Trignometry (3)

Geometry Level 4

If sin 4 x 2 + cos 4 x 3 = 1 5 \dfrac{\sin^4 x}2 + \dfrac{\cos^4 x}3 = \dfrac15 , find 81000 ( sin 8 x 8 + cos 8 x 27 ) 81000 \left( \dfrac{\sin^8 x}8 + \dfrac{\cos^8 x}{27} \right) .


The answer is 648.

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1 solution

Chew-Seong Cheong
Jan 20, 2016

sin 4 x 2 + cos 4 x 3 = 1 5 15 sin 4 x + 10 cos 4 x = 6 15 ( 1 cos 2 x ) 2 + 10 cos 4 x = 6 15 ( 1 2 cos 2 x + cos 4 x ) + 10 cos 4 x = 6 25 cos 4 x 30 cos 2 x + 9 = 0 ( 5 cos 2 x 3 ) 2 = 0 cos 2 x = 3 5 sin 2 x = 1 cos 2 x = 1 3 5 = 2 5 81000 ( sin 8 x 8 + cos 8 x 27 ) = 81000 ( 1 8 ( 2 5 ) 4 + 1 27 ( 3 5 ) 4 ) = 81000 ( 2 5 4 + 3 5 4 ) = 81000 5 3 = 648 \begin{aligned} \frac{\sin^4 x}{2} + \frac{\cos^4 x}{3} & = \frac{1}{5} \\ 15\sin^4 x + 10\cos^4 x & = 6 \\ 15(1-\cos^2 x)^2 + 10\cos^4 x & = 6 \\ 15(1-2\cos^2 x + \cos^4 x) + 10\cos^4 x & = 6 \\ 25 \cos^4 x - 30 \cos^2 x + 9 & = 0 \\ (5 \cos^2 x - 3)^2 & = 0 \\ \Rightarrow \cos^2 x & = \frac{3}{5} \\ \Rightarrow \sin^2 x & = 1 - \cos^2 x = 1 - \frac{3}{5} = \frac{2}{5} \\ & \\ \Rightarrow 81000 \left(\frac{\sin^8 x}{8} + \frac{\cos^8 x}{27} \right) & = 81000 \left(\frac{1}{8} \left( \frac{2}{5}\right)^4 + \frac{1}{27} \left( \frac{3}{5}\right)^4 \right) \\ & = 81000 \left(\frac{2}{5^4} + \frac{3}{5^4} \right) \\ & = \frac{81000}{5^3} \\ & = \boxed{648} \end{aligned}

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