Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 1 - Trigonometric Functions - Section 1.3 Trigonometric Functions - 1.3 Exercises - Page 28: 60

Answer

$sin ~\theta = \frac{\sqrt{3}}{2}$ $cos ~\theta = \frac{-1}{2}$ $tan ~\theta = -\sqrt{3}$ $csc ~\theta = \frac{2}{\sqrt{3}}$ $sec ~\theta = -2$ $cot ~\theta = \frac{-1}{\sqrt{3}}$

Work Step by Step

$\sqrt{3}~x+y = 0, x \leq 0$ $y = -\sqrt{3}~x$ $\frac{y}{x} = -\frac{\sqrt{3}}{1}$ Since $x\leq 0$, we can let $x = -1$ and $y = \sqrt{3}$ Then $r = \sqrt{(-1)^2+(\sqrt{3})^2} = 2$ We can find the values of the six trigonometric functions: $sin ~\theta = \frac{y}{r} = \frac{\sqrt{3}}{2}$ $cos ~\theta = \frac{x}{r} = \frac{-1}{2}$ $tan ~\theta = \frac{y}{x} = \frac{\sqrt{3}}{-1} = -\sqrt{3}$ $csc ~\theta = \frac{r}{y} = \frac{2}{\sqrt{3}}$ $sec ~\theta = \frac{r}{x} = \frac{2}{-1} = -2$ $cot ~\theta = \frac{x}{y} = \frac{-1}{\sqrt{3}}$
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