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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?

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  • $\begingroup$ It should at least have a $1/x^2$ term somewhere. Also note that $\frac{\cos^2x}{\sin^2x}=1-\frac1{\sin^2 x}$, which might be easier to deal with. $\endgroup$
    – Arthur
    Commented May 20, 2016 at 7:23
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    $\begingroup$ @user_of_math It is not $\cot^2x$. He is running $\cos^2x,\sin^2x$ side by side. $\endgroup$
    – almagest
    Commented May 20, 2016 at 7:24
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    $\begingroup$ @Thomas. Looks ok to me, although using $/$ to indicate two alternatives is a little confusing at first glance! $\endgroup$
    – almagest
    Commented May 20, 2016 at 7:25
  • $\begingroup$ @almagest Duh, did not notice the $\pm$. $\endgroup$ Commented May 20, 2016 at 7:26
  • $\begingroup$ @almagest Great, yes I'll edit / to \ :). $\endgroup$
    – Thomas
    Commented May 20, 2016 at 7:31

1 Answer 1

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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.

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    $\begingroup$ I agree that using the double-angle formula is cleverer. $\endgroup$ Commented May 20, 2016 at 9:30
  • $\begingroup$ Great work on showing how the Cauchy-product works in practise which I will review this instant!:) $\endgroup$
    – Thomas
    Commented May 20, 2016 at 9:44

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