Period of $\sin^2 x +\cos^4 x$? How do I approach this problem? I tried reducing it to $\sin ax + \cos bx$ form but can't. I converted it to 1+ $\sin^2 x \cos^2 x $, but I don't know if I can get the answer from it. Would someone please help me with this? The answer is $\pi$.
Since$$\sin\left(x+\frac\pi2\right)=\cos(x)\text{ and }\cos\left(x+\frac\pi2\right)=-\sin(x),$$it is clear that $\sin^2+\cos^4$ is periodic with period $\frac\pi2$. But is $\frac\pi2$ the smallest period? Yes, because\begin{align}(\sin^2+\cos^4)'(x)&amp;=2\sin(x)\cos(x)-4\cos^3(x)\sin(x)\\&amp;=2\sin(x)\cos(x)\times\bigl(1-2\cos^2(x)\bigr)\\&amp;=2\sin(x)\cos(x)\times\bigl(\sin^2(x)-\cos^2(x)\bigr)\\&amp;=-\sin(2x)\cos(2x)\\&amp;=-\frac{\sin(4x)}2\end{align}and so obviously $\frac\pi2$ is the smallest period of $(\sin^2+\cos^4)'$.
Use Euler's formula :
$$cos^4(x)=\displaystyle\left(\frac{e^{ix}+e^{-ix}}2\right)^4$$
And you can develop using Newton's binomial formula