Prove : cosx,cos2x,cos4x,cos6x

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August 5th 2009, 07:29 AM
dhiab

Prove : cosx,cos2x,cos4x,cos6x

Prove :
http://www.mathramz.com/xyz/latexren...9a39aa1ac7.png
August 5th 2009, 07:58 AM
Grandad

Hello dhiab

Quote:

Originally Posted by dhiab

Prove :
http://www.mathramz.com/xyz/latexren...9a39aa1ac7.png

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.

Grandad
August 5th 2009, 09:30 AM
red_dog

Another approach:

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

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

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

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

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

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

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