16N.1.AHL.TZ0.H_13

pestleMathematicsAIHLPaper 116N· ahl-3-8-unit-circle-pythag-identity-solving-trig-equations-graphicallysource ↗

Find the value of $\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}$ .

[2]

Show that $\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi$ where $k \in \mathbb{Z}$ .

[2]

Use the principle of mathematical induction to prove that

$\sin x + \sin 3x + \ldots + \sin (2n - 1)x = \frac{{1 - \cos 2nx}}{{2\sin x}},{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}x \ne k\pi$ where $k \in \mathbb{Z}$ .

[9]

Hence or otherwise solve the equation $\sin x + \sin 3x = \cos x$ in the interval $0 < x < \pi$ .

[6]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}$ (M1)A1

Note: Award M1 for 5 equal terms with \) + \) or $-$ signs.

[2 marks]

$\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \frac{{1 - (1 - 2{{\sin }^2}x)}}{{2\sin x}}$ M1

$\equiv \frac{{2{{\sin }^2}x}}{{2\sin x}}$ A1

$\equiv \sin x$ AG

[2 marks]

let ${\text{P}}(n):\sin x + \sin 3x + \ldots + \sin (2n - 1)x \equiv \frac{{1 - \cos 2nx}}{{2\sin x}}$

if $n = 1$

${\text{P}}(1):\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x$ which is true (as proved in part (b)) R1

assume ${\text{P}}(k)$ true, $\sin x + \sin 3x + \ldots + \sin (2k - 1)x \equiv \frac{{1 - \cos 2kx}}{{2\sin x}}$ M1

Notes: Only award M1 if the words “assume” and “true” appear. Do not award M1 for “let $n = k$ ” only. Subsequent marks are independent of this M1.

consider ${\text{P}}(k + 1)$ :

${\text{P}}(k + 1):\sin x + \sin 3x + \ldots + \sin (2k - 1)x + \sin (2k + 1)x \equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}$

$LHS = \sin x + \sin 3x + \ldots + \sin (2k - 1)x + \sin (2k + 1)x$ M1

$\equiv \frac{{1 - \cos 2kx}}{{2\sin x}} + \sin (2k + 1)x$ A1

$\equiv \frac{{1 - \cos 2kx + 2\sin x\sin (2k + 1)x}}{{2\sin x}}$

$\equiv \frac{{1 - \cos 2kx + 2\sin x\cos x\sin 2kx + 2{{\sin }^2}x\cos 2kx}}{{2\sin x}}$ M1

$\equiv \frac{{1 - \left( {(1 - 2{{\sin }^2}x)\cos 2kx - \sin 2x\sin 2kx} \right)}}{{2\sin x}}$ M1

$\equiv \frac{{1 - (\cos 2x\cos 2kx - \sin 2x\sin 2kx)}}{{2\sin x}}$ A1

$\equiv \frac{{1 - \cos (2kx + 2x)}}{{2\sin x}}$ A1

$\equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}$

so if true for $n = k$ , then also true for $n = k + 1$

as true for $n = 1$ then true for all $n \in {\mathbb{Z}^ + }$ R1

Note: Accept answers using transformation formula for product of sines if steps are shown clearly.

Note: Award R1 only if candidate is awarded at least 5 marks in the previous steps.

[9 marks]

EITHER

$\sin x + \sin 3x = \cos x \Rightarrow \frac{{1 - \cos 4x}}{{2\sin x}} = \cos x$ M1

$\Rightarrow 1 - \cos 4x = 2\sin x\cos x,{\text{ }}(\sin x \ne 0)$ A1

$\Rightarrow 1 - (1 - 2{\sin ^2}2x) = \sin 2x$ M1

$\Rightarrow \sin 2x(2\sin 2x - 1) = 0$ M1

$\Rightarrow \sin 2x = 0$ or $\sin 2x = \frac{1}{2}$ A1

$2x = \pi ,{\text{ }}2x = \frac{\pi }{6}$ and $2x = \frac{{5\pi }}{6}$

$\sin x + \sin 3x = \cos x \Rightarrow 2\sin 2x\cos x = \cos x$ M1A1

$\Rightarrow (2\sin 2x - 1)\cos x = 0,{\text{ }}(\sin x \ne 0)$ M1A1

$\Rightarrow \sin 2x = \frac{1}{2}$ of $\cos x = 0$ A1

$2x = \frac{\pi }{6},{\text{ }}2x = \frac{{5\pi }}{6}$ and $x = \frac{\pi }{2}$

THEN

$\therefore x = \frac{\pi }{2},{\text{ }}x = \frac{\pi }{{12}}$ and $x = \frac{{5\pi }}{{12}}$ A1

Note: Do not award the final A1 if extra solutions are seen.

[6 marks]

Examiners’ report

[N/A]