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pre-calculus

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Rewrite the expression so that it involves the sum or difference of only constants and sines and cosine to the 1st power
(sinx)^4

  • pre-calculus - ,

    sin^4
    (1-cos^2)^2
    (1-cos)^2(1+cos)^2
    (1-cos)(1-cos)(1+cos)(1+cos)

    sin^2(x) = (1-cos(2x))/2
    so,
    sin^4(x) = (1 - cos(2x))^2/4
    = (1 - 2cos(2x) + cos^2(2x))/4
    = (1 - 2cos(2x) + (1 - cos(4x))/2)/4
    = (1 - 2cos(2x) + 1/2 - cos(4x)/2)/4
    = (2 - 4cos(2x) + 1 - cos(4x))/8
    = (3 - 4cos(2x) - cos(4x))/8

    don't know which way you wanted to go there.

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