Miscellaneous · Q10: A $\tan^{-1}$ of square roots equals $\frac\pi4-\frac12\cos^{-1}x$

Basics you need

Solution

1. Choose θ as a principal value and find its interval

   Following the hint, let

   $$2\theta = \cos^{-1}x, \quad\text{so}\quad x = \cos 2\theta,\ \ \theta = \tfrac12\cos^{-1}x.$$

   $\cos^{-1}$ is decreasing, with $\cos^{-1}1 = 0$ and $\cos^{-1}\left(-\frac1{\sqrt2}\right) = \frac{3\pi}4$⁠. So for $-\frac1{\sqrt2} \le x \le 1$⁠,

   $$0 \le 2\theta \le \frac{3\pi}{4}, \qquad 0 \le \theta \le \frac{3\pi}{8}.$$

   In particular $\theta\in\left[0,\frac\pi2\right)$⁠, so $\cos\theta > 0$ and $\sin\theta \ge 0$⁠.

   The hint says "put $x = \cos2\theta$", but many angles have cosine $x$⁠. Choosing $2\theta = \cos^{-1}x$ picks one definite angle, makes $\frac12\cos^{-1}x$ equal to $\theta$ on the right side, and pins down an interval for $\theta$⁠. Every sign below comes from that interval.
   $\cos^{-1}$ has range $[0,\pi]$ (Basic 1) and is decreasing on $[-1,1]$⁠, because $\cos$ is decreasing on $[0,\pi]$⁠; the end value comes from Basic 5. Lesson: substitutions and simplest form

2. Rewrite 1 + x and 1 − x

   $$\begin{aligned} 1 + x &= 1 + \cos 2\theta = 2\cos^2\theta, \\ 1 - x &= 1 - \cos 2\theta = 2\sin^2\theta. \end{aligned}$$
   These are the two half-angle forms (Basic 2). They turn the expressions under the roots into perfect squares, which is the whole point of the substitution.
   $\cos 2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$ (Basic 2).

3. Take the square roots with the correct signs

   Since $\cos\theta > 0$ and $\sin\theta \ge 0$ (Step 1),

   $$\sqrt{1+x} = \sqrt2\cos\theta, \qquad \sqrt{1-x} = \sqrt2\sin\theta.$$
   $\sqrt{2\cos^2\theta} = \sqrt2\,|\cos\theta|$⁠, and similarly for $\sin$⁠. We can drop the absolute values only because Step 1 placed $\theta$ in the first quadrant (or at 0), where both are non-negative.
   $\sqrt{a^2} = |a|$ (Basic 3).

4. Turn the fraction into tan(π/4 − θ)

   Substitute and cancel the common factor $\sqrt2$⁠:

   $$\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}.$$

   The denominator is not zero: $\cos\theta > 0$ and $\sin\theta \ge 0$⁠, so $\cos\theta + \sin\theta > 0$⁠. Divide top and bottom by $\cos\theta > 0$⁠:

   $$\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \frac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\frac\pi4 - \theta\right).$$
   Dividing by $\cos\theta$ is allowed because it is positive. The result $\frac{1-\tan\theta}{1+\tan\theta}$ is the tan-of-a-difference formula with $\tan\frac\pi4 = 1$ (Basic 4).
   $\tan\left(\frac\pi4-\theta\right) = \frac{1-\tan\theta}{1+\tan\theta}$ (Basic 4).

5. Check the principal range and conclude

   From $0 \le \theta \le \frac{3\pi}{8}$⁠,

   $$-\frac\pi8 \le \frac\pi4 - \theta \le \frac\pi4,$$

   which lies inside $\left(-\frac\pi2,\frac\pi2\right)$⁠. Therefore

   $$\begin{aligned} \text{LHS} &= \tan^{-1}\left(\tan\left(\tfrac\pi4-\theta\right)\right) \\ &= \frac\pi4 - \theta \\ &= \frac\pi4 - \frac12\cos^{-1}x. \end{aligned}$$
   $\tan^{-1}(\tan\alpha) = \alpha$ only when $\alpha \in \left(-\frac\pi2,\frac\pi2\right)$⁠. We have just shown $\frac\pi4-\theta$ is in that interval for every allowed $x$⁠, so the last step is legal, and the proof covers every $x$ in $\left[-\frac1{\sqrt2}, 1\right]$⁠.
   $\tan^{-1}(\tan\alpha) = \alpha$ for $\alpha\in\left(-\frac\pi2,\frac\pi2\right)$ (Basic 1). Lesson: the folding rules

Every fact used, and where it comes from