How do you express #sin 3theta# in terms of trigonometric functions of #theta# ?

2 Answers
Jan 4, 2018

#sin3x=3sinxcos^2x-sin^3x#

Explanation:

.

#sin3x=sin(2x+x)=sin2xcosx+cos2xsinx#

We have double-angle identities:

#sin2x=2sinxcosx#

#cos2x=cos^2x-sin^2x#

#sin3x=2sinxcosxcosx+(cos^2x-sin^2x)sinx#

#sin3x=2sinxcos^2x+sinxcos^2x-sin^3x#

#sin3x=3sinxcos^2x-sin^3x#

Jan 4, 2018

#sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta = 3 sin theta - 4 sin^3 theta#

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#