Miscellaneous · Q9: A $\cot^{-1}$ of square roots equals $\frac x2$

Basics you need

Solution

1. Locate the half angle

   From $0 < x < \frac\pi4$⁠, dividing by 2,

   $$0 < \frac x2 < \frac\pi8.$$

   This is inside $\left(0,\frac\pi4\right)$⁠, so by Basic 3

   $$\cos\tfrac x2 > \sin\tfrac x2 > 0.$$
   The whole proof turns on square roots, and a square root depends on signs. Fixing the interval for $\frac x2$ first gives us every sign we will need.
   On $\left(0,\frac\pi4\right)$⁠, $\cos$ falls from 1 to $\frac{1}{\sqrt2}$ and $\sin$ rises from 0 to $\frac1{\sqrt2}$⁠, so $\cos > \sin > 0$ there (Basic 3).

2. Write 1 + sin x and 1 − sin x as squares

   To keep the lines short, write $c = \cos\frac x2$ and $s = \sin\frac x2$⁠. Then $1 = c^2 + s^2$ and $\sin x = 2sc$⁠, so

   $$\begin{aligned} 1+\sin x &= c^2 + s^2 + 2sc = (c+s)^2, \\ 1-\sin x &= c^2 + s^2 - 2sc = (c-s)^2. \end{aligned}$$
   The radicals are the obstacle. Writing what is under each root as a perfect square, using $(a\pm b)^2 = a^2 + b^2 \pm 2ab$⁠, is the standard way to remove them. The same factorisation appears in the textbook's Example 4.
   Pythagorean identity and the double-angle formula, both with angle $\frac x2$ (Basic 2).

3. Take the square roots, keeping the signs right

   By Step 1, $c > s > 0$⁠, so both $c + s$ and $c - s$ are positive. Hence

   $$\sqrt{1+\sin x} = c + s, \qquad \sqrt{1-\sin x} = c - s.$$
   $\sqrt{a^2} = |a|$⁠. We may drop the absolute value only because Step 1 showed each bracket is positive. If $\frac x2$ were above $\frac\pi4$⁠, we would have $c < s$⁠, and $\sqrt{1-\sin x}$ would be $s - c$ instead.
   $\sqrt{a^2} = a$ only for $a \ge 0$ (Basic 1).

4. Simplify the fraction

   Add and subtract the two roots:

   $$\begin{aligned} \text{top} &= (c+s) + (c-s) = 2c, \\ \text{bottom} &= (c+s) - (c-s) = 2s. \end{aligned}$$

   The bottom $2s$ is positive (Step 1), so the fraction is defined, and

   $$\frac{\text{top}}{\text{bottom}} = \frac{2c}{2s} = \frac{\cos\frac x2}{\sin\frac x2} = \cot\frac x2.$$
   In the sum the $s$ terms cancel; in the difference the $c$ terms cancel. The ratio of what is left is cosine over sine, which is the cotangent.
   $\cot\alpha = \frac{\cos\alpha}{\sin\alpha}$ when $\sin\alpha\ne 0$⁠. Basics: tan, cot, sec and cosec

5. Apply cot⁻¹ using its principal range

   The left side of the question is now $\cot^{-1}\left(\cot\frac x2\right)$⁠. Since $\frac x2\in\left(0,\frac\pi8\right)\subset(0,\pi)$⁠,

   $$\cot^{-1}\left(\cot\tfrac x2\right) = \frac x2.$$
   $\cot^{-1}$ returns the unique angle in $(0,\pi)$ with the given cotangent. Since $\frac x2$ is such an angle, it is the one returned.
   $\cot^{-1}(\cot\alpha) = \alpha$ for $\alpha\in(0,\pi)$ (Basic 4). Lesson: the folding rules

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