How do you prove cos 3x = 4 cos ^3 x - 3 cos x ?

Answer 1

cos (2x + x)
cos 2x cos x - sin 2x sin x
cos 2x cos x - 2 sin ² x cos x
cos 2x cos x - 2 cos x (1 - cos ² x)
(2 cos ² x - 1) (cos x) + 2(cos ² x - 1))(cos x)
2 cos ³ x - cos x + 2 cos ³ x - 2 cos x
4 cos ³ x - 3 cos x

Answer 2

Cos 3x Formula

Answer 3

Here is how I prove it.

(cos x + i sin x)^3 = cos (3x) + i sin(3x) by deMoivre's Theorem.

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

equating the real components of the equation (the ones without the i's in them) gives:
cos^3 x - 3 cos x sin^2 x = cos (3x)
cos^3 x - 3 cos x (1-cos^2 x) = cos(3x)
cos^3 x - 3 cos x + 3 cos^3 x = cos(3x)
cos(3x) = 4 cos^3 x - 3 cos x.

Or, you could do
cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x)
since cos(2x) = 2cos^2 x - 1 and sin(2x) = 2 sin x cos x:

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

Answer 4

i dont prove i aporve