Date | November 2016 | Marks available | 9 | Reference code | 16N.1.hl.TZ0.13 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Prove that | Question number | 13 | Adapted from | N/A |
Question
Find the value of \(\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}\).
Show that \(\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi \) where \(k \in \mathbb{Z}\).
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 \(k \in \mathbb{Z}\).
Hence or otherwise solve the equation \(\sin x + \sin 3x = \cos x\) in the interval \(0 < x < \pi \).
Markscheme
\(\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}\) (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 \(2x = \frac{{5\pi }}{6}\)
OR
\(\sin x + \sin 3x = \cos x \Rightarrow 2\sin 2x\cos x = \cos x\) 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]