How do you express (sin 4 theta)/(sin theta)sin4θsinθ in terms of sin thetasinθ and cos thetacosθ ?
1 Answer
(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=4cos3θ−4cosθsin2θ=4cosθ(cos2θ−sin2θ)
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)