Question

Show that sin 3x = 3 sin x - 4sin^3 x ?

Answer 1

`sin (A+-B) = sin A cos B +- cos A sin B`
`sin 2A = 2sin A cos A`
`cos 2A = cos^2 A - sin^2 A`
`sin^2 A + cos^2 A = 1`

`sin 3x = sin x cos 2x + cos x sin 2x = sin x(cos^2 x - sin^2 x) + 2sin x cos^2 x = 3 sin x cos^2 x - sin^3 x = 3 sin x (1-sin^2 x) - sin^3 x = 3 sin x - 4 sin^3 x`

Answer 2

Alternatively, use Euler's Formula, expand `cos(3x)+i sin(3x)` and equate the imaginary parts.

Explanation:

If you have encountered Euler's Formula then it works out like this:

Euler's Formula gives us: `cos x + i sin x = e^(ix)`

So:

`cos(3x)+i sin(3x) = e^(i3x) = (e^(ix))^3 = (cos(x)+i sin(x))^3`

`=cos^3(x)+3cos^2(x) i sin(x) + 3 cos(x) i^2 sin^2(x) + i^3 sin^3(x)`

`=(cos^3(x)-3 cos(x)sin^2(x)) + i(3 cos^2(x) sin(x)-sin^3(x))`

Equating the imaginary parts of both ends, we get:

`sin(3x) = 3 cos^2(x)sin(x)-sin^3(x)`

`= sin(x)(3 cos^2(x)-sin^2(x))`

`=sin(x)(3(1 - sin^2(x))-sin^2(x))` `color(green)([[ "using: " sin^2(x)+cos^2(x) = 1 ]])`

`=sin(x)(3-4sin^2(x))`

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

Equating the real parts of both ends, we also get:

`cos(3x) = cos^3(x)-3 cos(x)sin^2(x)`

`=cos(x)(cos^2(x)-3sin^2(x))`

`=cos(x)(cos^2(x)-3(1-cos^2(x)))`

`=cos(x)(4 cos^2(x)-3)`

`=4cos^3(x)-3cos(x)`