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Chapter 4: Problem 8

In Exercises \(1-8,\) a point on the terminal side of angle \(\theta\) is given. Find the exact value of each of the six trigonometric functions of \(\theta\). $$(-1,-3)$$

Short Answer

Expert verified
The exact values of the six trigonometric functions for the given point (-1,-3) are sin(theta) = -\(\sqrt{10}\)/10, cos(theta) = -\(\sqrt{10}\)/10, tan(theta) = 3, csc(theta) = -\(\sqrt{10}\), sec(theta) = -\(\sqrt{10}\), cot(theta) = 1/3.

Step by step solution

01

Identify the coordinates

The given point (-1,-3) represents the values x = -1 and y = -3.
02

Calculate the radius r using Pythagorean theorem

We can find r = \(\sqrt{x^2+y^2}\) = \(\sqrt{(-1)^2+(-3)^2}\) = \(\sqrt{1+9}\) = \(\sqrt{10}\).
03

Compute the six trigonometric functions

We know: \n - sin(theta) = y/r = -3/\sqrt{10} = -\sqrt{10}/10 after rationalizing the denominator,\n - cos(theta) = x/r = -1/\sqrt{10} = -\sqrt{10}/10 after rationalizing the denominator,\n - tan(theta) = y/x = -3/(-1) = 3,\n - csc(theta) = 1/sin(theta) = -10/\sqrt{10} = -\sqrt{10} after rationalizing the denominator,\n - sec(theta) = 1/cos(theta) = -10/\sqrt{10} = -\sqrt{10} after rationalizing the denominator, \n - cot(theta) = 1/tan(theta) = 1/3

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