Evaluate \(sin^{-1}(sin10)\)

Solution :

We know that \(sin^{-1}(sinx)\) = x, if \(-\pi\over 2\) \(\le\) x \(\le\) \(\pi\over 2\)

Here, x = 10 radians which does not lie between -\(\pi\over 2\) and \(\pi\over 2\)

But, \(3\pi\) – x i.e. \(3\pi\) – 10 lie between -\(\pi\over 2\) and \(\pi\over 2\)

Also, sin(\(3\pi\) – 10) = sin 10

\(\therefore\)  \(sin^{-1}(sin10)\) = \(sin^{-1}(sin(3\pi – 10)\) = (\(3\pi\) – 10)


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