Prove that:3sin−1⁡x=sin−1⁡3x−4x3,x∈−12,12

Let,RHS=sin−1⁡3x−4x3Put x=sin⁡θ in RHS, it followsRHS =sin−1⁡3sin⁡θ−4sin3⁡θ∴RHS =sin−1⁡(sin⁡3θ)Now,⇒−12≤x≤12⇒−12≤sin⁡θ≤12⇒−π6≤θ≤π6−π2≤3θ≤π2Hence, sin−1⁡(sin⁡3θ)=3θ as −π2≤3θ≤π2⇒ R.H.S. =3θ=3sin−1⁡x∵x=sin⁡θ⇒θ=sin−1⁡xTherefore, LHS= RHSTherefore, 3sin−1⁡x=sin−1⁡3x−4x3,x∈−12,12 has proven.

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Prove that:
$3 \sin^{- 1} x = \sin^{- 1} (3 x - 4 x^{3}), x \in [- \frac{1}{2}, \frac{1}{2}]$

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Detailed Solution

Let,
$RHS = \sin^{- 1} (3 x - 4 x^{3})$
Put $x = \sin θ$ in RHS, it follows
RHS $= \sin^{- 1} (3 \sin θ - 4 \sin^{3} θ)$
$∴ RHS$ $= \sin^{- 1} (\sin 3 θ)$
Now,
$\begin{array}{l} \Rightarrow - \frac{1}{2} \leq x \leq \frac{1}{2} \\ \Rightarrow - \frac{1}{2} \leq \sin θ \leq \frac{1}{2} \\ \Rightarrow - \frac{π}{6} \leq θ \leq \frac{π}{6} \\ - \frac{π}{2} \leq 3 θ \leq \frac{π}{2} \end{array}$
Hence, $\sin^{- 1} (\sin 3 θ) = 3 θ (as \frac{- π}{2} \leq 3 θ \leq \frac{π}{2})$
$\begin{array}{l} \Rightarrow R.H.S. = 3 θ \\ = 3 \sin^{- 1} x \\ [∵ x = \sin θ \Rightarrow θ = \sin^{- 1} x] \end{array}$
Therefore, LHS= RHS
Therefore, $3 \sin^{- 1} x = \sin^{- 1} (3 x - 4 x^{3}), x \in [- \frac{1}{2}, \frac{1}{2}]$ has proven.