$$, and $$a, b, c \in \mathbb{Z}$$   [2]
(b) Factorize f (x) [1]
(c) Solve $$f(x)=0 \text { for } 0^{\circ} \leq x \leq 360^{\circ}$$ [3]

Ans:
(a) $$2 \sin ^{2} x=1+\cos x \Leftrightarrow 2\left(1-\cos ^{2} x\right)=1+\cos x \Leftrightarrow 2-2 \cos ^{2} x=1+\cos x$$
$$\Rightarrow 2 \cos ^{2} x+\cos x-1=0$$
(b) $$2 \cos ^{2} x+\cos x-1=(2 \cos x-1)(\cos x+1)$$
(c) $$\cos x=\frac{1}{2} \Rightarrow x=60^{\circ}, \text { or } x=300^{\circ}$$
cos x = –l ⇒ x = 180°
Solutions ⇒ x = 60°, x = 180° , x = 300°

### Question

[Maximum mark: 10] [without GDC]
Consider the equation 3cos 2x + sinx = 1.
(a) Write this equation in the form $$f(x)=0, \text { where } f(x)=p \sin ^{2} x+q \sin x+r$$ and $$p, q, r \in \mathbb{Z}$$.  [2]
(b) Factorize f (x). [2]
(c) Write down the number of solutions of $$f(x)=0, \text { for } 0 \leq x \leq 2 \pi$$. [2]
Extra Questions:
(d) Find the two solutions in the 3rd and fourth quadrant [4]

Ans:
(a) $$3 \cos 2 x+\sin x=1 \Leftrightarrow 3\left(1-2 \sin ^{2} x\right)+\sin x=1$$
6 sinx2 – sinx – 2 = 0
(b) (3 sin x – 2)(2 sin x + 1)
(c) 4 solutions
Extra Question
(d) The last equation gives $$\sin x=\frac{2}{3} \text { or } \sin x=-\frac{1}{2}$$
In the 3rd and 4th quadrants sin x is negative. Hence $$\sin x=-\frac{1}{2}$$
$$x=\frac{\pi}{6}+\pi=\frac{7 \pi}{6} \quad \text { or } x=2 \pi-\frac{\pi}{6}=\frac{11 \pi}{6}$$

### Question

[Maximum mark: 7] [with / without GDC]
Solve $$\cos 2 x-3 \cos x-3-\cos ^{2} x=\sin ^{2} x, \text { for } 0 \leq x \leq 2 \pi$$.

Ans:
$$\cos 2 x-3 \cos x-3-\cos ^{2} x=\sin ^{2} x \Leftrightarrow \cos 2 x-3 \cos x-3=\cos ^{2} x+\sin ^{2} x$$
$$2 \cos ^{2} x-1-3 \cos x-3=1 \Leftrightarrow 2 \cos ^{2} x-3 \cos x-5=0$$
$$\cos x=\frac{5}{2}(\text { rejected }), \cos x=-1$$
$$x=\pi$$

### Question

[Maximum mark: 7] [without GDC]
(a) Show that $$4-\cos 2 \theta+5 \sin \theta \equiv 2 \sin ^{2} \theta+5 \sin \theta+3$$ [2]
(b) Hence, solve the equation $$4-\cos 2 \theta+5 \sin \theta=0 \text { for } 0 \leq \theta \leq 2 \pi \text {. }$$ [5]

Ans:
(a) $$4-\cos 2 \theta+5 \sin \theta=4-\left(1-2 \sin ^{2} \theta\right)+5 \sin \theta=2 \sin ^{2} \theta+5 \sin \theta+3$$
(b) $$\left.2 \sin ^{2} \theta+5 \sin \theta+3=0 \Leftrightarrow 2 \sin \theta+3\right)(\sin \theta+1)=0$$
$$\sin \theta=-1, ; \sin \theta=-\frac{3}{2} \text { (rejected) }$$
$$\theta=\frac{3 \pi}{2}$$

### Question

[Maximum mark: 4] [without GDC]
Solve the equation $$\sqrt{3} \cos x=\sin x$$ for x in the interval $$0^{\circ} \leq x \leq 360^{\circ}$$

Ans:
$$\sqrt{3} \cos x=\sin x \Rightarrow \tan x=\sqrt{3} \Rightarrow x=60^{\circ} \text { or } x=240^{\circ}$$

### Question

[Maximum mark: 6] [without GDC]
Let $$f(x)=\sqrt{3} \mathrm{e}^{2 x} \sin x+\mathrm{e}^{2 x} \cos x, \text { for } 0 \leq x \leq \pi$$. Given that $$\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$$, solve the equation f (x) = 0.

Ans:
$$\mathrm{e}^{2 x}(\sqrt{3} \sin x+\cos x)=0$$
$$\mathrm{e}^{2 x}=0 \text { not possible }$$
$$\sqrt{3} \sin x+\cos x=0: \Leftrightarrow \sqrt{3} \sin x=-\cos x: \Leftrightarrow \frac{\sin x}{-\cos x}=\frac{1}{\sqrt{3}} \Leftrightarrow \tan x=-\frac{1}{\sqrt{3}}$$
$$x=\frac{5 \pi}{6}$$

### Question

[Maximum mark: 6] [with GDC]
Solve the equations
(a) 3cosx = 5sinx , for x in the interval $$0^{\circ} \leq x \leq 360^{\circ}$$, (to the nearest degree) [3]
(b) $$\cos 3 x=\cos (0.5 x), \text { for } 0 \leq x \leq \pi$$ [3]
Solve also [without GDC] by using the following information
For (a) it is given that $$\tan ^{-1} 0.6=31^{\circ}$$
For (b) it is given that $$\cos A=\cos B \Rightarrow\left\{\begin{array}{l} A=B+2 k \pi \\ \text { or } \\ A=-B+2 k \pi \end{array} \quad \text { for some } k \in Z\right.$$

Ans:
(a) $$x=31^{\circ} \text { or } x=211^{\circ}$$
(b) x= 0, x = 1.80 , x = 2.51 [3 s.f.]
Extra Questions:
(a) $$3 \cos x=5 \sin x \Leftrightarrow \tan x=3 / 5\left(=0.6 \Leftrightarrow x=31^{\circ}+180^{\circ} k\right.$$
$$x=31^{\circ} \text { or } x=211^{\circ}$$
(b) $$\cos 3 x=\cos (0.5 x) \Leftrightarrow 3 x=0.5 x+2 k \pi \text { or } 3 x=-0.5 x+2 k \pi$$
$$3 x=0.5 x+2 k \pi \Leftrightarrow 2.5 x=2 k \pi \Leftrightarrow x=(4 / 5) k \pi \text {. So } x=0, x=(4 / 5) \pi$$
$$3 x=-0.5 x+2 k \pi \Leftrightarrow 3.5 x=2 k \pi \Leftrightarrow x=(4 / 7) k \pi \text {. So } x=0, x=(4 / 7) \pi$$
Solutions: x = 0, x = (4/5)π, x = (4/7)π

### Question

[Maximum mark: 4] [without GDC]
Solve the equation $$3 \sin ^{2} x=\cos ^{2} x, \text { for } 0^{\circ} \leq x \leq 180^{\circ} \text {. }$$
[ Please solve in 3 different ways! ]
By expressing the equation in terms of sin x only:
By expressing the equation in terms of cos x only:
By expressing the equation in terms of tan x only:

Ans:
– By expressing the equation in terms of sin x only:
$$3 \sin ^{2} x=\cos ^{2} x \Leftrightarrow 3 \sin ^{2} x=1-\sin ^{2} x \Leftrightarrow 4 \sin ^{2} x=1 \Leftrightarrow \sin ^{2} x=1 / 4 \Leftrightarrow \sin x=\pm 1 / 2$$
Solutions: x = 30° or x = 150°
– By expressing the equation in terms of cos x only:
$$\sin ^{2} x=\cos ^{2} x \Leftrightarrow 3\left(1-\cos ^{2} x\right)=\cos ^{2} x \Leftrightarrow 4 \cos ^{2} x=3 \Leftrightarrow \cos ^{2} x=3 / 4 \Leftrightarrow \cos x=\pm \sqrt{3} / 2$$
Solutions: x = 30° or x = 150°
– By expressing the equation in terms of tan x only:
$$\sin ^{2} x=\cos ^{2} x \Leftrightarrow \tan ^{2} x=\frac{1}{3} \Rightarrow \tan x=\pm \frac{1}{\sqrt{3}} \Rightarrow x=30^{\circ} \text { or } x=150^{\circ}$$

### Question

[Maximum mark: 11] [with GDC]
(a) Consider the equation $$x^{2}+k x+1=0$$. For what values of k does this equation have two equal roots? [3]
Let f be the function $$f(\theta)=2 \cos 2 \theta+4 \cos \theta+3, \text { for }-360^{\circ} \leq x \leq 360^{\circ}$$.
(b) Show that this function may be written as $$f(\theta)=4 \cos ^{2} \theta+4 \cos \theta+1$$  [1]
(c) Consider the equation $$f(\theta)=0 \text {, for }-360^{\circ} \leq x \leq 360^{\circ}$$.
(i) How many distinct values of cosθ satisfy this equation?
(ii) Find all values of θ which satisfy this equation. [5]
(d) Given that $$f(\theta)=c$$ is satisfied by only three values of θ, find the value of c.
[hint: look at the graph of the function f(θ) ]   [2]

Ans:
(a) $$\Delta=0 \Leftrightarrow k^{2}-4 \times 4 \times 1=0 \Leftrightarrow k^{2}=16 \Leftrightarrow k=4, k=-4$$
(b) using cos $$2 \theta=2 \cos ^{2} \theta-1$$
$$f(\theta)=4 \cos ^{2} \theta+4 \cos \theta+1$$
(c) (i) 1
(ii) METHOD 1
Solve for cos θ
$$\cos \theta=-\frac{1}{2}$$
θ = 240, 120, – 240, -120
METHOD 2
Directly by GDC: θ = 240, 120, -240, -120
(d) Using graph, c = 9

### Question

[Maximum mark: 10] [with GDC]
The diagram below shows a plan for a window in the shape of a trapezium.

Three sides of the window are 2 m long. The angle between the sloping sides of the window and the base is θ, where $$0<\theta<\frac{\pi}{2}$$.
(a) Show that the area of the window is given by $$y=4 \sin \theta+2 \sin 2 \theta$$. [5]
(b) Zoe wants a window to have an area of 5 m2. Find the two possible values of θ. [3]
(c) John wants two windows which have the same area A but different values of θ.
Find all possible values for A.
[hint: in other words, find the range y, except max/min values] [2]

(a) For the height h, $$\sin \theta=\frac{h}{2} \Leftrightarrow h=2 \sin \theta$$
For the base of triangle b, $$\cos \theta=\frac{b}{2} \Leftrightarrow b=2 \cos \theta$$
$$\text { Area } y=2\left(\frac{1}{2} \times 2 \cos \theta \times 2 \sin \theta\right)+2 \times 2 \sin \theta=4 \sin \theta \cos \theta+4 \sin \theta$$