How do you express (sin 4 theta)/(sin theta)sin4θsinθ in terms of sin thetasinθ and cos thetacosθ ?

1 Answer
Mar 11, 2017

(sin 4 theta) / (sin theta) = 4 cos^3 theta - 4 cos theta sin^2 theta = 4 cos theta(cos^2 theta - sin^2 theta)sin4θsinθ=4cos3θ4cosθsin2θ=4cosθ(cos2θsin2θ)

Explanation:

By de Moivre's formula we have:

cos n theta + i sin n theta = (cos theta + i sin theta)^ncosnθ+isinnθ=(cosθ+isinθ)n

So:

cos 4 theta + i sin 4 thetacos4θ+isin4θ

= (cos theta + i sin theta)^4=(cosθ+isinθ)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)