Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.4 Evaluate Inverse Trigonometric Functions - 13.4 Exercises - Quiz for Lessons 13.3-13.4 - Page 880: 3


$\sin \theta=\dfrac{2 \sqrt 5}{5}$ $\cos \theta=\dfrac{\sqrt 5}{5}$ $\tan \theta=\dfrac{8}{4}=2$ $\csc \theta=\dfrac{\sqrt 5}{2}$ $\sec \theta=\dfrac{4 \sqrt {5}}{4}=\sqrt 5$ $\cot \theta=\dfrac{1}{2}$

Work Step by Step

The Trigonometric Identities can be defined as: $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{8}{4 \sqrt {5}}=\dfrac{2 \sqrt 5}{5}$ $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{4}{4 \sqrt {5}}=\dfrac{\sqrt 5}{5}$ $\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{8}{4}=2$ $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{4 \sqrt {5}}{8}=\dfrac{\sqrt 5}{2}$ $\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{4 \sqrt {5}}{4}=\sqrt 5$ $\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{4}{8}=\dfrac{1}{2}$
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