Inverse Trigonometry Solutions: Exercise 2.2

This lesson solves the important questions of Exercise 2.2 on inverse trigonometric functions. The common strategy throughout is to substitute the variable as a suitable trig function (a sine, cosine, secant, or tangent) so that each expression reduces to a known angle and the identity falls out.

Proving $\sin^{-1}\!\big(2x\sqrt{1-x^2}\big) = 2\sin^{-1}x$

This holds for $-\tfrac{1}{\sqrt 2} \le x \le \tfrac{1}{\sqrt 2}$. Start from the left side and put $x = \sin\theta$, so that $\theta = \sin^{-1}x$.

$\text{LHS} = \sin^{-1}\!\big(2\sin\theta\,\sqrt{1-\sin^2\theta}\big) = \sin^{-1}\!\big(2\sin\theta\,\sqrt{\cos^2\theta}\big)$

Since $\sqrt{\cos^2\theta} = \cos\theta$ on this range,

$= \sin^{-1}\!\big(2\sin\theta\cos\theta\big) = \sin^{-1}\!\big(\sin 2\theta\big),$

using $2\sin\theta\cos\theta = \sin 2\theta$. This simplifies to $2\theta = 2\sin^{-1}x$, which is the right side.

The same expression equals $2\cos^{-1}x$

For $\tfrac{1}{\sqrt 2} \le x \le 1$, the same left side equals $2\cos^{-1}x$. Here we want the answer in terms of cosine, so put $x = \cos\theta$, giving $\theta = \cos^{-1}x$.

$\text{LHS} = \sin^{-1}\!\big(2\cos\theta\,\sqrt{1-\cos^2\theta}\big) = \sin^{-1}\!\big(2\cos\theta\,\sqrt{\sin^2\theta}\big)$

Since $1 - \cos^2\theta = \sin^2\theta$,

$= \sin^{-1}\!\big(2\sin\theta\cos\theta\big) = \sin^{-1}\!\big(\sin 2\theta\big) = 2\theta = 2\cos^{-1}x.$

Triple angle identities

$3\sin^{-1}x = \sin^{-1}\!\big(3x - 4x^3\big)$

We use the known identity $3\sin\theta - 4\sin^3\theta = \sin 3\theta$. Put $x = \sin\theta$, so $\theta = \sin^{-1}x$, and start from the right side.

$\text{RHS} = \sin^{-1}\!\big(3\sin\theta - 4\sin^3\theta\big) = \sin^{-1}\!\big(\sin 3\theta\big) = 3\theta = 3\sin^{-1}x = \text{LHS}.$

$3\cos^{-1}x = \cos^{-1}\!\big(4x^3 - 3x\big)$

For $x \in \left[\tfrac{1}{2}, 1\right]$, use $4\cos^3\theta - 3\cos\theta = \cos 3\theta$. Put $x = \cos\theta$ in the right side.

$\text{RHS} = \cos^{-1}\!\big(4\cos^3\theta - 3\cos\theta\big) = \cos^{-1}\!\big(\cos 3\theta\big) = 3\theta = 3\cos^{-1}x = \text{LHS}.$

Simplifying $\tan^{-1}\!\dfrac{\sqrt{1+x^2}-1}{x}$

For $x \neq 0$, put $x = \tan\theta$, so $\theta = \tan^{-1}x$.

$\tan^{-1}\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta} = \tan^{-1}\frac{\sqrt{\sec^2\theta}-1}{\tan\theta} = \tan^{-1}\frac{\sec\theta - 1}{\tan\theta},$

using $1 + \tan^2\theta = \sec^2\theta$. Rewriting in sines and cosines,

$= \tan^{-1}\frac{\tfrac{1}{\cos\theta} - 1}{\tfrac{\sin\theta}{\cos\theta}} = \tan^{-1}\frac{1 - \cos\theta}{\sin\theta}.$

Now use the half angle forms $1 - \cos\theta = 2\sin^2\tfrac{\theta}{2}$ and $\sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$:

$= \tan^{-1}\frac{2\sin^2\tfrac{\theta}{2}}{2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}} = \tan^{-1}\!\big(\tan\tfrac{\theta}{2}\big) = \frac{\theta}{2} = \frac{1}{2}\tan^{-1}x.$

Simplifying $\tan^{-1}\!\dfrac{1}{\sqrt{x^2-1}}$

For $|x| > 1$, put $x = \sec\theta$, so $\theta = \sec^{-1}x$, and use $\sec^2\theta - 1 = \tan^2\theta$.

$\tan^{-1}\frac{1}{\sqrt{\sec^2\theta - 1}} = \tan^{-1}\frac{1}{\sqrt{\tan^2\theta}} = \tan^{-1}\frac{1}{\tan\theta} = \tan^{-1}\!\big(\cot\theta\big).$

Writing $\cot\theta = \tan\!\big(\tfrac{\pi}{2} - \theta\big)$,

$= \tan^{-1}\tan\!\Big(\frac{\pi}{2} - \theta\Big) = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \sec^{-1}x.$

A half angle simplification

Simplify $\tan^{-1}\sqrt{\dfrac{1 - \cos x}{1 + \cos x}}$. Use $1 - \cos x = 2\sin^2\tfrac{x}{2}$ and $1 + \cos x = 2\cos^2\tfrac{x}{2}$:

$\tan^{-1}\sqrt{\frac{2\sin^2\tfrac{x}{2}}{2\cos^2\tfrac{x}{2}}} = \tan^{-1}\sqrt{\tan^2\tfrac{x}{2}} = \tan^{-1}\!\big(\tan\tfrac{x}{2}\big) = \frac{x}{2}.$

Simplifying $\tan^{-1}\!\dfrac{\cos x - \sin x}{\cos x + \sin x}$

This is for $-\tfrac{\pi}{4} < x < \tfrac{\pi}{4}$. Divide the numerator and denominator by $\cos x$:

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

Using the identity $\dfrac{1 - \tan x}{1 + \tan x} = \tan\!\big(\tfrac{\pi}{4} - x\big)$,

$= \tan^{-1}\tan\!\Big(\frac{\pi}{4} - x\Big) = \frac{\pi}{4} - x.$

Key takeaways

• Substituting $x = \sin\theta$, $\cos\theta$, $\tan\theta$, or $\sec\theta$ turns a messy inverse expression into a known angle.
• The double and triple angle identities ($\sin 2\theta$, $3\sin\theta - 4\sin^3\theta$, $4\cos^3\theta - 3\cos\theta$) are what let you collapse $\sin^{-1}(\sin k\theta)$ to $k\theta$.
• Half angle forms $1 - \cos x = 2\sin^2\tfrac{x}{2}$ and $1 + \cos x = 2\cos^2\tfrac{x}{2}$ simplify square root expressions to a clean $\tfrac{x}{2}$.