+289 Use the sum of cubes factorization to get:
Suppose that sin x = x and cos x = y.
x^6 + y^6 = (x^2 + y^2)((x^2 + y^2)^2 - 3x^2y^2)
x^2 + y^2 = 1 as sin^2 x + cos^2 x is always one by the pythagorean identity.
By sin(2x) = 2/3, we can find that 2sin(x)cos(x) = 2/3
Leaving us with xy = 1/3 which finally gives us x^2y^2 = 1/9
Substituing all values, we result in:
(1)(2/3) = \(\boxed{\frac23}\)
+289 Use the sum of cubes factorization to get:
Suppose that sin x = x and cos x = y.
x^6 + y^6 = (x^2 + y^2)((x^2 + y^2)^2 - 3x^2y^2)
x^2 + y^2 = 1 as sin^2 x + cos^2 x is always one by the pythagorean identity.
By sin(2x) = 2/3, we can find that 2sin(x)cos(x) = 2/3
Leaving us with xy = 1/3 which finally gives us x^2y^2 = 1/9
Substituing all values, we result in:
(1)(2/3) = \(\boxed{\frac23}\)
