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