Determining the amplitude of $x(t) = 3\cos^2(\omega t - \frac{\pi}{3})$ at $2\omega$ with fourier series

Question

The given function is $x(t) = 3\cos^2(\omega t - \frac{\pi}{3})$. I have to determine the amplitude of the component with frequency $2\omega$ in the fourier series of the function. I can only do it with calculating the fourier series of $x(t)$ with the integral formulas. The function is neither even nor odd, so I have to calculate both of the integrals. This takes a lot of time, and I'm wondering, if there is a faster way to calculate this amplitude.

Answer 1

More generally, any sum of powers of $\sin (\omega t-\phi)$ and $\cos (\omega t-\phi)$ can be rewritten as a sum of sine and cosine waves using trigonometric identities. In this particular case, the identity (given by Fantini) is square-of-cosine: $$3\cos^2 (\omega t-\pi/3) = \frac32 + \frac32 \cos (2\omega t - 2\pi/3)$$ This already gives the amplitude. But if you wanted the Fourier series written in the usual way, you can do that too with a different trig identity (cosine of difference): $$3\cos^2 (\omega t-\pi/3) = \frac32 + \frac32 (\cos (2\omega t) \cos( 2\pi/3)+\sin (2\omega t) \sin( 2\pi/3)) $$