Miscellaneous · Q8: $\tan^{-1}\sqrt x=\frac12\cos^{-1}\frac{1-x}{1+x}$

Basics you need

Solution

1. Both sides make sense for every x in [0, 1]

   Left side: $x \ge 0$⁠, so $\sqrt{x}$ is a real number, and $\tan^{-1}$ accepts every real number.

   Right side: $1 + x \ge 1 > 0$⁠, so the fraction is defined. Also $1 - x \le 1 + x$ (because $x \ge 0$⁠) and $1 - x \ge 0$ (because $x \le 1$⁠), so

   $$0 \le \frac{1-x}{1+x} \le 1,$$

   which is inside $[-1, 1]$⁠, the domain of $\cos^{-1}$⁠.

   An identity is a claim about every $x$ in the stated set. Before proving the two sides equal, we confirm that both sides are real numbers for each such $x$⁠; otherwise the statement would be meaningless at some points.
   $\cos^{-1}$ is defined only on $[-1, 1]$ (Basic 1).

2. Name the left side θ and locate θ

   Let $\theta = \tan^{-1}\sqrt{x}$⁠. By the definition of $\tan^{-1}$⁠,

   $$\tan\theta = \sqrt{x}, \qquad \theta \in \left(-\tfrac\pi2, \tfrac\pi2\right).$$

   Since $0 \le \sqrt{x} \le 1$ and $\tan$ is increasing on $\left(-\frac\pi2,\frac\pi2\right)$ with $\tan 0 = 0$⁠, $\tan\frac\pi4 = 1$⁠, we get

   $$0 \le \theta \le \frac\pi4.$$

   Squaring $\tan\theta = \sqrt{x}$ gives $x = \tan^2\theta$⁠.

   We do not "substitute" a new letter blindly. Defining $\theta$ as the left side guarantees that $\theta$ exists and is a principal value, and it tells us exactly which interval $\theta$ is in. That interval is what Step 4 needs.
   If $y = \tan^{-1}a$⁠, then $\tan y = a$ and $y\in\left(-\frac\pi2,\frac\pi2\right)$ (Basic 1, Basic 4). Lesson: substitutions and simplest form

3. Turn the fraction into cos 2θ

   Replace $x$ by $\tan^2\theta$⁠:

   $$\frac{1-x}{1+x} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta.$$

   So the right side of the question is $\frac12\cos^{-1}(\cos 2\theta)$⁠.

   The pattern $\frac{1-t^2}{1+t^2}$ with $t = \tan\theta$ is exactly the formula for $\cos 2\theta$ (Basic 3). Recognising it is the reason for choosing $\sqrt{x} = \tan\theta$⁠: with this choice, $x$ itself is $\tan^2\theta$⁠.
   $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$⁠, valid whenever $\cos\theta \ne 0$⁠, which is true here because $\theta\in\left[0,\frac\pi4\right]$ (Basic 3).

4. Check 2θ is in the principal range of cos⁻¹

   From Step 2, $0 \le \theta \le \frac\pi4$⁠. Doubling,

   $$0 \le 2\theta \le \frac\pi2, \quad\text{so}\quad 2\theta \in [0, \pi].$$

   Therefore $\cos^{-1}(\cos 2\theta) = 2\theta$⁠.

   $\cos^{-1}$ always returns an angle in $[0,\pi]$⁠. If $2\theta$ were outside that interval, $\cos^{-1}(\cos 2\theta)$ would be some other angle with the same cosine, and the proof would break. This one-line check is what makes the step legal.
   $\cos^{-1}(\cos\alpha) = \alpha$ for $\alpha\in[0,\pi]$ (Basic 2). Lesson: the folding rules

5. Halve and conclude

   $$\begin{aligned} \frac12\cos^{-1}\frac{1-x}{1+x} &= \frac12\cos^{-1}(\cos 2\theta) \\ &= \frac12 \cdot 2\theta = \theta \\ &= \tan^{-1}\sqrt{x}. \end{aligned}$$

   This holds for every $x \in [0, 1]$⁠, which proves the identity.

   The right side has become $\theta$⁠, and $\theta$ was defined as the left side. Because Step 2 started from an arbitrary $x$ in $[0,1]$⁠, the argument covers every such $x$⁠.
   Remark: the same argument works for every $x \ge 0$⁠, because then $\theta\in\left[0,\frac\pi2\right)$ and $2\theta \in [0,\pi)$⁠. The question only asks for $[0,1]$⁠.

Every fact used, and where it comes from