How to find $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}$? 
Let $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}=f(x)$, where $f(u)=c\sin u$.Find $c$.

Trial:$\sum_{k=1}^{\infty}\frac{c \sin (x+k\pi)}{2^k}=c\sin x$.Then $c=0$ is a solution. Is there any other solution of $c$? Mainly I am interested in the sum $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}$. please help.
 A: Using a little trigonometry and the sum of a convergent infinite geometric series:
$$\sin(x+k\pi)=\sin x\cos k\pi+\sin k\pi\cos x=\sin x\cos k\pi=(-1)^k\sin x\Longrightarrow$$
$$c\sin x=\sum_{k=1}^\infty\frac{c\sin(x+k\pi)}{2^k}=c\sin x\sum_{k=1}^\infty\left(-\frac{1}{2}\right)^k=c\sin x\frac{-\frac{1}{2}}{1+\frac{1}{2}}=$$
$$=-\frac{c\sin x}{3}\Longrightarrow c\sin x=-\frac{c\sin x}{3}\Longleftrightarrow c=0\,\,\,\vee\,\,\,\sin x=0\Longleftrightarrow$$
$$\Longleftrightarrow c=0\,\,\,\vee\,\,\,x=n\pi\,\,\,,n\in\Bbb Z$$
A: Note that $\sin(x+k\pi)=(-1)^k \sin x$ so that 
$$\sum_{k=1}^\infty \frac{\sin(x+k\pi)}{2^k}= \sin x \cdot \sum_{k=1}^\infty \frac{(-1)^k}{2^k}=(-1/3)\sin x,$$ on using the sum of a geometric series.
A: If $$T_k=\frac{f(x+k\pi)}{2^k}=\frac{c\sin(x+k\pi)}{2^k}$$
$$T_{k+1}=\frac{f\{x+(k+1)\pi\}}{2^k}=\frac{c\sin\{x+(k+1)\pi\}}{2^{k+1}}=-\frac{T_k}2$$
So, $T_2=-\frac{T_1}2,T_3=-\frac{T_2}2=\left(-\frac12\right)^2T_1$ and $T_1=\frac{c\sin(x+\pi)}2=-\frac{c\sin x}2$
$$\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}=T_1\sum_{k=0}^{\infty}\left(-\frac12\right)^k=\left(-\frac{c\sin x}2\right)\frac1{1-\left(-\frac12\right)}=-\frac c3\sin x$$
Now equate this with $f(x)=c\sin x$
