Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 264: 12


Using a calculator, we calculate $y=\displaystyle \tan^{-1}\frac{1}{a} $and add $\pi$

Work Step by Step

(*) lnverse Tangent Function: $y=\tan^{-1}x$ or $y=$ arctan $x$ means that $x=\tan y$, for $-\displaystyle \frac{\pi}{2} < y < \frac{\pi}{2}$. (**) lnverse Cotangent Function: $y=\cot^{-1}x$ or $y=$ arccot $x$ means that $x=\cot y$, for $ 0 < y < \pi$ ----------------- If a is negative, $y=\cot^{-1}a$ means that $a=\cot y$, where $\displaystyle \frac{\pi}{2} < y < \pi$ (second quadrant) Also, if a is negative $y=\displaystyle \tan^{-1}\frac{1}{a}$ that $\displaystyle \frac{1}{a}=\tan y$, for $-\displaystyle \frac{\pi}{2} < y < 0$ (fourth quadrant) Calculating $\displaystyle \tan^{-1}\frac{1}{a}$, we get a number from Q.IV. We need to find a number from Q.II which will be in the domain of $\cot^{-1}$. We know the period for tan and cot, $\tan y=\tan(y+\pi)$, $-\displaystyle \frac{\pi}{2} < y < 0\qquad /+\pi$ $\displaystyle \frac{\pi}{2} < y+\pi < \pi$ So, we calculate $y=\displaystyle \tan^{-1}\frac{1}{a}, $which is in quadrant IV, add $\pi$ to $y $so that $ y+\pi$ is in quadrant $II, $and $\cot(y+\pi)=a $, that is $\displaystyle \tan^{-1}\frac{1}{a}+\pi=\cot^{-1}a$
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