Trigonometry
posted by Shin .
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.

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
and
(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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