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

Use the principle of mathematical induction to prove that

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

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

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

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

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

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

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

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

OR

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

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

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

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

THEN

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

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

[6 marks]