Question

What is the period of `sin^2(x)` ?

Answer 1

Period `=( 2 pi )/ b = (2 pi )/ 2 = pi`

Explanation:

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

Equation is in the form
`a cos (bx + c) + d` where

Amplitude a = -1/2,

Period = `(2pi )/ b =( 2pi)/2 = pi`

Hence the answer.

Answer 2

See explanation...

Explanation:

Given that for any `x`:

`sin(x+pi) = - sin(x)`

we have:

`sin^2(x+pi) = (-sin(x))^2 = sin^2(x)`

So `sin^2(x)` has period `pi`