Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.3 - Definition I: Trigonometric Functions - 1.3 Problem Set - Page 33: 67

Answer

$\sin{\theta} =\dfrac{a}{\sqrt{a^2+b^2}} $ $\cos{\theta} = \dfrac{b}{\sqrt{a^2+b^2}} $ $\tan{\theta} =\dfrac{a}{b}$ $\csc{\theta} =\dfrac{\sqrt{a^2+b^2}}{a}$ $\sec{\theta} = \dfrac{\sqrt{a^2+b^2}}{b}$ $\cot{\theta} = \dfrac{b}{a}$

Work Step by Step

$\tan{\theta} = \dfrac{y}{x} = \dfrac{a}{b}$ $\therefore y = a \hspace{20pt} x = b $ $\because x $ and $y$ are both positive $ \therefore$ The terminal side of $\theta$ in standard position lies in $QI$ $r = \sqrt{x^2+y^2} = \sqrt{b^2+a^2} $ $\sin{\theta} = \dfrac{y}{r} = \dfrac{a}{\sqrt{a^2+b^2}} $ $\cos{\theta} = \dfrac{x}{r} = \dfrac{b}{\sqrt{a^2+b^2}} $ $\csc{\theta} = \dfrac{1}{\sin{\theta}} = \dfrac{\sqrt{a^2+b^2}}{a}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}} = \dfrac{\sqrt{a^2+b^2}}{b}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}} = \dfrac{b}{a}$
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