Inverse Trigonometric Functions • Topic 2 of 2

Properties of Inverse Trigonometric Functions

A small set of identities lets you simplify almost every inverse-trig expression in the syllabus. Each identity holds only on a stated domain — quoting the domain is part of a correct answer.

Complementary (co-function) identities

For all $x\in[-1,1]$:

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

Similarly $\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}$ for all $x\in\mathbb{R}$, and $\sec^{-1}x+\csc^{-1}x=\dfrac{\pi}{2}$ for $|x|\ge 1$.

Negative-argument identities

$\sin^{-1}(-x)=-\sin^{-1}x$, $\ \tan^{-1}(-x)=-\tan^{-1}x$ (odd functions).
$\cos^{-1}(-x)=\pi-\cos^{-1}x$, $\ \cot^{-1}(-x)=\pi-\cot^{-1}x$.

Sum identity for arctangent

$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\!\left(\frac{x+y}{1-xy}\right),\qquad xy<1.$$

When $xy>1$ (with $x,y>0$) add $\pi$; when $xy>1$ with $x,y<0$ subtract $\pi$. Forgetting this correction is the most common error in the chapter.

Double-argument forms

$2\tan^{-1}x=\tan^{-1}\dfrac{2x}{1-x^2}$ (for $|x|<1$), and $2\tan^{-1}x=\sin^{-1}\dfrac{2x}{1+x^2}$ (for $|x|\le 1$). These convert awkward rational arguments into a single inverse function.

Solved Examples

Worked Example

Evaluate $\sin^{-1}\!\left(\tfrac{1}{2}\right)+\cos^{-1}\!\left(\tfrac{1}{2}\right)$.

Solution

By the complementary identity $\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}$ for every $x\in[-1,1]$. Here $x=\tfrac12$, so the sum is $\dfrac{\pi}{2}$ — no need to compute each term.

Worked Example

Simplify $\cos^{-1}\!\left(-\tfrac{1}{2}\right)$ using a negative-argument identity.

Solution

$\cos^{-1}(-x)=\pi-\cos^{-1}x$. With $x=\tfrac12$: $\cos^{-1}\!\left(-\tfrac12\right)=\pi-\cos^{-1}\tfrac12=\pi-\tfrac{\pi}{3}=\dfrac{2\pi}{3}$.

Worked Example

Find $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$.

Solution

First $\tan^{-1}2+\tan^{-1}3$: here $xy=6>1$ with both positive, so we add $\pi$: $\tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}\dfrac{2+3}{1-6}=\pi+\tan^{-1}(-1)=\pi-\tfrac{\pi}{4}=\tfrac{3\pi}{4}$. Adding $\tan^{-1}1=\tfrac{\pi}{4}$ gives $\tfrac{3\pi}{4}+\tfrac{\pi}{4}=\pi$.

Worked Example

Write $2\tan^{-1}\!\left(\tfrac{1}{3}\right)$ as a single arctangent.

Solution

$2\tan^{-1}x=\tan^{-1}\dfrac{2x}{1-x^2}$ for $|x|<1$. With $x=\tfrac13$: $\dfrac{2\cdot\tfrac13}{1-\tfrac19}=\dfrac{\tfrac23}{\tfrac89}=\dfrac{2}{3}\cdot\dfrac{9}{8}=\dfrac{3}{4}$. So $2\tan^{-1}\tfrac13=\tan^{-1}\dfrac{3}{4}$.

Key Points

$\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}$ on $[-1,1]$; same complementary pattern for $\tan/\cot$ and $\sec/\csc$.
$\sin^{-1},\tan^{-1}$ are odd; $\cos^{-1}(-x)=\pi-\cos^{-1}x$, $\cot^{-1}(-x)=\pi-\cot^{-1}x$.
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\tfrac{x+y}{1-xy}$ only when $xy<1$; otherwise add/subtract $\pi$.
$2\tan^{-1}x=\tan^{-1}\tfrac{2x}{1-x^2}=\sin^{-1}\tfrac{2x}{1+x^2}$ on the stated domains.
Always state the domain on which an identity is applied.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.$\sin^{-1}x+\cos^{-1}x$ equals (for $x\in[-1,1]$):

Explanation: It is the constant $\tfrac{\pi}{2}$ for every valid $x$.

Q2.$\cos^{-1}(-x)=$

Explanation: $\cos^{-1}$ is not odd; $\cos^{-1}(-x)=\pi-\cos^{-1}x$.

Q3.$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\tfrac{x+y}{1-xy}$ holds provided:

Explanation: The plain identity requires $xy<1$; otherwise a $\pm\pi$ correction is needed.

Q4.$2\tan^{-1}\tfrac{1}{2}$ written as a single arctangent is:

Explanation: $\tfrac{2x}{1-x^2}=\tfrac{1}{1-1/4}=\tfrac{1}{3/4}=\tfrac{4}{3}$.

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