Exercise 2.2 · Q7: Simplest form of $\tan^{-1}\frac{3a^2x-x^3}{a^3-3ax^2}$

Basics you need

Solution

1. Derive the triple-angle formula for tangent

   Write $t=\tan\theta$⁠. First, $\tan2\theta = \dfrac{2t}{1-t^2}$ (the addition formula with $A=B=\theta$⁠). Now treat $3\theta$ as $2\theta+\theta$⁠:

   $$\begin{aligned} \tan3\theta &= \frac{\tan2\theta+\tan\theta}{1-\tan2\theta\tan\theta} \\[1mm] &= \frac{\dfrac{2t}{1-t^2}+t}{1-\dfrac{2t}{1-t^2}\cdot t}. \end{aligned}$$

   Multiply the top and the bottom by $1-t^2$⁠:

   $$\begin{aligned} \tan3\theta &= \frac{2t+t(1-t^2)}{(1-t^2)-2t^2} \\[1mm] &= \frac{3t-t^3}{1-3t^2}. \end{aligned}$$
   The fraction in the question has exactly the shape $\frac{3t-t^3}{1-3t^2}$ once we divide by $a^3$ (Step 3), so this is the formula we need. Deriving it from the addition formula, rather than quoting it, shows where it comes from. It is valid whenever $\tan\theta$⁠, $\tan2\theta$ and $\tan3\theta$ are all defined, which Step 3 confirms for our $\theta$⁠.
   $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ (Class XI, Chapter 3, Trigonometric Functions).

2. Substitute $x=a\tan\theta$ and pin down $\theta$

   Let

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

   This is always possible, since $\tan^{-1}$ is defined for every real number, and it gives $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠. Now use the condition. Dividing $-\frac{a}{\sqrt3}<x<\frac{a}{\sqrt3}$ by $a>0$⁠:

   $$-\frac{1}{\sqrt3}<\frac{x}{a}<\frac{1}{\sqrt3}.$$

   Apply the increasing function $\tan^{-1}$ to all three parts, using $\tan^{-1}\frac{1}{\sqrt3}=\frac{\pi}{6}$ and $\tan^{-1}\left(-\frac{1}{\sqrt3}\right)=-\frac{\pi}{6}$⁠:

   $$-\frac{\pi}{6}<\theta<\frac{\pi}{6}, \qquad\text{so}\qquad -\frac{\pi}{2}<3\theta<\frac{\pi}{2}.$$
   This is the heart of the problem. The answer will be $\tan^{-1}(\tan3\theta)$⁠, and that equals $3\theta$ only if $3\theta$ is in the principal range. Tripling the interval $\left(-\frac{\pi}{6},\frac{\pi}{6}\right)$ lands exactly on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠. That is why the book chose the limits $\pm\frac{a}{\sqrt3}$⁠.
   An increasing function keeps inequalities: $p<q \Rightarrow \tan^{-1}p<\tan^{-1}q$⁠. And $\tan\frac{\pi}{6}=\frac{1}{\sqrt3}$⁠. Basics: special angles

3. The fraction equals $\tan3\theta$

   Divide the numerator and the denominator by $a^3$ (allowed, $a\ne0$⁠), and write $t=\frac{x}{a}=\tan\theta$⁠:

   $$\begin{aligned} \frac{3a^2x-x^3}{a^3-3ax^2} &= \frac{3\frac{x}{a}-\left(\frac{x}{a}\right)^3}{1-3\left(\frac{x}{a}\right)^2} \\[1mm] &= \frac{3t-t^3}{1-3t^2} = \tan3\theta. \end{aligned}$$

   Nothing here divides by zero. The denominator is $a^3-3ax^2=a(a^2-3x^2)$⁠, and $x^2<\frac{a^2}{3}$ gives $a^2-3x^2>0$⁠, so it is positive. Also $\theta$⁠, $2\theta\in\left(-\frac{\pi}{3},\frac{\pi}{3}\right)$ and $3\theta$ all lie in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠, so $\tan\theta$⁠, $\tan2\theta$ and $\tan3\theta$ are all defined, as Step 1 required.

   Every formula has fine print. The triple-angle formula needs $1-3t^2\ne0$ and the double-angle step needs $1-t^2\ne0$⁠. Both hold here because $t^2<\frac13<1$⁠. Checking this is part of a complete answer.
   $\frac{3a^2x}{a^3}=3\frac{x}{a}$⁠, $\frac{x^3}{a^3}=\left(\frac{x}{a}\right)^3$⁠, $\frac{3ax^2}{a^3}=3\left(\frac{x}{a}\right)^2$⁠.

4. Cancel $\tan^{-1}$ against $\tan$

   $$\begin{aligned} \tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right) &= \tan^{-1}(\tan3\theta) \\ &= 3\theta, \end{aligned}$$

   and the last equality is justified because $3\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ (Step 2), the principal range of $\tan^{-1}$⁠.

   Without the range check this line would be a guess. $\tan^{-1}$ always returns an angle in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠. Since $3\theta$ is already in there, and tangent is one-one on that interval, $3\theta$ is the only possible answer.
   $\tan^{-1}(\tan\phi)=\phi$ for $\phi\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ (Section 2.3). Lesson: the folding rules

5. Back to $x$⁠, and why the condition matters

   Since $\theta=\tan^{-1}\frac{x}{a}$⁠,

   $$\tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right) = 3\tan^{-1}\frac{x}{a}.$$

   The condition is not decoration. Take $a=1$⁠, $x=1$⁠, which breaks $x<\frac{1}{\sqrt3}$⁠. The fraction is $\frac{3-1}{1-3}=-1$⁠, so the left side is $\tan^{-1}(-1)=-\frac{\pi}{4}$⁠. But $3\tan^{-1}1=\frac{3\pi}{4}$⁠. The two differ by $\pi$⁠: here $3\theta=\frac{3\pi}{4}$ is outside the principal range, and $\tan^{-1}$ returns $\frac{3\pi}{4}-\pi$ instead. At $x=\pm\frac{a}{\sqrt3}$ the denominator is zero and the expression is not defined at all.

   A counterexample just outside the domain shows that the range check in Step 2 was doing real work. A formula like this is only an identity on the stated interval.

Every fact used, and where it comes from