# 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

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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*}

$$\sin^2 x = 2x^2/2! - 8x^4/4! + 32x^6/6! - \cdots = \sum_{n=1}^\infty \frac {(-1)^{n+1}2^{2n-1}x^{2n}}{(2n)!}$$

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Thanks! So is using the Maclaurin series considered as using the Taylor series expansion? –  Polly Sep 27 '13 at 20:32
@Polly The M$^c$Laurin series is what you get when you expand a Taylor series around zero. In general Taylor series you have $(x-a)^n$ in your terms, in M$^c$Laurin series you have $x^n$. Some people think it's not worth the effort having the M$^c$Laurin term for a special case. I'm undecided at this point. Maths conventions can be tricky. For example, Lie groups are named after the mathematician of the same name but not everybody capitalises the L. Some people believe very strongly that you should. Similarly with Abelian, named after Abel. –  Geoff Pointer Sep 27 '13 at 23:11

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 $\cos z$.

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