prove that

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.