trig

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Prove that
cos4x=8cos^4x-8cos^2x+1

so I'm guessing you begin with the left side cos4x=cos2(2x) then im kinda lost

  • trig -

    You began the problem correctly. Continue to use the cos "double angle" formula, applying it twice.
    cos4x = cos[2(2x)] = 2 cos^2(2x) - 1
    = 2(2cos^2 x - 1)^2 -1
    = 2*[4 cos^4 x -4cos^2 x +1) -1
    = 8cos^2 x -8cos^2 x +1

  • trig -

    4cos^{2}2x-1=cos4x

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