Trigonometry! #22

Geometry Level 3

If 2 cos ⁡ 2 α − 1 = 0 2\cos ^{2} \alpha - 1 = 0 and α \alpha lies in the third quadrant, then sin ⁡ α tan ⁡ α \sin \alpha \tan \alpha =

This problem is part of the set Trigonometry .

− 1 -1 − 1 2 \frac {-1}{\sqrt{2}} 2 \sqrt{2} 1 2 \frac{1}{\sqrt{2}}

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Jared Low
Jan 26, 2015

We have cos ⁡ 2 α = 1 2 \cos^2\alpha=\frac{1}{2} . Since α \alpha is in the third quadrant, we have cos ⁡ α = − 1 2 \cos\alpha=-\frac{1}{\sqrt{2}} . From there, we have sin ⁡ α = − 1 2 \sin\alpha=-\frac{1}{\sqrt{2}} and tan ⁡ α = 1 \tan\alpha=1 .

Therefore, sin ⁡ α tan ⁡ α = − 1 2 \sin\alpha\tan\alpha=\boxed{-\frac{1}{\sqrt{2}}}

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