how can I find the period of $f=\sin^4(x)$? How can I find the period of $f=\sin^4(x)$?
$\sin^4(x)=\frac{3-4\cos2x+\cos4x}{8}$. I didn't manage to reduce $\cos 4x$ to $\cos 2x$.
 A: If $T$ is a period of $\sin^4{x},$ then $\forall x \in \mathbb{R}$
$$
\sin^4{(x+T)}-\sin^4{x}=0 ,\\
(\sin^2{(x+T)}-\sin^2{x})(\sin^2{(x+T)}+\sin^2{x})=0,\\
\dfrac{1-\cos(2(x+T))}{2}-\dfrac{1-\cos(2x)}{2}=0, \\
\cos(2x)-\cos{(2(x+T))}=0,\\
 -2\sin( 2x+T)\sin(-T)=0.  \\
$$
The last equation implies that $T=\pi.$
A: Note that the equation $$\sin^4 (x+y)=\sin^4 x$$
Implies $$\sin^4 (x+y)-\sin^4 x = \left(\sin (x+y)-\sin x\right)\left(\sin (x+y)+\sin x\right)\left(\sin^2 (x+y)+\sin^2 x\right)=0$$
The final factor is nonnegative, and can only be zero when $\sin x=0$ so contributes no value of $y$ which works for all $x$.
So we have $$\sin (x+y)=\sin x\cos y+\cos x \sin y=\pm \sin x$$ And this has to be true for all $x$ and constant $y$. Set $x=0$ to obtain $\sin y=0$ and $y=n\pi$. So the period is $\pi$ (after checking that these values of $y$ work).
A: $\dfrac{3}{8}$ is constant, $-\dfrac{1}{2}\cos 2x$ has period $\dfrac{2\pi}{2} = \pi$, and $\dfrac{1}{8}\cos 4x$ has period $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$. 
This should be enough to tell you that $\sin^4 x = \dfrac{3}{8} -\dfrac{1}{2}\cos 2x + \dfrac{1}{8}\cos 4x$ has period $\pi$. 
A: Note that 
$$\sin^4(x+\pi)=(-\sin(x))^4=\sin^4(x).$$
