Proof of the angle sum identity for cos using series Let $sin(x)=\sum_{n=0}^{\infty}{(-1)^n\frac{x^{2n+1}}{(2n+1)!}}$ and $cos(x)=\sum_{n=0}^{\infty}{(-1)^n\frac{x^{2n}}{(2n)!}}$.How using above definition prove that $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$?
When you multiply the series for $sin(x)$ by $i$, and add it to the series for $cos(x)$, you get the series for $exp(ix)$. In other words, $cos(x)$, is the real part of $exp(ix)$. 

Therefore, $cos(x+y)$ is the real part of $exp(i(x+y))$. By laws of exponents, this is $$real( exp(ix)exp(iy))$$

By the first observation, this is $$real( (cox(x)+i sin(x)) (cox(y)+i sin(y)))$$

Multiply it out and select the real part: $$cos(x)cos(y) - sin(x)sin(y)$$
Very bad answer.  One cannot use complex analysis to prove the sum of angles formula, one must in fact use the sum of angles formula to prove all of the useful properties of the "complex exponential" (which is based on the definitions of sine and cosine!).  Otherwise you are just playing with definitions and giving circular reasoning.  The proof should be self-contained, involving multiplying the partial sums of the series and re-arranging the terms in a clever way.