Show the energy of a function $f(x) = a_k\cos(kx)+b_k\sin(kx)$ is $a_k^2 + b_k^2$

Question

Recently I am introduced to Fourier series and the energy theorem. By the energy thm, $E = \frac{1}{\pi} \int_{-\pi}^{\pi} (f(x))^2dx$.
I tried to do my simplification with trig but did not seem to get the right result, are there any other way to approach this problem?
My simplification is attached as an image below:

Answer 1

Just observe that $$ \frac{1}{2\pi}\int_{-\pi}^\pi dx\cos^2(kx)=\frac{1}{2\pi}\int_{-\pi}^\pi dx\sin^2(kx)=1 $$ and $$ \int_{-\pi}^\pi dx\sin(2kx)=0 $$ and you are done.