# How to reperesent $\sin^{4}(x)$ byFourier series? [closed]

how to represent $\sin^{4}(x)$ by Fourier series? Obviously,$\sin^{4}(x)$ is an even function, so $b_n=0$. How can i get $a_n$ ?

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## closed as too localized by Andres Caicedo, Raskolnikov, Amzoti, 23rd, Lord_FarinApr 26 '13 at 7:04

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You can see a solution here math.stackexchange.com/questions/373154/… –  Amzoti Apr 26 '13 at 5:12

$$\sin^4{x}=\left(\dfrac{1-\cos{2x}}{2}\right)^2=\dfrac{1}{4}\left(1-2\cos{2x}+\cos^2{2x} \right)=\\ =\dfrac{1}{4}\left(1-2\cos{2x}+\dfrac{1+\cos{4x}}{2} \right)=\dfrac{3}{8}-\dfrac{1}{2}\cos{2x}+\dfrac{1}{8}\cos{4x}$$

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what i want is not this way rather than the way by fourier fomula--the way of integrating. –  park ning Apr 26 '13 at 5:17
It is an identity, so you can simply integrate the RHS multiplied by $\cos{kx}$ . –  M. Strochyk Apr 26 '13 at 5:21
:How to compute the integral a_n=\int_{-pi}^{pi}sin^4x cosnx dx? –  park ning Apr 26 '13 at 5:41
$a_k=\dfrac{1}{\pi}\int\limits_{-\pi}^{\pi}\sin^4{x}\, \cos{kx}\, dx=\dfrac{1}{\pi}\int\limits_{-\pi}^{\pi}{\left(\dfrac{3}{8}-\dfrac{1}{2}\cos{2x‌​} + \dfrac{1}{8}\cos{4x}\right)\, \cos{kx}\, dx}$ –  M. Strochyk Apr 26 '13 at 6:05

$$\sin^4 x=\frac 38 -\frac 12 \cos 2x+\frac 18\cos 4x$$

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i want to kown the process. –  park ning Apr 26 '13 at 5:12
You can use De Moivre's formula. –  Raskolnikov Apr 26 '13 at 5:13
This is a trig formula. You can derive it in many different ways. The easiest on is de Moivre's formula –  Artem Apr 26 '13 at 5:14

Without relying on the trigonometric formulas in full extent to simplify $\sin^4$, you *could try partial integration: \begin{align}\int \underbrace{\sin^4x}_u\;\underbrace{\cos nx}_{v'}\;\mathrm dx &= \sin^4x\cdot\frac1n\sin nx-\frac1n\int4\sin^3x\cos x\sin nx\,\mathrm dx\end{align}. Of course now you should at least make use of $2\sin x\cos x=\sin 2x$ and repeat until something nice enough remains. In the end, however, you will always need the formulas to convert products into sums of trig functions, so there's no argument at all against doing so right in the beginning (and getting the Fourier expansion for free, like M. Strochyk does in his solution)

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