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) .

1 Answer
Jun 2, 2018

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.