Finding the domain of a nested inverse trigonometric composition

Expert intro: This problem is a domain-checking exercise where each inverse trig layer imposes its own admissible interval.

Detailed walkthrough

Step 1: Start from the outer function

The function is

$\cos^{-1}\left(\frac{2\sin^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi}\right).$

For the outer $\cos^{-1}(u)$ to be defined, its input must satisfy

$-1\le u\le 1.$

So we need

$-1\le \frac{2\sin^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi}\le 1.$

Multiplying by $\pi/2$ gives

$-\frac{\pi}{2}\le \sin^{-1}\left(\frac{1}{4x^2-1}\right)\le \frac{\pi}{2}.$

But $\sin^{-1}(t)$ always lies in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ when it is defined, so the real restriction comes from the inner argument of $\sin^{-1}$.

Step 2: Apply the domain condition for arcsine

For $\sin^{-1}(z)$ to exist, we need

$-1\le z\le 1.$

Here,

$z=\frac{1}{4x^2-1}.$

So we require

$\left|\frac{1}{4x^2-1}\right|\le 1,$ which is equivalent to

$|4x^2-1|\ge 1.$

Now solve this inequality:

Case 1: $4x^2-1\ge 1$

$4x^2\ge 2 \Rightarrow x^2\ge \frac12 \Rightarrow |x|\ge \frac{1}{\sqrt2}.$

Case 2: $4x^2-1\le -1$

$4x^2\le 0 \Rightarrow x=0.$

We must also exclude points where $4x^2-1=0$, i.e. $x=\pm\frac12$, since the denominator would vanish.

Combining everything, the domain is

$(-\infty,-\tfrac{1}{\sqrt2}]\cup\{0\}\cup[\tfrac{1}{\sqrt2},\infty).$

Why this works

This type of problem is solved by moving from the outside inward. First check the admissible output of the outer inverse cosine, then enforce the input condition for arcsine, and finally exclude denominator zeros. The final set is the intersection of all these constraints.

A useful habit is to simplify absolute-value inequalities before splitting into cases. That keeps the domain work clean and reduces algebraic mistakes.

💡 Pitfall guide

Students often forget that the denominator $4x^2-1$ cannot be zero even if the inequality seems to allow nearby values. Another common mistake is stopping after writing $-1\le \frac{1}{4x^2-1}\le 1$ without converting it into an absolute-value condition. That can lead to missed intervals or incorrect extra points. Also, the outer $\cos^{-1}$ does not create a new restriction here beyond requiring its argument to stay in $[-1,1]$, so the real challenge is the nested arcsine and the denominator exclusion.

🔄 Real-world variant

If the inner expression were $\sin^{-1}\!\left(\frac{2}{4x^2-1}\right)$, the domain would change because the inequality would become $\left|\frac{2}{4x^2-1}\right|\le 1$, or $|4x^2-1|\ge 2$. That would produce a different threshold for $|x|$ and could eliminate the isolated point $x=0$. If the outer function were $\sin^{-1}$ instead of $\cos^{-1}$, the outer-step check would still require the same input interval $[-1,1]$, so the main work would remain unchanged.

🔍 Related terms

inverse sine domain, nested function composition, absolute value inequality