Posted by **Shin** on Monday, May 27, 2013 at 11:51am.

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 -
**Reiny**, Monday, May 27, 2013 at 12:33pm
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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