Inverse Trigonometric Functions — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 2.2, Q. 1 , Page 47
Question

$3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3}),\;x \in \left[ { - \frac{1}{2},\;\frac{1}{2}} \right]$

Step-by-step Solution

Put ${\sin ^{ - 1}}x = \theta .\;\;Then,\;\,x = \sin \theta$
Now, $\sin 3\theta = (3\sin \theta - 4{\sin ^3}\theta ) = (3x - 4{x^3})$

$\Rightarrow$ $3\theta = {\sin ^{ - 1}}(3x - 4{x^3})$

$\Rightarrow$ $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3})$
Hence, $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3})$

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Inverse Trigonometric Functions. Curated by Sachin Sharma. Free for all students.