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