Question

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

Answer 1

`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`

Answer 2

`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`