Simplifying Inverse Trigonometric Expressions

This lesson simplifies five inverse trigonometric expressions. The recurring idea is to pick a substitution or an algebraic step that turns the inside of the inverse function into a tangent (or cotangent) of a recognisable angle.

Dividing through by cosine

Simplify, for $-\tfrac{\pi}{4} < x < \tfrac{\pi}{4}$:

$\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$

Divide the numerator and denominator by $\cos x$:

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

Using the difference identity $\dfrac{1 - \tan x}{1 + \tan x} = \tan\!\left(\tfrac{\pi}{4} - x\right)$, the expression becomes

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

The domain keeps $\tfrac{\pi}{4} - x$ inside the principal range, so this is exact.

A sine substitution for a square-root expression

Simplify, for $|x| < a$:

$\tan^{-1}\left(\frac{x}{\sqrt{a^2 - x^2}}\right)$

Because an $a^2 - x^2$ appears, put $x = a\sin\theta$, so $\theta = \sin^{-1}\tfrac{x}{a}$. Then

$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = \sqrt{a^2\cos^2\theta} = a\cos\theta,$

and the expression reduces to

$\tan^{-1}\!\left(\frac{a\sin\theta}{a\cos\theta}\right) = \tan^{-1}(\tan\theta) = \theta = \sin^{-1}\frac{x}{a}.$

A triple-angle expression

Simplify, for $-\tfrac{a}{\sqrt 3} < x < \tfrac{a}{\sqrt 3}$:

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

Put $x = a\tan\theta$, so $\theta = \tan^{-1}\tfrac{x}{a}$. Substituting and taking out $a^3$ from top and bottom:

$\frac{3a^2(a\tan\theta) - (a\tan\theta)^3}{a^3 - 3a(a\tan\theta)^2} = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan 3\theta.$

This is exactly the triple-angle formula for tangent, so

$\tan^{-1}(\tan 3\theta) = 3\theta = 3\tan^{-1}\frac{x}{a}.$

A secant substitution

Simplify, for $x > 1$:

$\cot^{-1}\left(\frac{1}{\sqrt{x^2 - 1}}\right)$

Put $x = \sec\theta$, so $\theta = \sec^{-1} x$. Then $\sqrt{x^2 - 1} = \sqrt{\sec^2\theta - 1} = \tan\theta$, giving

$\cot^{-1}\!\left(\frac{1}{\tan\theta}\right) = \cot^{-1}(\cot\theta) = \theta = \sec^{-1} x.$

Half-angle identities

Simplify, for $-\tfrac{3\pi}{2} < x < \tfrac{\pi}{2}$:

$\tan^{-1}\left(\frac{\cos x}{1 - \sin x}\right)$

Write the numerator and denominator using half angles. With $\cos x = \cos^2\tfrac{x}{2} - \sin^2\tfrac{x}{2}$ and $1 - \sin x = \cos^2\tfrac{x}{2} + \sin^2\tfrac{x}{2} - 2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$:

$\frac{\cos x}{1 - \sin x} = \frac{\left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)\left(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\right)}{\left(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\right)^2} = \frac{\cos\tfrac{x}{2} + \sin\tfrac{x}{2}}{\cos\tfrac{x}{2} - \sin\tfrac{x}{2}}.$

Divide top and bottom by $\cos\tfrac{x}{2}$:

$\frac{1 + \tan\tfrac{x}{2}}{1 - \tan\tfrac{x}{2}} = \tan\!\left(\tfrac{\pi}{4} + \tfrac{x}{2}\right).$

Therefore

$\tan^{-1}\left(\frac{\cos x}{1 - \sin x}\right) = \frac{\pi}{4} + \frac{x}{2}.$

Key takeaways

• When $a^2 - x^2$ appears, substitute $x = a\sin\theta$; for $x^2 - 1$ under a root, substitute $x = \sec\theta$; for triple-angle patterns, substitute $x = a\tan\theta$.
• $\dfrac{1 - \tan x}{1 + \tan x} = \tan\!\left(\tfrac{\pi}{4} - x\right)$ and $\dfrac{1 + \tan x}{1 - \tan x} = \tan\!\left(\tfrac{\pi}{4} + x\right)$.
• Half-angle identities turn $\cos x$ and $1 - \sin x$ into perfect-square and difference-of-squares forms that cancel cleanly.