Question

In Exercises 23–34, find the exact value of each of the remaining trigonometric functions of θ. sec θ = -3, tan θ \u003e 0

Accepted Answer

Recall the definition of secant: $$\sec 	heta = \frac{1}{\cos 	heta}. $$Given $$\sec 	heta = -3$$, find $$\cos 	heta by $$taking the reciprocal: $$\cos 	heta = \frac{1}{\sec 	heta} = \frac{1}{-3} = -\frac{1}{3}. $$Determine the quadrant where $$	heta$$ lies using the signs of $$\sec 	heta$$ and $$	an 	heta. $$Since $$\sec 	heta = -3 ($$negative) and $$	an 	heta \u003e 0 ($$positive), recall that $$\sec 	heta$$ has the same sign as $$\cos 	heta. So$$ $$\cos 	heta is $$negative and $$	an 	heta is $$positive. This occurs in Quadrant III. Use the Pythagorean identity to find $$\sin 	heta$$: $$\sin^2$ 	heta + \cos^2$ 	heta = 1. $$Substitute $$\cos 	heta = -\frac{1}{3} to $$get $$\sin^2$ 	heta = 1 - \left(-\frac{1}{3}\$$right)^2$$ = 1 - \frac{1}{9} = \frac{8}{9}. $$Then, $$\sin 	heta = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3}. $$Since $$	heta is in $$Quadrant III, where sine is negative, choose $$\sin 	heta = -\frac{2\sqrt{2}}{3}. $$Find $$	an 	heta$$ using the definition $$	an 	heta = \frac{\sin 	heta}{\cos 	heta}. $$Substitute the values found: $$	an 	heta = \frac{-\frac{2\sqrt{2}}{3}}{-\frac{1}{3}}. $$Simplify the fraction to find $$	an 	heta. $$Calculate the remaining trigonometric functions using the relationships: $$\csc 	heta = \frac{1}{\sin 	heta}$$, $$\cot 	heta = \frac{1}{	an 	heta}$$, and verify the signs based on the quadrant to ensure consistency.

In Exercises 23–34, find the exact value of each of the remaining trigonometric functions of θ. sec θ = -3, tan θ > 0

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주요 개념

Reciprocal Trigonometric Functions

Sign of Trigonometric Functions in Quadrants

Pythagorean Identities