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Prove that:

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.

 

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