class 12 maths inverse trigonometric functions

The value of ${\sin ^{ - 1}}\left( {\sin \frac{{3\pi }}{5}} \right)$ is …………….

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📘 Inverse Trigonometric Functions NCERT,Ex.2.3,Q.39,Page.40 FillBlank

The value of ${\sin ^{ - 1}}\left( {\sin \frac{{3\pi }}{5}} \right)$ is …………….

Official Solution

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$- \frac{\pi }{2} \le {\sin ^{ - 1}}x \le \frac{\pi }{2}$

$therefore,$ ${\sin ^{ - 1}}\left( {\sin \frac{{3\pi }}{5}} \right) = {\sin ^{ - 1}}\sin \left( {\pi - \frac{{2\pi }}{5}} \right) = {\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{5}} \right) = \frac{{2\pi }}{5}$

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