Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 260: 91

Answer

$cos(tan^{-1}\frac{5}{12}-tan^{-1}\frac{3}{4}) = \frac{63}{65}$

Work Step by Step

Let $A = tan^{-1}\frac{5}{12}$ Then: $cos~A = \frac{12}{\sqrt{5^2+12^2}} = \frac{12}{13}$ $sin~A = \frac{5}{\sqrt{5^2+12^2}} = \frac{5}{13}$ Let $B = tan^{-1}\frac{3}{4}$ $cos~B = \frac{4}{\sqrt{3^2+4^2}} = \frac{4}{5}$ $sin~B = \frac{3}{\sqrt{3^2+4^2}} = \frac{3}{5}$ We can find $~~cos(tan^{-1}\frac{5}{12}-tan^{-1}\frac{3}{4})$: $cos(A-B) = cos~A~cos~B+sin~A~sin~B$ $cos(A-B) = (\frac{12}{13})~(\frac{4}{5})+(\frac{5}{13})~(\frac{3}{5})$ $cos(A-B) = \frac{48}{65}+\frac{15}{65}$ $cos(A-B) = \frac{63}{65}$ Therefore, $~~cos(tan^{-1}\frac{5}{12}-tan^{-1}\frac{3}{4}) = \frac{63}{65}$
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