Question

Question #6d8e6

Answer 1

We will make repeated use of the power reduction formulas:

• `sin^2x = (1-cos2x)/2`
• `cos^2x = (1+cos2x)/2`

Proceeding,

`sin^2thetacos^2theta = (1 - cos2theta)/2*(1+cos2theta)/2`

`= 1/4(1-cos^2 2theta)`

`= 1/4(1 -(1+cos(2(2theta))/2))`

`= 1/4(1-(1+cos4theta)/2)`

`= 1/4((1-cos4theta)/2)`

`= 1/8(1-cos4theta)`