Inverse Trignometric Function Examples

Example 1 : Find the value of \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\).

Solution : \(sin^{-1}({-\sqrt{3}\over 2})\) = – \(sin^{-1}({\sqrt{3}\over 2})\) = \(-\pi\over 3\)

\(cos^{-1}(cos({7\pi\over 6}))\) = \(cos^{-1}(cos({2\pi – {5\pi\over 6}}))\) = \(cos^{-1}(cos({5\pi\over 6}))\) = \(5\pi\over 6\)

Hence \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\) = \(-\pi\over 3\) + \(5\pi\over 6\) = \(\pi\over 2\)

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Example 2 : Prove that : \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(sin^{-1}{56\over 65}\)

Solution : We have, L.H.S. = \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(tan^{-1}{5\over 12}\) + \(tan^{-1}{3\over 4}\)

\(\because\)   [ \(cos^{-1}{12\over 13}\) = \(tan^{-1}{5\over 12}\) & \(sin^{-1}{3\over 5}\) = \(tan^{-1}{3\over 4}\) ]

L.H.S. = \(tan^{-1}({{{5\over 12} + {3\over 4}}\over {1 – {5\over 12}.{3\over 4}}})\) = \(tan^{-1}{56\over 33}\)

R.H.S. = \(sin^{-1}{56\over 65}\) = \(tan^{-1}{56\over 33}\)

L.H.S = R.H.S.   Hence Proved.

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Example 3 : 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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Example 4 : Prove that : \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over 3}\) + \(tan^{-1}{63\over 16}\) = \(\pi\)

Solution : We have, A = \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over 3}\) + \(tan^{-1}{63\over 16}\)

A = \(tan^{-1}{12\over 5}\) + \(tan^{-1}{3\over 4}\) + \(tan^{-1}{63\over 16}\)

\(\implies\) A = \(\pi\) + \(tan^{-1}({{{12\over 5} + {3\over 4}}\over {1 – {12\over 5} \times {3\over 4}}})\) + \(tan^{-1}{63\over 16}\)

\(\implies\) A = \(\pi\) + \(tan^{-1}{63\over (-16)}\) + \(tan^{-1}{63\over 16}\)

= \(\pi\) – \(tan^{-1}{63\over 16}\) + \(tan^{-1}{63\over 16}\)

= \(\pi\)

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Example 5 : Solve the equation : 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\)

Solution : Here, 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\)

cos(2\(tan^{-1}({2x+1})\)) = x       { We Know cos2x = \({1-tan^2x\over {1+tan^2x}}\)}

\(\therefore\)     \({{1-{(2x+1)}^2}\over {1-{(2x+1)}^2}}\) = x   \(\implies\)   (1 – 2x – 1)(1 + 2x + 1) = x(\(4x^2 + 4x + 2\))

\(\implies\)   -2x.2(x + 1) = 2x(\(2x^2 + 2x + 1\))   \(\implies\)   2x(\(2x^2 + 2x + 1 + 2x + 2\)) = 0

\(\implies\)   x = 0   or   \(2x^2 + 4x + 3\) = 0   { No Solution }

Verify       x = 0

\(2tan^{-1}(1)\) = \(cos^{-1}(1)\)   \(\implies\)   \(\pi\over 2\) = \(\pi\over 2\)

\(\therefore\)     x = 0 is only the solution.

Practice these given inverse trignometric function examples to test your knowledge on concepts of inverse trigonometric function.