Assuming that 
has a power-series representation in terms of powers of 
, verify that it has the form
![\[ \sin^2 x = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^{2n-1}}{(2n)!}x^{2n} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a2ebf2b2096190a007f06552716d0b9e_l3.png "Rendered by QuickLaTeX.com")
valid for all real .
Following the hint, we recall the trig identity
![\[ \cos 2x = 1 - 2 \sin^2 x \qquad \implies \qquad \sin^2 x = \frac{1}{2} \left( 1 - \cos (2x) \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ccb2c0895534a5b868dc80eeb032f366_l3.png "Rendered by QuickLaTeX.com")
Then, we know the expansion for 
is
![\[ \cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ccf343e043ce14cb4517ab3c69a5c11f_l3.png "Rendered by QuickLaTeX.com")
So, putting these together we have
![\begin{align*} \sin^2 x &= \frac{1}{2} - \frac{1}{2} \cos(2x) \\[9pt] &= \frac{1}{2} - \frac{1}{2} \left( \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!} \right)\\[9pt] &= \frac{1}{2} + \frac{1}{2} \left( \sum_{n=0}^{\infty} \frac{(-1)^{n+1}(2x)^{2n}}{(2n)!} \right) \\[9pt] &= \frac{1}{2} + \sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2^{2n-1}}{(2n)!} x^{2n} \\[9pt] &= \frac{1}{2} + \frac{-1}{2} + \sum_{n=1}^{\infty} \frac{(-1)^{n+1} 2^{2n-1}}{(2n)!} x^{2n} \\[9pt] &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} 2^{2n-1}}{(2n)!} x^{2n}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-52c483dfe04c278b079d851a460f133d_l3.png "Rendered by QuickLaTeX.com")
This is valid for all 
since the expansion for cosine is valid for all .