Trigonometric series

Let $a \in \mathbb{R}$ . Prove that

$\sum_{n=1}^\infty 2^{2n}\sin^4 \frac a{2^n}=a^2-\sin^2a$

Solution

First of all we note that

$\begin{align*} 2^{2n}\sin^4\frac a{2^n}&=2^{2n}\cdot\sin^2\frac a{2^n}\cdot\sin^2\frac a{2^n}\\ &=2^{2n}\cdot\sin^2\frac a{2^n}\cdot\left(1-\cos^2\frac a{2^n}\right)\\ &=2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n}\cdot\sin^2\frac a{2^n}\cos^2\frac a{2^n}\\ &=2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n-2}\sin^2\frac a{2^{n-1}} \end{align*}$

Hence,

$\begin{align*} \sum_{n=1}^{m}2^{2n}\sin^4 \frac{a}{2^n} &= \sum_{n=1}^{m} \left( 2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n-2}\sin^2\frac a{2^{n-1}} \right) \\ &= 2^{2m}\sin^2\frac{a}{2^m}-\sin^2a \end{align*}$

Letting $m \rightarrow +\infty$ we get the requested value.

Trigonometric series

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