Question

Question #1b893

Answer 1

see below

Explanation:

Use the following formulas:

1. `cos(A+B)=cosAcosB-sinAsinB`
2. `cos 2A=2cos^2 A -1`
3. `sin 2A=2sinAcosA`
4. `sin^2A+cos^2A=1`

Left Hand Side:

`cos 3 theta = cos (2theta+theta)`

`=cos2theta cos theta - sin 2 theta sin theta`

`=(2cos^2theta-1)cos theta-2sin theta cos theta*sin theta`

`=2cos^3 theta-cos theta-2cos theta sin^2 theta`

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

`=2cos^3 theta-cos theta-2cos theta+2cos^3 theta`

`=4cos^3 theta-3 cos theta`

`:.=` Right Hand Side

Answer 2

See below.

Explanation:

De Moivre's Formula:

`(cos x +i sin x )^n =\cos nx +i\sin nx`

So:

`(cos x +i sin x )^3 =\cos 3x +i\sin 3x qquad square`

But it is also true from a Binomial Expansion that:

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

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

Collecting real and imaginary terms:

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

Equating real terms in `triangle` and `square`:

`implies cos 3x = cos^3 x - 3 cos x sin^2 x`

Using the Pythagorean Identity:

`implies cos 3x = cos^3 x - 3 cos x color(red)((1 - cos^2 x))`

`implies cos 3x = cos^3 x - 3 cos x + 3cos^3 x`

`implies cos 3x = 4 cos^3 x - 3 cos x`

Ta da!!

Of course, replace `x` with `theta`