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 \(0 < x < \pi \).

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