Question

Find values of the sine and cosine functions for each angle measure.θ, given cos 2θ = 3/4 and θ terminates in quadrant III

Accepted Answer

Recall the double-angle identity for cosine: $$\cos 2	heta = 2\cos^2$ 	heta - 1. $$This relates $$\cos 2	heta to$$ $$\cos 	heta. $$Substitute the given value $$\cos 2	heta = \frac{3}{4}$$ into the identity: $$\frac{3}{4} = 2\cos^2$ 	heta - 1. $$Solve the equation for $$\cos^2$ 	heta$$: add 1 to both sides to get $$\frac{3}{4} + 1 = 2\cos^2$ 	heta$$, which simplifies to $$\frac{7}{4} = 2\cos^2$ 	heta. $$Then divide both sides by 2 to find $$\cos^2$ 	heta = \frac{7}{8}. $$Take the square root of both sides to find $$\cos 	heta = \pm \sqrt{\frac{7}{8}}. $$Since $$	heta$$ terminates in quadrant III, where cosine is negative, choose the negative root: $$\cos 	heta = -\sqrt{\frac{7}{8}}. $$Use the Pythagorean identity $$\sin^2$ 	heta + \cos^2$ 	heta = 1 to $$find $$\sin 	heta. $$Substitute $$\cos^2$ 	heta = \frac{7}{8} to $$get $$\sin^2$ 	heta = 1 - \frac{7}{8} = \frac{1}{8}. $$Then $$\sin 	heta = \pm \sqrt{\frac{1}{8}}. $$Since $$	heta is in $$quadrant III, where sine is negative, choose the negative root: $$\sin 	heta = -\sqrt{\frac{1}{8}}.$$

Find values of the sine and cosine functions for each angle measure.

θ, given cos 2θ = 3/4 and θ terminates in quadrant III

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Key Concepts

Double-Angle Identity for Cosine

Sign of Trigonometric Functions in Quadrants

Pythagorean Identity