Fourier Series Trig Functions

Question

I need assistance finding the fourier series for the following function:

$$ f(x)=3\cos^2(5x) $$

I know that

$$ a_0={1\over 2\pi}\int_{-\pi}^\pi 3\cos^2(5x)\,dx={3 \over 2} $$

and

$$ b_n={1\over \pi}\int_{-\pi}^\pi 3\cos^2(5x)\sin(nx)\,dx=0 $$

I'm having difficulty calculating $a_n$ because I keep getting zero.

$$ a_n={1\over \pi}\int_{-\pi}^\pi 3\cos^2(5x)\cos(nx)\,dx=? $$

Thanks for your help.

Answer 1

Recalling the identity

$$ \cos(2t) = 2\cos(t)^2-1 $$

which gives us the desired Fourier series

$$ 3\cos(5x)^2= \frac{3}{2}( 1+\cos(10 x) ). $$

Added: to get $a_n$ note that the function is even so you can do the following

$$ a_n = \frac{2}{\pi} \int_{0}^{\pi} 3\cos(5x)^2 \cos(nx) dx = \frac{6}{\pi}\,{\frac {\sin \left( \pi \,n \right) \left( {n}^{2}-50 \right) }{n \left( {n}^{2}-100 \right) }} . $$

Answer 2

Hint: $\cos^2(2x) = \frac{1+\cos(4x)}{2}$, and you don't need to do any integration.

Answer 3

$$u=5x\implies \frac{du}5=dx\implies$$

$$\int\limits_{-\pi}^\pi\cos^25x\,dx=\frac15\int\limits_{-5\pi}^{5\pi}\cos^2u\,du=\frac15\left.\frac{u+\cos u\sin u}2\right|_{-5\pi}^{5\pi}=\pi\neq 0$$