How do the double and treble angle formulas for #sin# and #cos# relate to powers of #cos theta + i sin theta# ?

1 Answer
Aug 17, 2015

#(cos theta + i sin theta)^2 = (cos^2 theta - sin^2 theta) + 2i cos theta sin theta#
#(cos theta + i sin theta)^3 = (cos ^3 theta - 3 cos theta sin^2 theta) + i(3 cos^2 theta sin theta - sin^3 theta)#

Explanation:

#e^(i theta) = cos theta + i sin theta#

Double angle

#cos 2 theta + i sin 2 theta = e^(2 i theta) = (e^(i theta))^2 = (cos theta + i sin theta)^2#

#= cos ^2 theta + 2i cos theta sin theta + i^2 sin^2 theta#

#= (cos^2 theta - sin^2 theta) + 2i cos theta sin theta#

Equating real and imaginary parts we find:

#cos 2 theta = cos^2 theta - sin^2 theta#

#sin 2 theta = 2 cos theta sin theta#

Treble angle

#cos 3 theta + i sin 3 theta = e^(3 i theta) = (e^(i theta))^3 = (cos theta + i sin theta)^3#

#= cos^3 theta + 3i cos ^2 theta sin theta + 3i^2 cos theta sin^2 theta + i^3 sin^3 theta#

#= (cos^3 theta - 3 cos theta sin^2 theta) + i(3 cos^2 theta sin theta - sin^3 theta)#

Equating real and imaginary parts we find:

#cos 3 theta = cos^3 theta - 3 cos theta sin^2 theta#

#sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta#

These formulae can be re-expressed using the identity #sin^2 theta + cos^2 theta = 1# as follows:

#cos 3 theta = cos^3 theta - 3 cos theta sin^2 theta#

#=cos theta(cos^2 theta - 3(1 - cos^2 theta))#

#=cos theta(4 cos^2 theta - 3)#

#sin 3 theta = 3cos^2 theta sin theta - sin^3 theta#

#=sin theta(3(1 - sin^2 theta) - sin^2 theta)#

#=sin theta(3 - 4sin^2 theta)#