Trigonometry! #84

Geometry Level 3

Simplify the following expression to a numerical value. 3 ( sin ⁡ 4 ( 3 π 2 − α ) + sin ⁡ 4 ( 3 π + α ) ) − 2 ( sin ⁡ 6 ( π 2 + α ) + sin ⁡ 6 ( 5 π − α ) ) 3\left(\sin^{4} \left(\frac{3\pi}{2} - \alpha \right) + \sin^{4}(3\pi+\alpha)\right) - 2 \left(\sin^{6} \left(\frac{\pi}{2} + \alpha \right) + \sin^{6}(5\pi-\alpha)\right)

This problem is part of the set Trigonometry .


The answer is 1.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

Nihar Mahajan
Sep 11, 2015

By All Silver Tea Cup rule , we have the following relations:

sin ⁡ ( 3 π 2 − α ) = cos ⁡ α , sin ⁡ ( 3 π + α ) = sin ⁡ α , sin ⁡ ( π 2 + α ) = cos ⁡ α , sin ⁡ ( 5 π + α ) = sin ⁡ α \sin\left(\dfrac{3\pi}{2}-\alpha\right) = \cos\alpha \ , \ \sin(3\pi + \alpha) = \sin\alpha \ , \ \sin\left(\dfrac{\pi}{2}+\alpha\right) = \cos\alpha \ , \ \sin(5\pi + \alpha) = \sin\alpha

Thus , by substituting the above things in give expression we get:

3 cos ⁡ 4 α + 3 sin ⁡ 4 α − 2 c o s 6 α − 2 sin ⁡ 6 α = 3 sin ⁡ 4 α − 2 sin ⁡ 6 α + 3 cos ⁡ 4 α − 2 c o s 6 α = sin ⁡ 4 α ( 3 − 2 sin ⁡ 2 α ) + cos ⁡ 4 α ( 3 − 2 cos ⁡ 2 α ) = sin ⁡ 4 α ( 1 + 2 cos ⁡ 2 α ) + cos ⁡ 4 α ( 1 + sin ⁡ 2 α ) = sin ⁡ 4 α + cos ⁡ 4 α + 2 sin ⁡ 4 α sin ⁡ 2 α + 2 cos ⁡ 4 α cos ⁡ 2 α = sin ⁡ 4 α + cos ⁡ 4 α + 2 sin ⁡ 2 α cos ⁡ 2 α ( cos ⁡ 2 α + sin ⁡ 2 α ) = sin ⁡ 4 α + cos ⁡ 4 α + 2 sin ⁡ 2 α cos ⁡ 2 α = ( cos ⁡ 2 α + sin ⁡ 2 α ) 2 = 1 2 = 1 3\cos^4\alpha + 3\sin^4\alpha-2cos^6\alpha-2\sin^6\alpha \\ = 3\sin^4\alpha-2\sin^6\alpha+3\cos^4\alpha-2cos^6\alpha \\ = \sin^4\alpha(3-2\sin^2\alpha) + \cos^4\alpha(3-2\cos^2\alpha) \\ = \sin^4\alpha(1+2\cos^2\alpha) + \cos^4\alpha(1+\sin^2\alpha) \\ = \sin^4\alpha + \cos^4\alpha + 2\sin^4\alpha\sin^2\alpha+2\cos^4\alpha\cos^2\alpha \\ = \sin^4\alpha + \cos^4\alpha + 2\sin^2\alpha\cos^2\alpha(\cos^2\alpha + \sin^2\alpha) \\ = \sin^4\alpha + \cos^4\alpha + 2\sin^2\alpha\cos^2\alpha \\ = (\cos^2\alpha + \sin^2\alpha)^2 \\ = 1^2 \\ = \Large\boxed{1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Deepak Kumar
Sep 11, 2015

Simply put alpha 0 as the expression is always defined .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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