My task is this:
Find the taylor-series of $\cos^2x$ and $\sin^2x$.
My work so far:
We know that $\cos^2x \backslash \sin^2x = \frac{1\pm \cos 2x}{2}$, and the series for $\cos x = \sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!} \implies \cos 2x = \sum_{n=0}^\infty\frac{(-1)^n(2x)^{2n}}{(2n)!}$. Which leads us to $$\cos^2x \backslash\sin^2x = \frac{1\pm\sum_{n=0}^\infty\frac{(-1)^n(2x)^{2n}}{(2n)!}}{2} = \frac{1}{2}\left(1\pm\sum_{n=0}^\infty\frac{(-1)^n(2x)^{2n}}{(2n)!}\right).$$
Is it correct?
Best Answer
Since you know the Taylor expansion of $\sin x$ and $\sin^2x = (\sin x)^2$ you can use the Cauchy product: \begin{align*} \sin^2x &= \left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1}\right)^2 \\ &= \sum_{k=0}^\infty c_k, \end{align*} where \begin{align*} c_k &= \sum_{i=0}^k \frac{(-1)^i}{(2i+1)!}\frac{(-1)^{k-i}}{(2k - 2i + 1)!} x^{2i+1}x^{2k-2i + 1} \\ & = (-1)^kx^{2k+2}\sum_{i=0}^k \frac{1}{(2i+1)!(2k+2-(2i+1))!} \\ & = \frac{(-1)^kx^{2k+2}}{(2k+2)!}\sum_{i=0}^k \frac{(2k+2)!}{(2i+1)!(2k+2-(2i+1))!} \\ & = \frac{(-1)^kx^{2k+2}}{(2k+2)!}\sum_{i=0}^k \binom{2k+2}{2i+1} \\ & = \frac{(-1)^kx^{2k+2}}{(2k+2)!} 2^{2k+1} =\frac12 \frac{(-1)^k(2x)^{2k+2}}{(2k+2)!} \end{align*} Putting it together yields \begin{align*} \sin^2x &= \frac12\sum_{k=0}^\infty \frac{(-1)^k (2x)^{2k+2}}{(2k+2)!} \end{align*} By shifting the summation index down one, we arrive at the very same thing that you derived (maybe a bit cleverer than me).
Analog $\cos^2x$ can be done.