Frequency of a trigonometric function - Where is my mistake? I need to find the frequency of the following trigonometric function.$$y=\sin^4(x)+\cos^4(x)$$
The "answers" section says the answer is: $$F_y=\frac{\pi}{2}$$
This is what i did:
Finding $\sin(x)^4$ frequency (I'll call it F1):
$$\cos(2x)=1-\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$\sin^4(x)=\frac{\cos^2(2x)-2\cos(2x)+1}{4}=\frac{cos^2(2x)+4\sin^2(x)-1}{4}$$
Finding $\cos(2x)^2$ frequency:
$$\cos(4x)=2\cos^2(2x)-1$$
$$\cos^2(2x)=\frac{\cos(4x)+1}{2}$$
$$f_1=\frac{2\pi}{4}=\frac{\pi}{2}$$
Finding $\sin(x)^2$ frequency:
$$\cos(2x)=1-2\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$f_2=\frac{2\pi}{2}=\pi$$
$$F_1: \frac{f_1}{f_2}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_1=\frac{\pi}{2}\times2=\pi$$
Finding $cos(x)^4$ frequency (I'll call it F2):
$$\cos(2x)=2\cos^2(x)-1$$
$$\cos^2(x)=\frac{\cos(2x)+1}{2}$$
$$\cos^4(x)=\frac{\cos^2(2x)+2\cos(2x)+1}{4}$$
Finding $\cos(2x)$ frequency (we already have $\cos(2x)^2$ frequency - f1):
$$f_3=\frac{2\pi}{2}=\pi$$
$$F_2: \frac{f_1}{f_3}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_2=\frac{\pi}{2}\times2=\pi$$
Finding $y$'s frequency:
$$F_y: \frac{F_1}{F_2}=\frac{\pi}{\pi}=\frac{1}{1}$$
$$F_y=\pi\times1=\pi$$

You might consider using Euler's formula, which can be used to obtain
$$ \cos(x) = \frac{e^{ix} + e^{-ix}}{2} $$ and
$$ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}. $$
Put $z = e^{ix}$, so that
$$\cos(x) =\frac{z+1/z}{2}$$
and
$$\sin(x) = \frac{z-1/z}{2i}$$
This gives
$$\cos(x)^4+\sin(x)^4 = 
\left(\frac{z+1/z}{2}\right)^4 + \left(\frac{z-1/z}{2i}\right)^4 =
\frac{z^4}{8} + 3/4 + \frac{1/z^4}{8}$$
which is
$$\frac{3}{4} + \frac{1}{4} \cos(4x).$$
So your frequency is $$ \frac{2\pi}{4} = \frac{\pi}{2}.$$
