Question 1:
Prove that \(3\sin^{-1}x=\sin^{-1}(3x-4x^3)\), \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\).
Answer 1:
Let \[ \sin^{-1}x=\theta \] then \[ x=\sin\theta. \]
We have,
\[ \begin{aligned} \text{RHS} &=\sin^{-1}(3x-4x^3)\\ &=\sin^{-1}(3\sin\theta-4\sin^3\theta)\\ &=\sin^{-1}(\sin3\theta)\\ &=3\theta\\ &=3\sin^{-1}x\\ &=\text{LHS}. \end{aligned} \]
Hence, \[ \boxed{3\sin^{-1}x=\sin^{-1}(3x-4x^3)} \] for \[ x\in\left[-\frac12,\frac12\right]. \]
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