Cfrgtkky

let $$C=frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...$$
$$S=frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...$$
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
$$C+iS=(frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...)+i(frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...)$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(cos2theta+isin2theta)+frac{1}{8}(cos3theta+isin3theta) + ...$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(costheta+isintheta)^2+frac{1}{8}(costheta+isintheta)^3 + ...$$

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:
$$frac{1}{2}[e^{itheta}-frac{1}{2}(e^{itheta})^2+frac{1}{4}(e^{itheta})^3...]$$ But I'm still not sure where to continue.

HINT

We have that

$$C+iS=sum_{k=1}^inftyfrac{(-1)^{k+1}}{2^k}e^{left(ikthetaright)}
=-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k$$

then refer to geometric series which holds also for $r$ complex $|r|<1$.

Hint:

Using Intuition behind euler's formula and the special case $e^{ipi}=-1$

$$dfrac{(cos t+isin t)^n(-1)^{n-1}}{2^n}=-dfrac{e^{int}(e^{ipi})^n}{2^n}=-left(dfrac{e^{i(t+pi)}}2right)^n$$

Now for the common ratio$(r)$ of Geometric Series,

$|r|=left|dfrac{e^{i(t+pi)}}2right|=dfrac12<1$

Hint: Consider geometric series
$$sum_{ngeq0}(-frac{z}{2})^n=dfrac{1}{1+frac{z}{2}}$$
about $z=0$.