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Problem 1140

National olympiad, first round

Algebra Difficulty 6.2 Prove it

11.31. Prove that

$\begin{aligned} \frac{\sin (2 k+1) \varphi}{\sin \varphi}= & (-4)^{k}\left(\sin ^{2} \varphi-\sin ^{2} \frac{\pi}{2 k+1}\right) \times \\ & \times\left(\sin ^{2} \varphi-\sin ^{2} \frac{2 \pi}{2 k+1}\right) \ldots\left(\sin ^{2} \varphi-\sin ^{2} \frac{k \pi}{2 k+1}\right) \end{aligned}$

## 11.7. Auxiliary Trigonometric Functions

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

11.31. According to problem 11.30 b) $\frac{\sin (2 k+1) \varphi}{\sin \varphi}=P_{k}\left(\sin ^{2} \varphi\right)$ , where $P_{k}$ is a polynomial of degree $k$ . Further, if $\varphi=\frac{l \pi}{2 k+1}$ , then $\sin (2 k+1) \varphi=$ $=0$ . Therefore,

$P_{k}(x)=\lambda\left(x-\sin ^{2} \frac{\pi}{2 k+1}\right)\left(x-\sin ^{2} \frac{2 \pi}{2 k+1}\right) \ldots\left(x-\sin ^{2} \frac{k \pi}{2 k+1}\right)$

where $\lambda$ is some number. To compute $\lambda$ , let $x=0$ . Clearly, $P(0)=\lim _{\varphi \rightarrow 0} \frac{\sin (2 k+1) \varphi}{\sin \varphi}=2 k+1$ . Moreover, according to
problem 23.7 v)

$\sin ^{2} \frac{\pi}{2 k+1} \sin ^{2} \frac{2 \pi}{2 k+1} \ldots \sin ^{2} \frac{k \pi}{2 k+1}=\frac{2 k+1}{4^{k}}$

Thus, $\lambda=(-4)^{k}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.