Trigonometric functions

f ( θ )= 8sin^(4) θ - 2sin^(2) θ - 2
1. Show that f (θ) = cos4 θ - 3cos 2 θ

8 sin⁴(x) - 2 sin²(x) - 2 =
= 1 - 8 sin²(x) + 8 sin⁴(x) - 3 + 6 sin²(x) =
= 1 - 8 sin²(x) + 8 sin⁴(x) - 3[1 - 2 sin²(x)] =
= 1 - 8 sin²(x) + 8 sin⁴(x) - 3 cos(2x) =
= 1 - 8 sin²(x) [1 - sin²(x)] - 3 cos(2x) =
= 1 - 8 sin²(x) cos²(x) - 3 cos(2x) =
= 1 - 2 [2 sin(x) cos(x)]² - 3 cos(2x) =
= 1 - 2 sin²(2x) - 3 cos(2x) =
= cos(4x) - 3 cos(2x)
Q.E.D.

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