Question

Question #9c7b3

Answer 1

`sin^3x=(3sinx-sin(3x))/4`

Explanation:

If you know the de Moivre's identity it is quite direct.

`e^(i x) = cosx+isinx`

so

`e^(3ix)= (cosx + isinx)^3=cos(3x)+isin(3x)`

or

`cos^3x+3icos^2xsinx-3cosxsin^2x-isin^3x=cos(3x)+isin(3x)`

grouping the real and the imaginary parts,

`cos^3x-3cosxsin^2x=cos(3x)` and
`3cos^2xsinx-sin^3x=sin(3x)`

now, using the identity

`cos^2x+sin^2x=1` we will get at

`sin(3x)=3sinx(1-sin^2x)-sin^3x` or

`sin^3x=(3sinx-sin(3x))/4`

Answer 2

Expand `sin^3 x`

Explanation:

There is another simpler method.
Use the trig identity:
`sin 3x = 3sin x - 4sin^3 x`
By transposing terms, we get:
`4sin^3 x = 3sin x - sin 3x`
`sin^3 x = (3sin x - sin 3x)/4`