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Trig please help!

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sin^6 x + cos^6 x=1 - (3/4)sin^2 2x

work on one side only!

  • Trig please help! - ,

    LS looks like the sum of cubes
    sin^6 x + cos^6 x
    = (sin^2x)^3 + (cos^2x)^3
    = (sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + sin^4x)
    = (1)(sin^4x - (sin^2x)(cos^2x) + sin^4x)

    Now let's do some "aside"
    (sin^2x + cos^2)^2 would be
    sin^4x + 2(sin^2x)(cos^2x) + cos^4x

    we 'almost' have that above, differing only by the coefficient of the middle term. We can fix that by saying
    (sin^4x - (sin^2x)(cos^2x) + sin^4x)
    = sin^4x + 2(sin^2x)(cos^2x) + cos^4x - 4(sin^2x)(cos^2x)
    = (sin^2x + cos^2x) - 3(sin^2x)(cos^2x)
    = 1 - 3(sin^2x)(cos^2x)
    almost there!
    recall sin 2A = 2(sinA)(cosA)
    so 3(sin^2x)(cos^2x)
    = 3(sinxcosx)^2
    = 3((1/2)sin 2x)^2
    = (3/4)sin^2 2x

    so
    1 - 3(sin^2x)(cos^2x)
    = 1 - (3/4)sin^2 2x
    = RS !!!!!!

    Q.E.D.

  • Trig please help! - ,

    (sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + sin^4x) I DON'T AGREE WITH THIS

    SHOULDNT IT BE
    (sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + COS^4 X ??

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