let $$C=\frac{\cos\theta}{2}-\frac{\cos2\theta}{4}+\frac{\cos3\theta}{8}+…$$
$$S=\frac{\sin\theta}{2}-\frac{\sin2\theta}{4}+\frac{\sin3\theta}{8}+…$$
I want to find the sum of the series $C+iS$ and thus find expressions for $C$ and $S$.
so the sum of $C+iS$ is
$$C+iS=(\frac{\cos\theta}{2}-\frac{\cos2\theta}{4}+\frac{\cos3\theta}{8}+…)+i(\frac{\sin\theta}{2}-\frac{\sin2\theta}{4}+\frac{\sin3\theta}{8}+…)$$
$$=\frac{1}{2}(cos\theta+i\sin\theta)-\frac{1}{4}(cos2\theta+i\sin2\theta)+\frac{1}{8}(cos3\theta+i\sin3\theta) + …$$
$$=\frac{1}{2}(cos\theta+i\sin\theta)-\frac{1}{4}(cos\theta+i\sin\theta)^2+\frac{1}{8}(cos\theta+i\sin\theta)^3 + …$$
So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?
So I've continued and got the series to here:
$$\frac{1}{2}[e^{i\theta}-\frac{1}{2}(e^{i\theta})^2+\frac{1}{4}(e^{i\theta})^3…]$$ But I'm still not sure where to continue.
Best Answer
Hint:
Using Intuition behind euler's formula and the special case $e^{i\pi}=-1$
$$\dfrac{(\cos t+i\sin t)^n(-1)^{n-1}}{2^n}=-\dfrac{e^{int}(e^{i\pi})^n}{2^n}=-\left(\dfrac{e^{i(t+\pi)}}2\right)^n$$
Now for the common ratio$(r)$ of Geometric Series,
$|r|=\left|\dfrac{e^{i(t+\pi)}}2\right|=\dfrac12<1$