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:
$$F_y: \frac{F_1}{F_2}=\frac{\pi}{\pi}=\frac{1}{1}$$
$$F_y=\pi\times1=\pi$$
 A: You have proved that $\pi$ is a period, but you have not shown that it is the smallest period. 
I would tackle the problem in more or less the same way that you did, using double angle identities, but the algebra can be simplified. Note that
$$1=(\cos^2 x+\sin^2 x)^2=\cos^4 x+\sin^4 x +2\cos^2 x\sin^2 x.$$
It follows that our function is equal to
$$1-2 \cos^2 x\sin^2 x.$$
The only interesting part is $2\cos^2 x\sin^2 x$, which is $\frac{1}{2}\sin^2 2x$.
It is clear that this has period $\dfrac{\pi}{2}$. 
If you wish, you can use the trigonometric identity $\cos 2u=1-2\sin^2 u$ to express our function in terms of $\cos 4x$. We get $\frac{1}{2}\sin^2 2x=\frac{1}{4}(1-\cos 4x)$, and therefore
$$\cos^4 x+\sin^4 x=\frac{3}{4}+\frac{1}{4}\cos 4x.$$
A: $$\sin^4\left(x+\frac{\pi}{2}\right)=\left(\sin x\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\cos x\right)^4=\cos^4x$$
$$\cos^4\left(x+\frac{\pi}{2}\right)=\left(\cos x\cos\left(\frac{\pi}{2}\right)-\sin x\sin\left(\frac{\pi}{2}\right)\right)^4=(-\sin x)^4=\sin^4 x$$
Thus, the period of $\,\sin^4 x+\cos^4x\,$ indeed is not more than $\,\pi/2\,$
A: I won't pretend that I would have thought of this without seeing the answer but we can look at the problem geometrically. This is more of a nice illustration than a genuine 'proof by pictures'.
We know firstly that the curve $x^4+y^4=a$ is symmetric about the $x$- and $y$-axes:

Suppose that $f(\theta)=\sin^4\theta+\cos^4\theta=a$ for some $a$ in the range of $f$. Now we know that $\sin^2\theta+\cos^2\theta$ is equal to one. Hence we can place the point $(\cos\theta,\sin\theta)$ on the intersection of the unit circle and the curve $x^4+y^4=a$ as shown:

Now as the curve has $\pi/2$-rotation symmetry, the point $(\cos(\theta+\pi/2),\sin(\theta+\pi/2))$ is also on the curve $x^4+y^4=a$ and we are done.
A: 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 $$ \frac{2\pi}{4} = \frac{\pi}{2}.$$
