Why is the average of $\sin^2 wt = 1/2$ and $\cos^2 wt = 1/2$

Question

Why is the average of both $\sin^2 = \frac{1}{2}$ and $\cos^2 = \frac{1}{2}$

I was revising Simple Harmonic motion notes and in the average of Kinetic energy derivation

$$KE = \frac12 k A^2 \cos^2(\omega t)$$

And the solution is given as $\frac{1}{4} kA^2$

I haven't gone that deep in integration help

Answer 1

Set $\text{avg}(\cos^2(x))=a$ and $ \text{avg}(\sin^2(x))=b$. The two functions are horisontal translations of one another, so the averages are the same. $a=b$

It is known that $\forall x:\sin^2(x)+\cos^2(x)=1$, so $\text{avg}\big(\sin^2(x)\big)+\text{avg}\big(\cos^2(x)\big)=1$, which means that $a+b=1$ so $2a=1$ so $a=\frac12$

$$\text{avg}(\cos^2(x))=\text{avg}(\sin^2(x))=\frac12$$

Q.E.D