Inverse trigonometric function
- \(\sin^{–1} x\) should not be confused with \(\sin^{–1} x\). In fact \(\sin^{–1} x = \dfrac{1}{\sin x}\) and similarly for other trigonometric functions.
- Whenever no branch of an inverse trigonometric functions is mentioned, we mean the principle value branch of that function.
- The value of an inverse trigonometric functions which lies in the range of principle branch is called the principle value of that inverse trigonometric functions.
Inverse Trigonometric Functions | Domain | Range |
|---|---|---|
\(\sin^{-1}\) | \([-1,1]\) | \(\Bigg[-\dfrac{\pi}{2},\dfrac{\pi}{2}\Bigg]\) |
\(\cos^{-1}\) | \([-1,1]\) | \([0,\pi]\) |
\(\mathrm{cosec}^{-1}\) | \(R-(-1,1)\) | \(\Bigg[-\dfrac{\pi}{2},\dfrac{\pi}{2}\Bigg]-\Bigg\{\dfrac{\pi}{2}\Bigg\}\) |
\(\sec^{-1}\) | \(R-(-1,1)\) | \((0,\pi)-\Bigg\{\dfrac{\pi}{2}\Bigg\}\) |
\(\tan^{-1}\) | \(R\) | \(\Bigg(-\dfrac{\pi}{2},\dfrac{\pi}{2}\Bigg)\) |
\(\cot^{-1}\) | \(R\) | \((0,\pi)\) |
Example 1:
Find the principle value of \(\sin^{-1}\dfrac{1}{\sqrt 2}\).
Solution:
Let \(\sin^{-1}\Bigg(\dfrac{1}{\sqrt 2}\Bigg)=y\).Then, \(\sin y=\dfrac{1}{\sqrt 2}\)
we know the range of principle value of \(\sin^{-1}\) is
\(\Bigg[-\dfrac{\pi}{2},\dfrac{\pi}{2}\Bigg]\) and \(\sin\Bigg[\dfrac{\pi}{4}\Bigg]=\dfrac{1}{\sqrt2}\)
Therefore,principle value of \(\sin^{1}\Bigg(\dfrac{1}{\sqrt 2}\Bigg)\) is \(\dfrac{\pi}{4}\)
Example 2:
Find the principle value of \(\cot^{-1}\dfrac{-1}{\sqrt 3}\).
Solution:
Find the principle value of \(\cot^{-1}\dfrac{-1}{\sqrt 3}\).
Solution:
Let \(\cot^{-1}\Bigg(\dfrac{1}{\sqrt 3}\Bigg)=y\).Then,
\(\cot y=\dfrac{-1}{\sqrt 3}=-\cot \dfrac{\pi}{3}\)\(=\cot \Bigg(\pi-\dfrac{\pi}{3}\Bigg)=\cot \Bigg( \dfrac{2\pi}{3} \Bigg)\)
we know the range of principle value of \(\cot^{-1}\) is \((0,\pi)\) and \(\cot \Bigg( \dfrac{2\pi}{3} \Bigg)=\dfrac{-1}{\sqrt 3}\)
Hence the principle value of \(\cot^{-1}\dfrac{1}{\sqrt 2}\) is \(\dfrac{2\pi}{3}\).
