trig

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sin^4 x = (3-4cos 2x + cos4x)/8

prove on one side please!

  • trig -

    just simplify both sides! easy as that!!!

  • trig -

    how?

  • trig -

    RS
    = (3 - 4cos(2x) + cos^2(2x) - sin^2(2x))/8
    = (3 - 4cos(2x) + cos^2(2x) - (1 - cos^2(2x)))/8
    = (2cos^2(2x) - 4cos(2x) + 2)/8
    = (2(cos^2(2x) - 2cos(2x) + 1))/8
    = (2(cos(2x) - 1)^2)/8
    = (2(1 - 2sin^2(x) - 1)^2)/8
    = (2(-2sin^2(x))^2)/8
    = (8sin^4(x))/8
    = sin^4(x)
    = LS

    hope I made no typos this time

  • trig -

    1-4cos2x+cos4x/8=sin4x

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