$\sin: \mathbb{R} \rightarrow [-1,1]$

$x \rightarrow \sin x \in [-1,1]$

Can we define the inverse function?

${\displaystyle y\in [-1,1],}$will there exist a unique ${\displaystyle x\in ℝ}$

such that

$\sin x = y?$

Let, ${\displaystyle f:ℝ\to {ℝ}^{+}}$

${\displaystyle f\left(x\right)=x^2}$

Let, $f’: R^+ \rightarrow R^+$ ${\displaystyle \mathrm{f’}\left(x\right)=x^2}$

And, taking ${\displaystyle g:{ℝ}^{+}\to {ℝ}^{+}}$ ${\displaystyle g\left(y\right)=\sqrt{y}}$ ,where, ${\displaystyle g\equiv f{\prime }^{-1}}$

${\displaystyle f{\prime }^{-1}:{ℝ}^{+}\to {ℝ}^{+}}$

${\displaystyle f{\prime }^{-1}\left(y\right)=\sqrt{y}}$ .

${\displaystyle {\sin }^{-1}\left(\frac{-1}{2}\right)=\frac{-\pi }{6}}$

${\displaystyle \sin \left(\frac{-\pi }{6}\right)=\frac{-1}{2}}$

${\displaystyle {\sin }^{-1}:[-1,1]\to [-\frac{\pi }{2},\frac{\pi }{2}]}$

${\displaystyle \sin \left(\frac{-\pi }{3}\right)=\frac{-\sqrt{3}}{2}}$

${\displaystyle {\sin }^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{-\pi }{3}}$

${\displaystyle {\sin }^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{-\pi }{4}}$

${\displaystyle \sin \left(\frac{-\pi }{2}\right)=-1}$

${\displaystyle {\sin }^{-1}(-1)=\frac{-\pi }{2}}$

cos$^{-1} : [-1,1] \rightarrow [0,π]$

${\displaystyle \cos \pi =-1}$

${\displaystyle {\cos }^{-1}(-1)=\pi }$

${\displaystyle \cos \left(\frac{3\pi }{4}\right)=-\frac{1}{\sqrt{2}}}$

${\displaystyle {\cos }^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3\pi }{4}}$

${\displaystyle {\tan }^{-1}:ℝ\to (\frac{-\pi }{2},\frac{\pi }{2})}$

${\displaystyle \tan \left(\frac{\pi }{4}\right)=1}$

${\displaystyle {\tan }^{-1}\left(1\right)=\frac{\pi }{4}}$

${\displaystyle {\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}}$

tan(0)=0

${\displaystyle {\tan }^{-1}\left(0\right)=0}$

$\cot: R-$ $\{n\pi | n \in Z\}$ $\rightarrow R$

${\displaystyle \cot \left(x\right)=\frac{1}{\tan x}}$

${\displaystyle \cot \left(x\right)=\frac{\cos x}{\sin x}}$

${\displaystyle {\cot }^{-1}:ℝ\to (0,\pi )}$

${\displaystyle {\cot }^{-1}(-1)=\frac{3\pi }{4}}$

${\displaystyle \cot \left(\frac{\pi }{3}\right)=\frac{1}{\sqrt{3}}}$

${\displaystyle {\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{3}}$

${\displaystyle \cot \left(\frac{\pi }{2}\right)=0}$

${\displaystyle {\cot }^{-1}\left(0\right)=\frac{\pi }{2}}$

$cot^{-1}(\sqrt{3}) = \frac{π}{6}$

$cosec: R-$ $\{n\pi | n \in Z\}$ $\rightarrow R - (-1,1)$

${\displaystyle \mathrm{cosec}\left(x\right)=\frac{1}{\sin x}}$

${\displaystyle \cos e{c}^{-1}:ℝ-(-1,1)\to [-\frac{\pi }{2},\frac{\pi }{2}]-\left\{0\right\}}$

${\displaystyle \cos ec\left(\frac{-\pi }{6}\right)=-2}$

${\displaystyle \cos e{c}^{-1}(-2)=\frac{-\pi }{6}}$

${\displaystyle \cos ec\left(\frac{-\pi }{2}\right)=-1}$

${\displaystyle \cos e{c}^{-1}(-1)=\frac{-\pi }{2}}$

$\operatorname{cosec}\left(\frac{-\pi}{3}\right)=\frac{-2}{\sqrt{3}}\qquad\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\frac{-\pi}{3}$

$\sec: R- \{(n+\frac{1}{2})\pi | n \in Z\}\rightarrow R - (-1,1)$

${\displaystyle \sec \left(x\right)=\frac{1}{\cos x}}$

${\displaystyle {\sec }^{-1}:ℝ-(-1,1)\to [0,\pi ]-\left\{\frac{\pi }{2}\right\}}$