A random trig question?

Geometry Level 3

Given that sin ⁡ x + sin ⁡ 2 x + sin ⁡ 3 x = 1 \sin x + \sin^2 x + \sin^3 x=1 , find the value of cos ⁡ 6 x − 4 cos ⁡ 4 x + 8 cos ⁡ 2 x . \cos ^6 x - 4\cos ^4 x +8 \cos ^2 x.

5 5 3 3 4 4 1 1 2 2

ChengYiin Ong by ChengYiin Ong , 17, Malaysia

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1 solution

ChengYiin Ong
Jan 14, 2021

The answer is 4 . \boxed{4}. By manipulating the given equation, we have sin ⁡ x + sin ⁡ 2 x + sin ⁡ 3 x = 1 sin ⁡ x ( 1 + sin ⁡ 2 x ) = 1 − sin ⁡ 2 x = cos ⁡ 2 x sin ⁡ x ( 2 − cos ⁡ 2 x ) = cos ⁡ 2 x sin ⁡ 2 x ( 2 − cos ⁡ 2 x ) 2 = cos ⁡ 4 x ( 1 − cos ⁡ 2 x ) ( 2 − cos ⁡ 2 x ) 2 = cos ⁡ 4 x 4 − 4 cos ⁡ 2 x + cos ⁡ 4 x − 4 cos ⁡ 2 x + 4 cos ⁡ 4 x − cos ⁡ 6 x = cos ⁡ 4 x cos ⁡ 6 x − 4 cos ⁡ 4 x + 8 cos ⁡ 2 x = 4. \begin{aligned} \sin x+\sin^2 x+\sin^3 x =& 1 \\ \sin x(1+\sin^2 x) =& 1-\sin^2 x=\cos^2 x \\ \sin x(2-\cos^2 x) =& \cos ^2 x \\ \sin^2 x(2-\cos^2 x)^2 =& \cos ^4 x \\ (1-\cos^2 x)(2-\cos^2 x)^2 =& \cos^4 x \\ 4-4\cos^2 x+\cos^4 x-4\cos^2 x+4\cos^4 x-\cos^6 x=& \cos^4 x \\ \boxed{\cos^6 x - 4\cos^4 x +8\cos^2 x = 4.} \end{aligned}

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