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: 68

Answer

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

Work Step by Step

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