Question:
Prove that
$\sin ^{-1}\left(3 x-4 x^{3}\right)=3 \sin ^{-1} x,|x| \leq \frac{1}{2}$
Solution:
To Prove: $\sin ^{-1}\left(3 x-4 x^{3}\right)=3 \sin ^{-1} x$
Formula Used: $\sin 3 A=3 \sin A-4 \sin ^{3} A$
Proof:
LHS $=\sin ^{-1}\left(3 x-4 x^{3}\right) \ldots$ (1)
Let $x=\sin A \ldots(2)$
Substituting (2) in (1),
$\mathrm{LHS}=\sin ^{-1}\left(3 \sin \mathrm{A}-4 \sin ^{3} \mathrm{~A}\right)$
$=\sin ^{-1}(\sin 3 \mathrm{~A})$
$=3 \mathrm{~A}$
From $(2), A=\sin ^{-1} x$
$3 A=3 \sin ^{-1} x$
$=$ RHS
Therefore, LHS = RHS
Hence proved.
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