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# trig question

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If sin(2x) = 2/3, then find sin(x)^6 + cos(x)^6.

Aug 5, 2022

#1
+193
+1

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) = $$\boxed{\frac23}$$

Aug 5, 2022

#1
+193
+1

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) = $$\boxed{\frac23}$$

Voldemort Aug 5, 2022