How do you express #sin 3theta# in terms of trigonometric functions of #theta# ?
2 Answers
Explanation:
.
We have double-angle identities:
Explanation:
We can use de Moivre's formula:
#(cos theta + i sin theta)^n = cos n theta + i sin n theta#
as follows:
#cos 3 theta + i sin 3 theta#
#= (cos theta + i sin theta)^3#
#= cos^3 theta + 3i cos^2 theta sin theta - 3 cos theta sin^2 theta - i sin^3 theta#
#= (cos^3 theta - 3 cos theta sin^2 theta) + i (3 cos^2 theta sin theta - sin^3 theta)#
Then 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) :}#
We can also use the trigonometric version of Pythagoras' theorem:
#cos^2 theta + sin^2 theta = 1#
to refactor these formulae as follows:
#cos 3 theta = cos^3 theta - 3 cos theta sin^2 theta#
#color(white)(cos 3 theta) = cos^3 theta - 3 cos theta (1-cos^2 theta)#
#color(white)(cos 3 theta) = 4cos^3 theta - 3 cos theta#
#sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta#
#color(white)(sin 3 theta) = 3 (1-sin^2 theta) sin theta - sin^3 theta#
#color(white)(sin 3 theta) = 3 sin theta - 4 sin^3 theta#