Question:
Prove that:
$\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$
Solution:
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.
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