# Prove : cosx,cos2x,cos4x,cos6x

• August 5th 2009, 07:29 AM
dhiab
Prove : cosx,cos2x,cos4x,cos6x
• August 5th 2009, 07:58 AM
Hello dhiab
Quote:

Originally Posted by dhiab

Begin with $\cos6x + 6\cos4x+15\cos2x$, and use the formulae for the cosines of multiple angles in terms of powers of $\cos x$ (for instance, here). This expression immediately reduces to $32\cos^6x - 10$ and the result then follows.

• August 5th 2009, 09:30 AM
red_dog
Another approach:

$\cos^6x=(\cos^2x)^3=\left(\frac{1+\cos 2x}{2}\right)^3=$

$=\frac{1}{8}+\frac{3}{8}\cos 2x+\frac{3}{8}\cos^22x+\frac{1}{8}\cos^32x=$

$=\frac{1}{8}+\frac{3}{8}\cos 2x+\frac{3}{8}\cdot\frac{1+\cos 4x}{2}+\frac{1}{8}\cdot\frac{1+\cos 4x}{2}\cdot\cos 2x=$

$=\frac{1}{8}+\frac{3}{8}\cos 2x+\frac{3}{16}+\frac{3}{16}\cos 4x+\frac{1}{16}\cos 2x+\frac{1}{16}\cos 4x\cos 2x=$

$=\frac{5}{16}+\frac{7}{16}\cos 2x+\frac{3}{16}\cos 4x+\frac{1}{32}\cos 6x+\frac{1}{32}\cos 2x=$

$=\frac{5}{16}+\frac{5}{32}\cos 2x+\frac{3}{16}\cos 4x+\frac{1}{32}\cos 6x$