Q.$3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$, $x \in \left[\frac{1}{2}, 1\right]$ Write the following functions in the simplest form:

The key idea is the triple angle identity for cosine: $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.

Let $x = \cos\theta$, where $\theta \in [0, \pi]$ so that $\cos^{-1}x = \theta$ is well-defined. Since $x \in \left[\frac12, 1\right]$, we have $\theta \in \left[0, \frac{\pi}{3}\right]$.

Now compute:

and

Because $\theta \in \left[0, \frac{\pi}{3}\right]$, we have $3\theta \in [0, \pi]$, which lies in the principal range of $\cos^{-1}$. Hence $\cos^{-1}(\cos 3\theta) = 3\theta$.

Therefore, the two sides are equal for all $x$ in the given interval.

The problem asks us to simplify the expression $3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$ for $x$ in the interval $\left[\frac12, 1\right]$. At first glance, the right-hand side looks like a messy cubic in $x$, but the form $4x^3 - 3x$ is a dead giveaway — it matches the triple-angle formula for cosine.

Why this works: The inverse cosine function $\cos^{-1}$ returns an angle whose cosine is the given number. If we can rewrite $4x^3 - 3x$ as $\cos(3\theta)$ where $\theta = \cos^{-1} x$, then the right-hand side becomes $\cos^{-1}(\cos 3\theta)$. The simplification then depends on whether $3\theta$ falls inside the principal branch of $\cos^{-1}$, which is $[0, \pi]$. The given domain $x \in [\frac12, 1]$ ensures exactly that.

Let’s walk through it step by step.

1. Set up the substitution.

   Let $\theta = \cos^{-1} x$. Then by definition, $x = \cos \theta$, and $\theta \in [0, \pi]$.

   Since $x \in [\frac12, 1]$, we have $\cos \theta \in [\frac12, 1]$, which means $\theta \in [0, \frac{\pi}{3}]$ (because cosine decreases from $1$ to $\frac12$ as $\theta$ goes from $0$ to $\frac{\pi}{3}$).

2. Apply the triple-angle identity.

   The standard identity is:

Substituting $x = \cos\theta$ gives:

3. Rewrite the right-hand side. The original equation becomes:

Now, $\cos^{-1}(\cos \alpha)$ simplifies to $\alpha$ only if $\alpha \in [0, \pi]$. Otherwise, we need to adjust using the periodic and symmetric properties of cosine.

4. Check the range of $3\theta$. From step 1, $\theta \in [0, \frac{\pi}{3}]$. Multiplying by 3:

This is exactly the principal range of $\cos^{-1}$. So for every $x$ in $[\frac12, 1]$, the angle $3\theta$ lies in $[0, \pi]$, and therefore:

5. Substitute back. Since $\theta = \cos^{-1} x$, we have:

which is precisely the given equation. So the expression is already in its simplest form — it’s an identity that holds for $x \in [\frac12, 1]$.

Method: Simplifying an inverse-cosine multiple-angle expression

Apply this to expressions of the form $\cos^{-1}(4x^3 - 3x)$.

Steps

Step 1: Substitute $x = \cos\theta$

Put $\theta = \cos^{-1}x$, so $\theta \in [0,\pi]$. The triple-angle identity gives $4x^3 - 3x = 4\cos^3\theta - 3\cos\theta = \cos 3\theta$.

Step 2: Reduce the inverse cosine

The expression becomes $\cos^{-1}(\cos 3\theta)$, which equals $3\theta$ ONLY when $3\theta \in [0,\pi]$.

Step 3: Check the domain forces the branch

For $x \in \left[\frac{1}{2}, 1\right]$, $\theta \in \left[0, \frac{\pi}{3}\right]$, so $3\theta \in [0,\pi]$ and the expression simplifies cleanly to $3\cos^{-1}x$.

Common Mistakes

Mistake 1: Writing $\cos^{-1}(\cos 3\theta) = 3\theta$ without checking $3\theta \in [0,\pi]$

Why it's wrong: the identity $\cos^{-1}(\cos\alpha) = \alpha$ holds only on $[0,\pi]$; elsewhere you must use $2\pi - \alpha$ or a shift. Correct approach: confirm $x \in \left[\frac{1}{2}, 1\right]$ keeps $3\theta$ in $[0,\pi]$.

Mistake 2: Misremembering the cosine triple-angle identity

Why it's wrong: $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$; swapping it with the sine form $3\sin\theta - 4\sin^3\theta$ ruins the match with $4x^3 - 3x$. Correct approach: pair $4x^3 - 3x$ with the cosine identity.