Taylor Series Expansion for $\sin^2(\omega t)$ What are the first few terms for the Taylor Series Expansion for $\sin^2(\omega t)$? $(\omega$=$2\pi f$)
If you could show some working, that would be helpful
 A: Hint: We have $\cos 2x=1-2\sin^2 x$.  Thus
$$\sin^2 (\omega t)=\frac{1}{2}(1-\cos(2\omega t)).$$
Now use the series expansion of $\cos z$. 
A: Sometimes a little sledgehammering is good practice. I prefer Andrés solution, which is elegant, but this can also be a good exam question to test various Taylor expansion skills by insisting that no short cuts be used.
Applying basic Taylor expansion
Consider $f(x) = \sin^2 x$:
\begin{align*}
 f'(x) &= 2\sin x\cos x = \sin2x\\
 f''(x) &= 2\cos^2 x - 2\sin^2 x  = 2\cos2x\\
 f'''(x) &= -4\cos x\sin x - 4\sin x\cos x = -8 \sin x\cos x = -4\sin2x\\
 f^{iv}(x) &= -8\cos^2x + 8\sin^2x = -8\cos2x\\
\cdots &= \cdots
\end{align*}
There's a definite pattern emerging here, but in fact after $f'(x) = \sin2x$, the long way isn't really necessary. A shortened version is:
\begin{align*}
f(x) &= \sin^2x \\
f'(x) &= 2\sin x\cos x \\
f'(x) &= \sin2x \\
f''(x) &= 2\cos2x \\
f'''(x) &= -4\sin2x \\
f^{iv}(x) &= -8\cos2x \\
\cdots &= \cdots
\end{align*}
Now expand around $0$ for the M$^c$Laurin Series
\begin{align*}
f(0) &= 0 \\
f'(0) &= 0 \\
f''(0) &= 2 \\
f'''(0) &= 0 \\
f^{iv}(0) &= -8\\
f^{v}(0) &= 0 \\
f^{vi}(0) &= 32 \\
\cdots &= \cdots
\end{align*}
$$\sin^2 x = 2x^2/2! - 8x^4/4! + 32x^6/6! - \cdots = \sum_{n=1}^\infty \frac {(-1)^{n+1}2^{2n-1}x^{2n}}{(2n)!}$$
