Question

How do you express `(sin 4 theta)/(sin theta)` in terms of `sin theta` and `cos theta` ?

Answer 1

`(sin 4 theta) / (sin theta) = 4 cos^3 theta - 4 cos theta sin^2 theta = 4 cos theta(cos^2 theta - sin^2 theta)`

Explanation:

By de Moivre's formula we have:

`cos n theta + i sin n theta = (cos theta + i sin theta)^n`

So:

`cos 4 theta + i sin 4 theta`

`= (cos theta + i sin theta)^4`

`= ((4),(0)) cos^4 theta + ((4),(1)) i cos^3 theta sin theta + ((4),(2)) i^2 cos^2 theta sin^2 theta + ((4),(3)) i^3 cos theta sin^3 theta + ((4),(4)) i^4 sin^4 theta`

`= cos^4 theta + 4 i cos^3 theta sin theta -6 cos^2 theta sin^2 theta -4i cos theta sin^3 theta + sin^4 theta`

`= (cos^4 theta -6 cos^2 theta sin^2 theta + sin^4 theta) + i(4 cos^3 theta sin theta -4 cos theta sin^3 theta)`

Equating imaginary parts, we find:

`sin 4 theta = 4 cos^3 theta sin theta - 4 cos theta sin^3 theta`

So:

`(sin 4 theta) / (sin theta) = 4 cos^3 theta - 4 cos theta sin^2 theta`

`color(white)((sin 4 theta) / (sin theta)) = 4 cos theta(cos^2 theta - sin^2 theta)`