Q2:

Prove the following: $3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$, $x \in [\frac{1}{2}, 1]$

Step 1: Establishing the Substitution

To prove the identity $3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$, we begin by substituting $x$ with a trigonometric function that simplifies the expression on the right-hand side. Let:

$x = \cos \theta$

Consequently, by the definition of the inverse cosine function, we have:

$\theta = \cos^{-1} x$ [Domain constraint: $x \in [\frac{1}{2}, 1]$ implies $\theta \in [0, \frac{\pi}{3}]$]

Step 2: Utilizing the Triple Angle Identity

Substitute $x = \cos \theta$ into the right-hand side (RHS) of the equation:

RHS $= \cos^{-1} (4\cos^3 \theta - 3\cos \theta)$

Recall the standard trigonometric identity for the cosine of a triple angle [Per De Moivre’s Theorem/Trigonometric Identity]:

$\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$

Substituting this identity into our expression, we obtain:

RHS $= \cos^{-1} (\cos 3\theta)$

Step 3: Verification of the Principal Value Branch

The expression $\cos^{-1} (\cos 3\theta)$ simplifies to $3\theta$ if and only if $3\theta$ lies within the principal value branch of the inverse cosine function, which is $[0, \pi]$.

• Given $x \in [\frac{1}{2}, 1]$
• $\cos \theta \in [\frac{1}{2}, 1]$
• $\theta \in [0, \frac{\pi}{3}]$
• Multiplying by 3, we get $3\theta \in [0, \pi]$

Since $3\theta \in [0, \pi]$, the property $\cos^{-1}(\cos \alpha) = \alpha$ holds for $\alpha = 3\theta$ [By the property of inverse trigonometric functions].

Step 4: Final Derivation

RHS $= 3\theta$

Substitute $\theta = \cos^{-1} x$ back into the equation:

RHS $= 3\cos^{-1} x$

Therefore, LHS = RHS.

Final Solution: $3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$ is proven for $x \in [\frac{1}{2}, 1]$.