Question

What is `4sin^5theta` in terms of non-exponential trigonometric functions?

Answer 1

Use this identity ...

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

Explanation:

`4sin^5x=4xx(sin^2x)xx(sin^2x)xxsinx`

`=4xx(1-cos2x)/2 xx (1-cos2x)/2 xx sinx`

`=(1-cos2x)(1-cos2x)(sinx)`

Hope that helped