University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 50: 76


Details of proof is shown below. - Do direct calculation when $x=-1$, $x=0$ and $x=1$. - To check on $(-1,0)$, replace $x=-a$ with $a\gt0$ then use the known identities to calculate.

Work Step by Step

$$\sin^{-1}x+\cos^{-1}x=\pi/2$$ We need to check this identity on $[-1,1]$ 1) For $x=-1$: $\sin^{-1}(-1)+\cos^{-1}(-1)=-\frac{\pi}{2}+\pi=\frac{\pi}{2}$ 2) For $x=0$: $\sin^{-1}0+\cos^{-1}0=0+\frac{\pi}{2}=\frac{\pi}{2}$ 3) For $x=1$ $\sin^{-1}1+\cos^{-1}1=\frac{\pi}{2}+0=\frac{\pi}{2}$ So the identity has been proved at $x=-1$, $x=0$ and $x=1$. 4) Prove the identity on $(-1,0)$ The identity has already been proved on $x\in(0,1)$. So if we take $x=-a$, $a\gt0$, then: - $\sin^{-1}(-a)=-\sin^{-1}a$ (shown in Eqs.2) - $\cos^{-1}(-a)=\pi-\cos^{-1}a$ (shown in Eqs.4) So, $\sin^{-1}(-a)+\cos^{-1}(-a)=-\sin^{-1}a+\pi-\cos^{-1}a=-(\sin^{-1}a+\cos^{-1}a)+\pi$ - Since $a\gt0$, $\sin^{-1}a+\cos^{-a}=\frac{\pi}{2}$ has already been proved. Therefore, $$\sin^{-1}(-a)+\cos^{-1}(-a)=-\frac{\pi}{2+\pi}=\frac{\pi}{2}$$ So the identity is also true as $x\lt0$, or $x\in(-1,0)$ since $x\in[-1,1]$. In conclusion, the identity $\sin^{-1}x+\cos^{-1}x=\pi/2$ has been shown on $[-1,1]$.
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