Question

How would you find the period of `sin^2theta`?

I don't get how we would use `"period"/k` for a square or cubic function where `k` is the co-efficient of the argument i.e `sin(ktheta)` .

Answer 1

`sin^2 theta = 1/2(1- cos(2 theta))` so has period `pi`.

Explanation:

We know the answer is `pi.` Let's see why.

The period of `sin(x)` is `2pi` and the period of `sin (kx)` is thus `{2pi}/k.` That doesn't tell us the period of `sin^2 x` until we apply one of the double angle formulas for cosine:

`cos(2 x) = 1 - 2 sin ^2 x`

`sin^2 x= 1/2(1- cos(2 x))`

The period of `cos 2 x` is `pi` by the rule, and the addition and multiplication by the constants don't change that.

In general the odd powers of `sin x` will have a have a period of `2pi` and the even ones a period of `pi.`