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Given that sin(2θ)=2/3, the value of
sin^6θ+cos^6θ can be written as a/b with a and b as coprime positive integers. Find a+b.

  • Trigonometry -

    sin^6 θ+cos^6 θ
    = (sin^2 Ø)^3 + (cos^2 Ø)^3 ---- the sum of cubes
    = (sin^2 Ø + cos^2 Ø)( (sin^2 Ø)^2 - (sin^2 Ø)(cos^2 Ø) + (cos^2 Ø)^2 )
    = (1)( (sin^2 Ø)^2 - (sin^2 Ø)(cos^2 Ø) + (cos^2 Ø)^2 )

    now, if sin 2Ø = 1/3
    2sinØcosØ = 2/3
    sinØcosØ = 1/3
    (sin^2 Ø)^2 + (cos^2 Ø)^2
    = (sin^2 Ø + cos^2 Ø)^2 - 2(sin^2 Ø)(cos^2 Ø)
    = (1 - 2(sinØcosØ)^2)
    = 1 - 2(1/9) = 7/9

    then (1)( (sin^2 Ø)^2 - (sin^2 Ø)(cos^2 Ø) + (cos^2 Ø)^2 )
    = (1)(7/9 - 1/9)
    = 6/9
    = 2/3

    so for you a+b stuff, a=2 and b=3
    and a+b= 5

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