Exercise 2.2 · Q6: Simplest form of $\tan^{-1}\frac{x}{\sqrt{a^2-x^2}}$

Basics you need

Solution

1. What the condition $|x|<a$ says

   Since $|x|\ge 0$ always, the inequality $|x|<a$ can only hold if $a>0$⁠. So throughout, $a$ is positive. Dividing $|x|<a$ by the positive number $a$ gives

   $$-1 < \frac{x}{a} < 1.$$

   Also $a^2-x^2>0$⁠, because $x^2=|x|^2<a^2$⁠. So the square root is a positive real number and the fraction is defined.

   The book does not say "$a>0$" separately, but it is hidden in $|x|<a$⁠. We need it twice: to divide the inequality by $a$ without flipping it, and later to write $\sqrt{a^2\cos^2\theta}=a\cos\theta$ (a negative $a$ would bring in a minus sign).
   Dividing an inequality by a positive number keeps its direction; dividing by a negative number reverses it.

2. Substitute $x=a\sin\theta$ and locate $\theta$

   Let

   $$\theta = \sin^{-1}\frac{x}{a}, \qquad\text{so}\qquad x = a\sin\theta.$$

   This is legitimate because $\frac{x}{a}$ lies in $(-1,1)$⁠, inside the domain $[-1,1]$ of $\sin^{-1}$⁠. Since $\sin^{-1}$ takes values in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠, and $\frac{x}{a}$ is never $\pm1$⁠, the endpoints are excluded:

   $$-\frac{\pi}{2} < \theta < \frac{\pi}{2}.$$

   (The endpoints would need $\sin\theta=\pm1$⁠, that is $x=\pm a$⁠, which $|x|<a$ rules out.)

   Defining $\theta$ as $\sin^{-1}\frac{x}{a}$⁠, rather than just saying "put $x=a\sin\theta$", fixes which angle we mean. Infinitely many angles have sine $\frac{x}{a}$⁠; the principal value is the one in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠, and knowing that range is what makes Steps 3 and 4 work.
   $\sin^{-1}:[-1,1]\to\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (table of principal value branches, Section 2.2). Basics: domain and range of sin and cos

3. Simplify the square root

   $$\begin{aligned} a^2-x^2 &= a^2 - a^2\sin^2\theta \\ &= a^2\left(1-\sin^2\theta\right) \\ &= a^2\cos^2\theta. \end{aligned}$$

   So $\sqrt{a^2-x^2} = \sqrt{a^2\cos^2\theta}$⁠, which is $|a|\,|\cos\theta|$⁠. Now $a>0$ (Step 1), and $\cos\theta>0$ because $\theta$ is strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ (Step 2), where cosine is positive. Hence

   $$\sqrt{a^2-x^2} = a\cos\theta \quad (>0).$$
   This is the step where careless solutions go wrong. $\sqrt{a^2\cos^2\theta}$ is $|a\cos\theta|$⁠; dropping the absolute value is only allowed after checking that $a\cos\theta$ is positive. Both facts we need came from the given condition.
   On $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ the point at angle $\theta$ on the unit circle is to the right of the vertical axis, so its $x$-coordinate $\cos\theta$ is positive. Basics: sin and cos on the unit circle

4. The fraction is $\tan\theta$⁠, and $\tan^{-1}$ undoes it

   $$\frac{x}{\sqrt{a^2-x^2}} = \frac{a\sin\theta}{a\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta.$$

   (Cancelling $a$ is fine because $a\ne 0$⁠.) Therefore

   $$\tan^{-1}\frac{x}{\sqrt{a^2-x^2}} = \tan^{-1}(\tan\theta) = \theta,$$

   and the last equality holds because $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠, which is exactly the principal range of $\tan^{-1}$⁠.

   $\tan^{-1}(\tan\theta)$ means "the angle in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ whose tangent is $\tan\theta$". If $\theta$ itself is in that interval, it is that angle, because tangent is one-one there. If $\theta$ were outside, the answer would be $\theta$ shifted by a multiple of $\pi$ instead.
   $\tan^{-1}(\tan\theta)=\theta$ for $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠: the tangent version of the property in Section 2.3. Lesson: the folding rules

5. Back to $x$

   By definition (Step 2), $\theta = \sin^{-1}\frac{x}{a}$⁠. So

   $$\tan^{-1}\frac{x}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}, \qquad |x|<a.$$
   "Simplest form" means a single inverse function of a simple argument, with no square roots left. $\sin^{-1}\frac{x}{a}$ is that. The triangle shows why it is natural: the angle whose tangent is "opposite over adjacent" is the same angle whose sine is "opposite over hypotenuse".
   The triangle picture is a quick way to guess the answer; the substitution in Steps 2 to 4 is the proof, and it also covers $x\le 0$⁠, which no triangle can show. Lesson: the triangle trick

Every fact used, and where it comes from