mean value of cosine with another trigonometric function inside it

Question

I have to evaluate to mean value of the following expression: $$\cos(4\pi \cdot 50t+8\pi \sin(5\cdot 10^3t))$$ At first I wanted to say the mean value of a cosine function along a period equals to zero, but the sine function inside my cosine function confused me, any ideas of how do I solve it?

Answer 1

Let $f(x)=\cos(4\pi \cdot 50t+8\pi \sin(5\cdot 10^3t))$.

Note that $8\pi \sin(5\cdot 10^3t)$ has a period of $\dfrac{\pi}{2500}$ but $\cos(4\pi \cdot 50t)$ has a period of $\dfrac{1}{100}$.

The period ratio is irrational, so the average value are not necessary zero.

Your conjecture seems to fail when I tried to do numerical integration over different intervals:

\begin{array}{|c|c|c|} \hline a & b & \displaystyle \int_{a}^{b} f(x) \, dx \\ \hline -0.020 & -0.015 & +0.000,017,867 \\ -0.015 & -0.010 & -0.000,012,196 \\ -0.010 & -0.005 & +0.000,006,081 \\ -0.005 & -0.000 & -0.000,003,895 \\ +0.000 & +0.005 & -0.000,003,895 \\ +0.005 & +0.015 & +0.000,006,081 \\ +0.010 & +0.015 & -0.000,012,196 \\ +0.015 & +0.020 & +0.000,017,867 \\ \hline \end{array}