Exercise 2.2 · Q4: Simplest form of $\tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}$

Basics you need

Solution

1. Check the expression is defined

   For $0 \lt x \lt \pi$ we have $-1 \lt \cos x \lt 1$⁠. So $1 + \cos x \gt 0$ (no division by zero) and $1 - \cos x \gt 0$⁠. The fraction is positive, its square root is a real number, and $\tan^{-1}$ accepts every real number.

   Before simplifying, make sure there is something to simplify. At $x = \pi$ the denominator $1 + \cos\pi$ would be 0, which is one reason the interval stops short of $\pi$⁠.

2. Derive the half-angle formulas

   Put $u = \frac{x}{2}$ in $\cos 2u = 1 - 2\sin^2u$ and in $\cos 2u = 2\cos^2u - 1$⁠:

   $$\cos x = 1 - 2\sin^2\frac{x}{2} \;\Rightarrow\; 1 - \cos x = 2\sin^2\frac{x}{2},$$ $$\cos x = 2\cos^2\frac{x}{2} - 1 \;\Rightarrow\; 1 + \cos x = 2\cos^2\frac{x}{2}.$$
   Each version of $\cos 2u$ is chosen so that the constant 1 cancels: $1 - \cos x$ needs the "$1 - 2\sin^2$" form, $1 + \cos x$ needs the "$2\cos^2 - 1$" form. What is left is a single squared term.
   Double-angle formulas (Basic 1), Class XI trigonometry.

3. Simplify the fraction

   $$\frac{1-\cos x}{1+\cos x} = \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan^2\frac{x}{2}.$$

   Here $\cos\frac{x}{2} \ne 0$⁠, because $\frac{x}{2} \in \left(0, \frac{\pi}{2}\right)$⁠.

4. Take the square root, with the right sign

   $$\sqrt{\frac{1-\cos x}{1+\cos x}} = \sqrt{\tan^2\frac{x}{2}} = \left|\tan\frac{x}{2}\right|.$$

   Since $0 \lt \frac{x}{2} \lt \frac{\pi}{2}$⁠, $\tan\frac{x}{2} \gt 0$⁠, so $\left|\tan\frac{x}{2}\right| = \tan\frac{x}{2}$⁠.

   $\sqrt{\tan^2\frac{x}{2}}$ is not automatically $\tan\frac{x}{2}$⁠: a root is never negative, while $\tan\frac{x}{2}$ can be. Dividing the interval $0 \lt x \lt \pi$ by 2 shows $\frac{x}{2}$ is in the first quadrant, where $\tan$ is positive.
   $\sqrt{a^2} = |a|$ (Basic 2); sign of $\tan$ (Basic 3).

5. Apply $\tan^{-1}$ after checking the range

   Now the function is $\tan^{-1}\left(\tan\frac{x}{2}\right)$⁠, and $\frac{x}{2} \in \left(0, \frac{\pi}{2}\right) \subset \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. So

   $$\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right) = \tan^{-1}\left(\tan\frac{x}{2}\right) = \frac{x}{2}.$$
   $\tan^{-1}$ returns the angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ whose tangent is the input. Since $\frac{x}{2}$ is already such an angle, the answer is $\frac{x}{2}$ itself.
   $\tan^{-1}(\tan y) = y$ for $y\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (Section 2.3). Lesson: sin⁻¹(sin x) and the folding rules

6. What happens outside $0 \lt x \lt \pi$

   The working of Steps 2 and 3 holds whenever $\cos x \ne -1$⁠. Only the sign checks change.

   • $-\pi \lt x \lt 0$⁠: $\frac{x}{2} \in \left(-\frac{\pi}{2}, 0\right)$⁠, so $\tan\frac{x}{2} \lt 0$ and the root is $-\tan\frac{x}{2} = \tan\left(-\frac{x}{2}\right)$⁠. The value is $-\frac{x}{2}$⁠, not $\frac{x}{2}$⁠. For instance at $x = -\frac{\pi}{2}$ the value is $\tan^{-1}1 = \frac{\pi}{4}$⁠, while $\frac{x}{2} = -\frac{\pi}{4}$⁠.
   • $\pi \lt x \lt 2\pi$⁠: $\frac{x}{2} \in \left(\frac{\pi}{2}, \pi\right)$⁠, so $\tan\frac{x}{2} \lt 0$⁠, the root is $\tan\left(\pi - \frac{x}{2}\right)$ with $\pi - \frac{x}{2} \in \left(0, \frac{\pi}{2}\right)$⁠, and the value is $\pi - \frac{x}{2}$⁠. At $x = \frac{3\pi}{2}$ the value is $\frac{\pi}{4}$⁠, not $\frac{3\pi}{4}$⁠.
   • $x = \pm\pi$⁠: $1 + \cos x = 0$⁠, so the expression is undefined.

   So "$\frac{x}{2}$" is correct on the stated interval (and also at $x = 0$⁠, where both sides are 0), but wrong for every negative $x$ and every $x$ beyond $\pi$⁠. The expression can never be negative, while $\frac{x}{2}$ is negative for $x \lt 0$⁠.

   Seeing the failures makes clear what the interval in the question is doing: it is exactly what makes both sign checks (Steps 4 and 5) come out in our favour.
   $\tan(\pi - y) = -\tan y$ and $\tan(-y) = -\tan y$⁠. Basics: special angles and reference angles

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