Inverse Trigonometric Functions
A function has an inverse only if it is one-one and onto. \(\sin x\) is neither on \(\mathbb{R}\), but restricted to \([-\frac\pi2,\frac\pi2]\) it becomes a bijection onto \([-1,1]\). Its inverse, \(\sin^{-1}x\) (also written \(\arcsin x\)), answers the question: which angle in \([-\frac\pi2,\frac\pi2]\) has sine equal to \(x\)?
Each trigonometric function gets its own standard interval – its principal value branch. The value of the inverse function in this branch is the principal value.
| Function | Domain | Range (principal branch) |
|---|---|---|
| \(\sin^{-1}x\) | \([-1,1]\) | \([-\frac{\pi}{2},\frac{\pi}{2}]\) |
| \(\cos^{-1}x\) | \([-1,1]\) | \([0,\pi]\) |
| \(\tan^{-1}x\) | \(\mathbb{R}\) | \((-\frac{\pi}{2},\frac{\pi}{2})\) |
| \(\cot^{-1}x\) | \(\mathbb{R}\) | \((0,\pi)\) |
| \(\sec^{-1}x\) | \(\mathbb{R}-(-1,1)\) | \([0,\pi]-\{\frac{\pi}{2}\}\) |
| \(\csc^{-1}x\) | \(\mathbb{R}-(-1,1)\) | \([-\frac{\pi}{2},\frac{\pi}{2}]-\{0\}\) |
Example: \(\cos^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)=\pi-\tfrac\pi6=\tfrac{5\pi}{6}\), which lies in \([0,\pi]\).
Caution: \(\sin^{-1}x\) is not \(\frac{1}{\sin x}\). The reciprocal is \((\sin x)^{-1}=\csc x\).
| Function | Domain | Range (principal branch) |
|---|---|---|
| \(\sin^{-1}x\) | \([-1,1]\) | \([-\frac{\pi}{2},\frac{\pi}{2}]\) |
| \(\cos^{-1}x\) | \([-1,1]\) | \([0,\pi]\) |
| \(\tan^{-1}x\) | \(\mathbb{R}\) | \((-\frac{\pi}{2},\frac{\pi}{2})\) |
| \(\cot^{-1}x\) | \(\mathbb{R}\) | \((0,\pi)\) |
| \(\sec^{-1}x\) | \(\mathbb{R}-(-1,1)\) | \([0,\pi]-\{\frac{\pi}{2}\}\) |
| \(\csc^{-1}x\) | \(\mathbb{R}-(-1,1)\) | \([-\frac{\pi}{2},\frac{\pi}{2}]-\{0\}\) |
\[\sin^{-1}(-x)=-\sin^{-1}x,\quad \tan^{-1}(-x)=-\tan^{-1}x,\quad \csc^{-1}(-x)=-\csc^{-1}x\]\[\cos^{-1}(-x)=\pi-\cos^{-1}x,\quad \cot^{-1}(-x)=\pi-\cot^{-1}x,\quad \sec^{-1}(-x)=\pi-\sec^{-1}x\]
| x | 0 | 1/2 | 1/√2 | √3/2 | 1 |
|---|---|---|---|---|---|
| \(\sin^{-1}x\) | 0 | \(\frac\pi6\) | \(\frac\pi4\) | \(\frac\pi3\) | \(\frac\pi2\) |
| \(\cos^{-1}x\) | \(\frac\pi2\) | \(\frac\pi3\) | \(\frac\pi4\) | \(\frac\pi6\) | 0 |
\(\tan^{-1}1=\frac\pi4\), \(\tan^{-1}\sqrt3=\frac\pi3\), \(\tan^{-1}\frac{1}{\sqrt3}=\frac\pi6\).
\(\sin^{-1}(\sin\frac{5\pi}{6})=\frac\pi6\), because \(\frac{5\pi}{6}\notin[-\frac\pi2,\frac\pi2]\) and \(\sin\frac{5\pi}6=\sin\frac\pi6\).
Solve \(-1\le g(x)\le 1\). For \(\sec^{-1}(g(x))\) solve \(|g(x)|\ge1\).
\(\cos^{-1}(-\tfrac12)\neq-\tfrac\pi3\). The range of \(\cos^{-1}\) is \([0,\pi]\), so the answer is \(\pi-\tfrac\pi3=\tfrac{2\pi}3\).
\(\sin^{-1}x\neq\dfrac{1}{\sin x}\). \(\sin^{-1}\) is an angle; \(\frac1{\sin x}=\csc x\).
\(\tan^{-1}(\tan\frac{3\pi}4)=-\frac\pi4\), not \(\frac{3\pi}4\): the answer must lie in \((-\frac\pi2,\frac\pi2)\).
Evaluate \(\tan^{-1}(\sqrt3)+\cos^{-1}(-\tfrac12)+\sin^{-1}(-\tfrac1{\sqrt2})\).
\(\frac\pi3+\frac{2\pi}3-\frac\pi4=\pi-\frac\pi4=\frac{3\pi}4\).
Find the domain of \(\sin^{-1}(2x-3)\).
\(-1\le2x-3\le1\Rightarrow 2\le2x\le4\Rightarrow x\in[1,2]\).
A fresh set of questions from this topic. Change answers freely – solutions appear after you submit.
Start PracticeLearn the six domains and ranges. A principal value must lie in its branch; negative arguments move to the other side of 0 for sin⁻¹, tan⁻¹, cosec⁻¹ and to π − (…) for cos⁻¹, cot⁻¹, sec⁻¹.
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