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Math Help - Prove : cosx,cos2x,cos4x,cos6x

  1. #1
    Super Member dhiab's Avatar
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    Prove : cosx,cos2x,cos4x,cos6x

    Prove :
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello dhiab
    Quote Originally Posted by dhiab View Post
    Prove :
    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
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  3. #3
    MHF Contributor red_dog's Avatar
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    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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