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INVERSE TRIGNOMETRIC FUNCTIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINVERSE TRIGNOMETRIC FUNCTIONS2 Marks
Question
Prove that:
$3\sin^{-1}\text{x}=\sin^{-1}(3\text{x}-4\text{x}^3),\text{x}\in\Big[-\frac{1}{2},\frac{1}{2}\Big]$
✓

Answer

$3 \sin^{-1}\text{x}=\sin^{-1}(3\text{x}-4\text{x}^3)$
$\text{Let}\ \sin^{-1}\text{x}=\theta$
$\text{x}=\sin\theta$
$\therefore\sin3\theta=3\sin\theta-4\sin^3\theta$
$\sin3\theta=3\text{x}-4\text{x}^3$
Case I$\text{ When}\frac{-1}{2}\leq\text{x}\leq\frac{1}{2}$
$\frac{-1}{2}\leq\sin\theta\leq\frac{1}{2}$
$\frac{-\pi}{6}\leq\theta\leq\frac{\pi}{6}$
$\frac{-\pi}{2}\leq3\theta\leq\frac{\pi}{2}$
$\text{Also}\ \frac{-1}{2}\leq\text{x}\leq\frac{1}{2}\Rightarrow-1\leq3\text{x}-4\text{x}^3\leq1$
$\sin3\theta=3\text{x}-4\text{x}^3$
$3\theta=\sin^{-1}(3\text{x}-4\text{x}^3)$
$3\sin^{-1}\text{x}=\sin^{-1}(3\text{x}-4\text{x}^3)$

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