Proving that the period of $\cos^2x$ is half the period of $\cos x$

Question

For example, I know that the period of $\cos^2x$ is simply $$\frac{\text{period of }\cos x}{2}$$ That is: $\pi$. Could you provide me with both a visual intuitive explanation and an algebraic proof?

Answer 1

For all $x$ we have $\cos^2 (x+\pi)=$ $(\cos (x+\pi))^2=(-\cos x)^2=(\cos x)^2=$ $=\cos^2 x.$

If $k/\pi\not \in \Bbb Z$ then let $k=k'\pi +k''$ with $k'\in \Bbb Z$ and $0<k''<\pi.$ Then $\cos^2(0+k)=\cos^2k=\cos^2 k''\ne 1=\cos^2 0,$ so $k$ is not a period of $\cos^2 x.$

Answer 2

Note that:

$$f(x)=\cos^2x=\frac 12(1+\cos2x)$$

Period of $\cos x$ is $2\pi$ so $\cos2x$ has period $\pi$ and that is also the period of $f(x)$.

Answer 3

We want period of $cos^2 (x)$ we know that $\cos^2(X)=\frac12(1+\cos2x)$ now consider only $\cos2x$ as the remaining are constants then it is in the form $\cos(ax+b)$ that is $\cos(2x+0)$ we know that period of $\cos x=\frac{2\pi}{|a|}$ now substitute $2$ in place of $a$ that is $\frac{2\pi}2=\pi$.
Then the period of $\cos^2(x)=\pi$