Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 77


$\arccos \dfrac{3}{\sqrt {x^2-2x+10}}=\arcsin \dfrac{|x-1|}{\sqrt {x^2-2x+10}}$

Work Step by Step

In order to compute the length of the opposite we will use the Pythagorean Theorem. Consider $\theta=\arccos \dfrac{3}{\sqrt {x^2-2x+10}}$ $h=\sqrt {x^2+y^2} \implies (\sqrt {x^2-2x+10})^2=\sqrt {(x-1)^2+9^2 }=|x-1|$ We can see that $\dfrac{3}{\sqrt {x^2-2x+10}}$ is a positive quantity, and $\theta$ must lie in the first quadrant. So, we have: $\sin \theta=\dfrac{|x-1|}{\sqrt {x^2-2x+10}} $ and $\arccos \dfrac{3}{\sqrt {x^2-2x+10}}=\arcsin \dfrac{|x-1|}{\sqrt {x^2-2x+10}}$
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