Step By Step Calculus » 6.5 - Inverse Trigonometric Identities and Applications

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Synopsis
Identities Involving Inverse Trigonometric Functions: Expressing \sin^{-1} x\sin^{-1} x and \cos^{-1}x\cos^{-1}x in a right angle triangle, we get the following identities:
\displaystyle \left.\begin{array}{ll} \textrm{(i) (Inverse Sin/Cos Identity)} & \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\\ \textrm{(ii) (Inverse Cos Symmetric Identity)} & \cos^{-1}x+\cos^{-1}(-x)=\pi \\ \textrm{(iii) (Inverse Sin Symmetric Identity)} & \sin^{-1}x+\sin^{-1}(-x)=0 \end{array}\right\} \forall x\in[-1,1]
\displaystyle \left.\begin{array}{ll} \textrm{(i) (Inverse Sin/Cos Identity)} & \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\\ \textrm{(ii) (Inverse Cos Symmetric Identity)} & \cos^{-1}x+\cos^{-1}(-x)=\pi \\ \textrm{(iii) (Inverse Sin Symmetric Identity)} & \sin^{-1}x+\sin^{-1}(-x)=0 \end{array}\right\} \forall x\in[-1,1]
In the identity (f\circ g)^{-1}=f^{-1}\circ g^{-1}(f\circ g)^{-1}=f^{-1}\circ g^{-1} we can set f(u)=\frac{1}{u}f(u)=\frac{1}{u} and g(x)=\cos(x)g(x)=\cos(x) and (recalling that \sec(x)=\frac{1}{\cos(x)}\sec(x)=\frac{1}{\cos(x)}) derive the identity
\displaystyle \sec^{-1}x=\cos^{-1}\left(\frac1x\right) \displaystyle \sec^{-1}x=\cos^{-1}\left(\frac1x\right)
(1)
Similarly,
\displaystyle \csc^{-1}x=\sin^{-1}\left(\frac1x\right). \displaystyle \csc^{-1}x=\sin^{-1}\left(\frac1x\right).
(2)
By replacing xx by \frac1x\frac1x in Identities (i-iii) and using Equations (1) and (2), we obtain the following identities:
\displaystyle \left.\begin{array}{ll} \textrm{(Inverse Sec/Csc Identity)} & \csc^{-1}x+\sec^{-1}x=\frac{\pi}{2}\\ \textrm{(Inverse Sec Symmetric Identity)} & \sec^{-1}x+\sec^{-1}(-x)=\pi \\ \textrm{(Inverse Csc Symmetric Identity)} & \csc^{-1}x+\csc^{-1}(-x)=0 \end{array}\right\} \forall x\in(-\infty, -1]\cup[1,\infty).
\displaystyle \left.\begin{array}{ll} \textrm{(Inverse Sec/Csc Identity)} & \csc^{-1}x+\sec^{-1}x=\frac{\pi}{2}\\ \textrm{(Inverse Sec Symmetric Identity)} & \sec^{-1}x+\sec^{-1}(-x)=\pi \\ \textrm{(Inverse Csc Symmetric Identity)} & \csc^{-1}x+\csc^{-1}(-x)=0 \end{array}\right\} \forall x\in(-\infty, -1]\cup[1,\infty).
Expressing \tan^{-1} x\tan^{-1} x and \cot^{-1}x\cot^{-1}x in a right angle triangle, we get three more identities:
\displaystyle \left.\begin{array}{ll} \textrm{(Inverse Tan/Cot Identity)} & \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\\ \textrm{(Inverse Cot Symmetric Identity)} & \cot^{-1}x+\cot^{-1}(-x)=\pi \\ \textrm{(Inverse Tan Symmetric Identity)} & \tan^{-1}x+\tan^{-1}(-x)=0 \end{array}\right\} \forall x\in(-\infty,\infty).
\displaystyle \left.\begin{array}{ll} \textrm{(Inverse Tan/Cot Identity)} & \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\\ \textrm{(Inverse Cot Symmetric Identity)} & \cot^{-1}x+\cot^{-1}(-x)=\pi \\ \textrm{(Inverse Tan Symmetric Identity)} & \tan^{-1}x+\tan^{-1}(-x)=0 \end{array}\right\} \forall x\in(-\infty,\infty).
Composites with Inverses: Later in our course we will have to deal with trigonometric functions of the form f\circ g^{-1}f\circ g^{-1}, specifically:
\newcommand{\T}{\rule{0pt}{3.3ex}} \newcommand{\B}{\rule[-2.4ex]{0pt}{0pt}} \begin{tabular}{|c|c|c|c|c|}\hline \T\B$f$ & $g$ & $f\circ g^{-1}$ &\textrm{Domain of $f\circ g^{-1}$}&\textrm{Range of $f\circ g^{-1}$}\\ \hline \T\B$\sin$ & $\cos$ & $\sin(\cos^{-1}x)=\sqrt{1-x^2}$ &$[-1,1]$ & $[0,1]$\\ \hline \T\B$\cos$ & $\sin$ & $\cos(\sin^{-1}x)=\sqrt{1-x^2}$ &$[-1,1]$ & $[0,1]$\\ \hline \T\B$\sin$ & $\tan$ & $\displaystyle \sin(\tan^{-1}x)=\dfrac{x}{\sqrt{1+x^2}}$ &$\mathbb{R}$ & $(-1,1)$\\ \hline \T\B$\cos$ & $\tan$ & $\displaystyle \cos(\tan^{-1}x)=\dfrac{1}{\sqrt{1+x^2}}$ &$\mathbb{R}$ & $(0,1]$\\ \hline \T\B$\sec$ & $\tan$ & $\displaystyle \sec(\tan^{-1}x)=\sqrt{1+x^2}$ &$\mathbb{R}$ &$[1,\infty)$ \\ \hline \end{tabular}