Sum of series $\sum_{n=1}^{\infty}\sin(\frac{1}{2^n})\cos(\frac{3}{2^n})$

Question

How can we get sum of this series? $$\sum_{n=1}^{\infty}\sin(\frac{1}{2^n})\cos(\frac{3}{2^n})$$

I think we must apply this theorem.

if in $\sum_{n=1}^\infty a_{n}$, $a_n = b_{n+1}-b_n$ and $\lim_{n\to\infty} b_n = b$; $ \implies $ $\sum_{n=1}^\infty a_{n}=b-b_1$

But how?

Answer 1

The identity $\sin x\cos 3x=\frac{1}{2}(\sin 4x-\sin 2x)$ gives$$\sum_{n\ge 1}\sin\frac{1}{2^n}\cos\frac{4}{2^n}=\frac{1}{2}\sum_{n\ge 1}(\sin\frac{4}{2^n}-\sin\frac{2}{2^n})=\frac{1}{2}\sin 2.$$

Answer 2

Hint: Use identity $$\sin a\cos b=\dfrac12(\sin(a+b)+\sin(a-b))$$ and then telescopic property of series.

Answer 3

Hint:

Compute first $\;\sum_{n=1}^{\infty}\sin \frac{1}{2^n}\, \cos \frac{3}{2^n}$ and use the linearisation formula $$2\sin a \,\cos b=\sin(a+b)+\sin(a-b).$$