## Derivation of Power Series

How do I write a derivation of the cosine power series?

(I understand and can derive it, but it takes much space and is disjointed! ; how do you write the shortest and fastest derivation for it--briefly and fluently??)
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 Of course, that would be much faster--->just differentiate the sum formula for sin(x). But suppose we can't do that!-->what would be another fast way to derive cos(x)? Currently I start with cos(x), and then explain that as a Taylor series, cos(x-a) in this case is really the McLaurin series for cos (x), where a=0. Then, I write cos(x) = $$c_0 + c_1 x + c_2 x^2 + c_3 x^3 +...$$ Next, I express each derivative of cos (x) as the same series, except I reduce the powers of x and subunits of the constants appropriately. Then, I substitute x=0 (b/c it is McLaurin) and express each derivative of cos(0) as the constant "c" with the appropriate subunits. Next, I show that the odd derivatives of cos(0) are zero, and explain why the sum formula includes $$(-1)^n$$ and $$x^{2n}$$ Then, I show why, as a power series, the factorial (2n)! is needed in the denominator so that the terms will match the derivatives when multiplied out. Finally, I write the formula (I wish I knew LaTex...better!) as $$cos(x) = \sum\limits_{n = 0}^\infty {\frac{{\left( {-1} \right)^n x^{2n} }}{{\left( {2n} \right)!}}}$$ My question is: HOW to constrict/shorten this procedure?? Are there parts here I can connect or skip??
 Do you know Euler's identity, $$e^{i\theta} = \cos \theta + i \sin \theta$$ ? If so, the Taylor series for $e^x$ is obviously really easy to derive. Just substitute in $x = i \theta$, and note that $$\cos \theta = \mbox{Re} \left[ e^{i \theta} \right] = \frac{ e^{i \theta} + e^{-i \theta}}{2}$$ from which you can just take the series for each of those and add them.