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Chapter 4: Problem 15

The point is on the terminal side of an angle in standard position. Determine the exact values of the six trigonometric functions of the angle. $$(-5,-2)$$

Short Answer

Expert verified
The six trigonometric function values are \( \sin(\theta) = -\frac{2\sqrt{29}}{29} \), \( \cos(\theta) = -\frac{5\sqrt{29}}{29} \), \( \tan(\theta) = \frac{2}{5} \), \( \csc(\theta) = -\frac{\sqrt{29}}{2} \), \( \sec(\theta) = -\frac{\sqrt{29}}{5} \), \( \cot(\theta) = \frac{5}{2} \).

Step by step solution

01

Calculate Radius

Calculate the radius \( r \) using Pythagorean theorem: \( r = \sqrt{x^2 + y^2}\). Here, \( x = -5 \) and \( y = -2 \). So, \( r = \sqrt{(-5)^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29}\).
02

Calculate Six Trigonometric Functions

Use the definitions of the trigonometric functions in terms of \( r, x, \) and \( y \). Here, we have: 1. \( \sin(\theta) = \frac{y}{r} = \frac{-2}{\sqrt{29}} = -\frac{2\sqrt{29}}{29} \),2. \( \cos(\theta) = \frac{x}{r} = \frac{-5}{\sqrt{29}} = -\frac{5\sqrt{29}}{29} \),3. \( \tan(\theta) = \frac{y}{x} = \frac{-2}{-5} = \frac{2}{5} \),4. \( \csc(\theta) = \frac{r}{y} = \frac{\sqrt{29}}{-2} = -\frac{\sqrt{29}}{2} \),5. \( \sec(\theta) = \frac{r}{x} = \frac{\sqrt{29}}{-5} = -\frac{\sqrt{29}}{5} \),6. \( \cot(\theta) = \frac{x}{y} = \frac{-5}{-2} = \frac{5}{2} \).

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