Inverse Trigonometric Functions
Definition and Principal Values
The trigonometric functions are not one-one on their natural domains — $\sin 30^\circ = \sin 150^\circ = \tfrac12$ — so they are not invertible as they stand. To define an inverse we first restrict the domain to an interval on which the function is one-one and still covers the entire range. That restricted interval is called the principal value branch.
The standard principal branches
On these branches each inverse function returns the single principal value:
| Function | Domain | Principal value branch (range) |
|---|---|---|
| $\sin^{-1}x$ | $[-1,1]$ | $\left[-\tfrac{\pi}{2},\ \tfrac{\pi}{2}\right]$ |
| $\cos^{-1}x$ | $[-1,1]$ | $[0,\ \pi]$ |
| $\tan^{-1}x$ | $\mathbb{R}$ | $\left(-\tfrac{\pi}{2},\ \tfrac{\pi}{2}\right)$ |
| $\csc^{-1}x$ | $|x|\ge 1$ | $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$ |
| $\sec^{-1}x$ | $|x|\ge 1$ | $[0,\pi]\setminus\{\tfrac{\pi}{2}\}$ |
| $\cot^{-1}x$ | $\mathbb{R}$ | $(0,\ \pi)$ |
Reading the symbol correctly
$\sin^{-1}x$ is the angle whose sine is $x$ — it is not $\dfrac{1}{\sin x}$. So $\sin^{-1}x = \theta$ means $\sin\theta = x$ and $\theta$ lies in the principal branch $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. The two conditions together pin down a unique answer.
A common trap: $\sin^{-1}\!\left(\sin\tfrac{2\pi}{3}\right)\ne \tfrac{2\pi}{3}$, because $\tfrac{2\pi}{3}$ is outside $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. You must bring the angle back into the branch (here the answer is $\tfrac{\pi}{3}$).
We need $\theta\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ with $\sin\theta=-\tfrac12$. Since $\sin\!\left(-\tfrac{\pi}{6}\right)=-\tfrac12$ and $-\tfrac{\pi}{6}$ lies in the branch, $\sin^{-1}\!\left(-\tfrac12\right)=-\dfrac{\pi}{6}$.
We need $\theta\in[0,\pi]$ with $\cos\theta=-\tfrac12$. Since $\cos\tfrac{2\pi}{3}=-\tfrac12$ and $\tfrac{2\pi}{3}\in[0,\pi]$, the value is $\cos^{-1}\!\left(-\tfrac12\right)=\dfrac{2\pi}{3}$.
$\tan^{-1}(1)=\tfrac{\pi}{4}$ (since $\tan\tfrac{\pi}{4}=1$ and $\tfrac{\pi}{4}\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$). $\cos^{-1}\!\left(\tfrac12\right)=\tfrac{\pi}{3}$. Sum $=\tfrac{\pi}{4}+\tfrac{\pi}{3}=\dfrac{3\pi+4\pi}{12}=\dfrac{7\pi}{12}$.
$\tfrac{2\pi}{3}\notin\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, so the answer is not $\tfrac{2\pi}{3}$. Use $\sin\tfrac{2\pi}{3}=\sin\!\left(\pi-\tfrac{2\pi}{3}\right)=\sin\tfrac{\pi}{3}$, and $\tfrac{\pi}{3}$ is in the branch. Hence the value is $\dfrac{\pi}{3}$.
- Trig functions are made invertible by restricting to a principal value branch where they are one-one.
- $\sin^{-1}$ and $\tan^{-1}$ have range centred on $0$: $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ and $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$.
- $\cos^{-1}$ and $\cot^{-1}$ have range $[0,\pi]$ and $(0,\pi)$.
- $\sin^{-1}x$ is an angle, not $\tfrac{1}{\sin x}$.
- For $\sin^{-1}(\sin\theta)$, first reduce $\theta$ into the principal branch.
Domains and Ranges
Each inverse function has a fixed domain and range: sin⁻¹ and cos⁻¹ take inputs in [−1, 1]; tan⁻¹ and cot⁻¹ take all reals; sec⁻¹ and cosec⁻¹ take |x| ≥ 1.
The ranges: tan⁻¹ ∈ (−π/2, π/2), cot⁻¹ ∈ (0, π), sec⁻¹ ∈ [0, π] \ {π/2}.
- sin⁻¹/cos⁻¹ need x ∈ [−1, 1]; sec⁻¹/cosec⁻¹ need |x| ≥ 1.
- tan⁻¹/cot⁻¹ accept every real number.
Properties and Identities
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.
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.
$\cos^{-1}(-x)=\pi-\cos^{-1}x$. With $x=\tfrac12$: $\cos^{-1}\!\left(-\tfrac12\right)=\pi-\cos^{-1}\tfrac12=\pi-\tfrac{\pi}{3}=\dfrac{2\pi}{3}$.
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 $\tfrac{3\pi}{4}+\tfrac{\pi}{4}=\pi$.
$2\tan^{-1}x=\tan^{-1}\dfrac{2x}{1-x^2}$ 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 $2\tan^{-1}\tfrac13=\tan^{-1}\dfrac{3}{4}$.
- $\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}$ 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.