Question
For the principal values, evaluate the following:
$\sin^{-1}[\cos\{2\text{cosec}^{-1}(-2)\}]$
$\sin^{-1}[\cos\{2\text{cosec}^{-1}(-2)\}]$
✓
Answer
$\text{cosec}^{-1}\text{x}$ represents an angle in $\Big[-\frac{\pi}{2},0\Big)\cup\Big(0,\frac{\pi}{2}\Big]$ whose cosecant is x. Let $\text{x}=\text{cosec}^{-1}(-2)$ $\Rightarrow\text{cosec x}=-2=\text{cosec c}\Big(-\frac{\pi}{6}\Big)$ $\Rightarrow\text{x}=-\frac{\pi}{6}$ $\sin^{-1}[\cos\{2\text{cosec}^{-1}(-2)\}]=\sin^{-1}\Big[\cos\Big\{2\times\Big(-\frac{\pi}{6}\Big)\Big\}\Big]$ $=\sin^{-1}\Big[\cos\Big(-\frac{\pi}{3}\Big)\Big]=\sin^{-1}\Big[\frac{1}{2}\Big]$ $\sin^{-1}\text{x}$ represents an angle in $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$ whose sin is x.Let $\text{x}=\sin^{-1}\Big[\frac{1}{2}\Big]$
$\Rightarrow\sin\text{x}=\frac{1}{2}=\sin\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=\frac{\pi}{6}$
$\Rightarrow\sin\text{x}=\frac{1}{2}=\sin\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=\frac{\pi}{6}$
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