Prove that:

To Prove: $\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$

Formula Used: $\cos 2 \mathrm{~A}=2 \cos ^{2} \mathrm{~A}-1$

Proof:

$\mathrm{LHS}=\cos ^{-1}\left(2 \mathrm{x}^{2}-1\right) \ldots(1)$

$\operatorname{Let} x=\cos A \ldots(2)$

Substituting (2) in (1),

$\mathrm{LHS}=\cos ^{-1}\left(2 \cos ^{2} \mathrm{~A}-1\right)$

$=\cos ^{-1}(\cos 2 \mathrm{~A})$

$=2 \mathrm{~A}$

From $(2), A=\cos ^{-1} x$

$=\mathrm{RHS}$

Therefore, LHS = RHS

Hence proved.