Exercise 2.2 · Q5: Simplest form of $\tan^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}$

Basics you need

Solution

1. Write $\cos x \pm \sin x$ as single functions

   Using $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt2}$ and the compound-angle formulas:

   $$\begin{aligned} \sqrt2\sin\left(x + \tfrac{\pi}{4}\right) &= \sqrt2\left(\tfrac{\sin x}{\sqrt2} + \tfrac{\cos x}{\sqrt2}\right) \\ &= \cos x + \sin x, \end{aligned}$$ $$\begin{aligned} \sqrt2\cos\left(x + \tfrac{\pi}{4}\right) &= \sqrt2\left(\tfrac{\cos x}{\sqrt2} - \tfrac{\sin x}{\sqrt2}\right) \\ &= \cos x - \sin x. \end{aligned}$$
   $\cos x + \sin x$ and $\cos x - \sin x$ (coefficients of equal size) are each $\sqrt2$ times one shifted sine or cosine. Doing it this way avoids dividing by $\cos x$⁠, which is 0 at $x = \frac{\pi}{2}$⁠, a point inside our interval (see Step 4).
   $\sin(A+B)$ and $\cos(A+B)$ with $A = x$⁠, $B = \frac{\pi}{4}$ (Basics 1, 2).

2. Turn the fraction into a tangent

   For $-\frac{\pi}{4} \lt x \lt \frac{3\pi}{4}$ we have $0 \lt x + \frac{\pi}{4} \lt \pi$⁠, so $\sin\left(x + \frac{\pi}{4}\right) \gt 0$ and the denominator is never 0. Then

   $$\begin{aligned} \frac{\cos x - \sin x}{\cos x + \sin x} &= \frac{\cos\left(x + \frac{\pi}{4}\right)}{\sin\left(x + \frac{\pi}{4}\right)} \\ &= \cot\left(x + \tfrac{\pi}{4}\right). \end{aligned}$$

   By the co-function rule,

   $$\cot\left(x + \tfrac{\pi}{4}\right) = \tan\left(\tfrac{\pi}{2} - x - \tfrac{\pi}{4}\right) = \tan\left(\tfrac{\pi}{4} - x\right).$$
   The $\sqrt2$ cancels and the fraction becomes cos over sin, which is a cotangent. Since the answer must be "$\tan^{-1}$ of a tangent", we convert $\cot$ to $\tan$ with $\cot\alpha = \tan\left(\frac{\pi}{2} - \alpha\right)$⁠. The endpoints $-\frac{\pi}{4}$ and $\frac{3\pi}{4}$ are exactly where $\cos x + \sin x = 0$⁠, which is why they are excluded.
   Co-function rule (Basic 3). Basics: tan, cot, sec and cosec

3. Check the principal range and conclude

   From $-\frac{\pi}{4} \lt x \lt \frac{3\pi}{4}$⁠, multiply by $-1$ (the inequalities reverse) and add $\frac{\pi}{4}$⁠:

   $$-\frac{3\pi}{4} \lt -x \lt \frac{\pi}{4} \;\Rightarrow\; -\frac{\pi}{2} \lt \frac{\pi}{4} - x \lt \frac{\pi}{2}.$$

   So $\frac{\pi}{4} - x$ is in the principal range of $\tan^{-1}$⁠, and

   $$\begin{aligned} \tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right) &= \tan^{-1}\left(\tan\left(\tfrac{\pi}{4} - x\right)\right) \\ &= \frac{\pi}{4} - x. \end{aligned}$$
   This step is where the stated interval earns its place: it is precisely the condition $-\frac{\pi}{2} \lt \frac{\pi}{4} - x \lt \frac{\pi}{2}$⁠, rewritten in terms of $x$⁠.
   $\tan^{-1}(\tan y) = y$ for $y\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (Section 2.3). Lesson: sin⁻¹(sin x) and the folding rules

4. The divide-by-$\cos x$ route, and the point $x = \frac{\pi}{2}$

   The usual shortcut divides top and bottom by $\cos x$ and uses $\tan\frac{\pi}{4} = 1$⁠:

   $$\begin{aligned} \frac{1 - \tan x}{1 + \tan x} &= \frac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4}\tan x} \\ &= \tan\left(\tfrac{\pi}{4} - x\right). \end{aligned}$$

   This reaches the same place quickly, but it needs $\cos x \ne 0$⁠, and $x = \frac{\pi}{2}$ lies inside our interval with $\cos\frac{\pi}{2} = 0$⁠. Check that point directly:

   $$\frac{\cos\frac{\pi}{2} - \sin\frac{\pi}{2}}{\cos\frac{\pi}{2} + \sin\frac{\pi}{2}} = \frac{0 - 1}{0 + 1} = -1,$$ $$\tan^{-1}(-1) = -\frac{\pi}{4} = \frac{\pi}{4} - \frac{\pi}{2}.$$

   So the answer $\frac{\pi}{4} - x$ holds there too. The route of Steps 1 and 2 never divides by $\cos x$⁠, so it covers $x = \frac{\pi}{2}$ automatically.

   Whenever you divide by something, ask where it is zero. Here the shortcut's only gap is one point, and checking it by hand closes the gap.
   $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$ with $A = \frac{\pi}{4}$⁠, $B = x$ (Basic 1).

5. What goes wrong outside the interval

   The fraction equals $\tan\left(\frac{\pi}{4} - x\right)$ wherever it is defined, but $\tan^{-1}$ returns an angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, and $\tan$ repeats every $\pi$⁠. For $\frac{3\pi}{4} \lt x \lt \frac{7\pi}{4}$ we get $\frac{\pi}{4} - x \in \left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right)$⁠, and adding $\pi$ brings it into the principal range:

   $$\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right) = \frac{5\pi}{4} - x.$$

   This differs from $\frac{\pi}{4} - x$ by $\pi$⁠, so the formula fails everywhere there. Example: at $x = \pi$ the fraction is $\frac{-1 - 0}{-1 + 0} = 1$ and the value is $\tan^{-1}1 = \frac{\pi}{4}$⁠, while $\frac{\pi}{4} - \pi = -\frac{3\pi}{4}$⁠. Similarly, for $-\frac{5\pi}{4} \lt x \lt -\frac{\pi}{4}$ the value is $-\frac{3\pi}{4} - x$⁠.

   The output of $\tan^{-1}$ can never leave $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, while $\frac{\pi}{4} - x$ runs over all real numbers. They can agree only on the piece of the line inside that strip, which is the given interval.
   $\tan$ has period $\pi$⁠: $\tan(y + \pi) = \tan y$⁠. Lesson: sin⁻¹(sin x) and the folding rules

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