Chapter 2 · Inverse Trigonometric Functions

2.1 Inverse Trigonometric Functions and Principal Value Branches

Sine, cosine and tangent repeat their values forever, so they cannot be run backwards on all of ℝ. Restrict each one to a branch on which it is one-one and onto, and it gets an inverse: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ and their three cousins.

A wheelchair ramp is $6$ m long and rises $0.5$ m. What angle does it make with the ground? You know the sine of the angle — $\tfrac{0.5}{6}=\tfrac1{12}$ — and you want the angle. Every scientific calculator has a key for this, marked $\sin^{-1}$, and it answers $4.78^\circ$ (about $0.0834$ radian). But think about what the key is being asked to do. The equation $\sin\theta=\tfrac1{12}$ has infinitely many solutions: $0.0834$, $\pi-0.0834$, $2\pi+0.0834$, and so on for ever. A key can give only one answer. Which one, and why?

This section answers that question for all six trigonometric functions. The answer uses exactly the idea of Section 1.2: a function can be run backwards only if it is one-one and onto. Sine is neither on $\mathbb R$, so we restrict it to an interval where it is both, and invert it there. The interval we choose is called the principal value branch, and the answer the calculator gives is the principal value.

Why Sine Must Be Restricted before It Can Be Inverted

Recall the picture behind an inverse. If $f:X\to Y$ is one-one and onto, every $y\in Y$ is the image of exactly one $x\in X$ (Definition 1.8 and the horizontal line test of Section 1.2). So the rule “send $y$ back to that $x$” is itself a function.

📘 Definition 2.1 — Inverse of a one-one, onto function
Let $f:X\to Y$ be one-one and onto. The inverse function $f^{-1}:Y\to X$ is defined by $$f^{-1}(y)=x\ \Longleftrightarrow\ f(x)=y\qquad(x\in X,\ y\in Y).$$ The domain of $f^{-1}$ is $Y$ (the range of $f$) and its range is $X$ (the domain of $f$). A function that is not one-one or not onto has no inverse.

Now look at $\sin:\mathbb R\to\mathbb R$ (angles in radians, as always in this chapter; from Class 11 Mathematics, Ch. 3 — Trigonometric Functions).

Both faults can be repaired. Shrinking the codomain to $[-1,1]$ makes sine onto. Shrinking the domain to an interval on which sine is strictly increasing or strictly decreasing makes it one-one. On $\left[-\tfrac\pi2,\tfrac\pi2\right]$ sine increases steadily from $-1$ to $1$, taking every value in $[-1,1]$ exactly once. So $$\sin:\left[-\tfrac\pi2,\tfrac\pi2\right]\to[-1,1]\ \text{ is one-one and onto.}$$ Other intervals would also work — $\left[\tfrac\pi2,\tfrac{3\pi}2\right]$, $\left[-\tfrac{3\pi}2,-\tfrac\pi2\right]$, … — and each gives a different branch of the inverse. Mathematicians agreed to use the branch through $0$, the one that contains the small positive angles of right-angled triangles.

📘 Definition 2.2 — Inverse sine
The inverse sine function $\sin^{-1}:[-1,1]\to\left[-\tfrac\pi2,\tfrac\pi2\right]$ is the inverse of sine restricted to $\left[-\tfrac\pi2,\tfrac\pi2\right]$: $$y=\sin^{-1}x\ \Longleftrightarrow\ \sin y=x\ \text{ and }\ -\tfrac\pi2\le y\le\tfrac\pi2.$$ In words, $\sin^{-1}x$ is the angle between $-\tfrac\pi2$ and $\tfrac\pi2$ whose sine is $x$. Its domain is $[-1,1]$; the interval $\left[-\tfrac\pi2,\tfrac\pi2\right]$ is its principal value branch.
⚠️ Read the notation carefully

$\sin^{-1}x$ does not mean $\dfrac1{\sin x}$. The “$-1$” is the inverse-function sign of Definition 2.1, not a power: $\sin^{-1}x\ne(\sin x)^{-1}=\operatorname{cosec}x$. For example $\sin^{-1}1=\tfrac\pi2$, while $\dfrac1{\sin1}\approx1.188$. (Contrast $\sin^2x$, which does mean $(\sin x)^2$.) If you need the reciprocal, write $(\sin x)^{-1}$ or $\operatorname{cosec}x$.

Interactive 1 — Restrict-the-Domain Lab

The faint curve is $y=\sin x$ on $\mathbb R$. Slide the window to choose a piece of the domain and pick its length. The lab tests the piece: two red dots on one dashed horizontal line mean not one-one; the green bar on the $y$-axis is $[-1,1]$ and any part of it not reached is marked as a gap (not onto). When the piece passes, slide reflect (or press Play) to flip it in the mirror line $y=x$ — the flipped curve is the graph of an inverse. Play also hunts through the windows of length $\pi$ by itself.

window—
one-one?—
onto $[-1,1]$?—
verdict—
✏️ Example 1 — Principal values of the inverse sine
Find (a) $\sin^{-1}\tfrac12$  (b) $\sin^{-1}\!\left(-\tfrac{\sqrt3}2\right)$  (c) $\sin^{-1}1$  (d) $\sin^{-1}\!\left(-\tfrac1{\sqrt2}\right)$  (e) $\sin^{-1}0$.
Method For each $x$ we need the one angle $y$ with $\sin y=x$ and $-\tfrac\pi2\le y\le\tfrac\pi2$. Negative $x$ gives an angle in $\left[-\tfrac\pi2,0\right)$, because $\sin(-\alpha)=-\sin\alpha$.
(a) $\sin\tfrac\pi6=\tfrac12$ and $\tfrac\pi6\in\left[-\tfrac\pi2,\tfrac\pi2\right]$, so $\sin^{-1}\tfrac12=\tfrac\pi6$. (The angle $\tfrac{5\pi}6$ also has sine $\tfrac12$, but it is outside the branch.)
(b) $\sin\tfrac\pi3=\tfrac{\sqrt3}2$, so $\sin\!\left(-\tfrac\pi3\right)=-\tfrac{\sqrt3}2$, and $-\tfrac\pi3$ is in the branch: $\sin^{-1}\!\left(-\tfrac{\sqrt3}2\right)=-\tfrac\pi3$.
(c) $\sin\tfrac\pi2=1$: $\sin^{-1}1=\tfrac\pi2$ (an endpoint of the branch — allowed, the interval is closed).
(d), (e) $\sin\!\left(-\tfrac\pi4\right)=-\tfrac1{\sqrt2}$, so $\sin^{-1}\!\left(-\tfrac1{\sqrt2}\right)=-\tfrac\pi4$; and $\sin0=0$ gives $\sin^{-1}0=0$.
Check All five answers lie between $-\tfrac\pi2$ and $\tfrac\pi2$, and a calculator in radian mode gives $0.5236$, $-1.0472$, $1.5708$, $-0.7854$, $0$ — that is $\tfrac\pi6$, $-\tfrac\pi3$, $\tfrac\pi2$, $-\tfrac\pi4$, $0$. ✓

Inverse Cosine and Inverse Tangent

The same repair works for cosine and tangent; only the interval changes, because each function must be one-one on its branch.

📘 Definition 2.3 — Inverse cosine
The inverse cosine function $\cos^{-1}:[-1,1]\to[0,\pi]$ is the inverse of cosine restricted to $[0,\pi]$: $$y=\cos^{-1}x\ \Longleftrightarrow\ \cos y=x\ \text{ and }\ 0\le y\le\pi.$$ Its principal value branch is $[0,\pi]$.
📘 Definition 2.4 — Inverse tangent
The inverse tangent function $\tan^{-1}:\mathbb R\to\left(-\tfrac\pi2,\tfrac\pi2\right)$ is the inverse of tangent restricted to $\left(-\tfrac\pi2,\tfrac\pi2\right)$: $$y=\tan^{-1}x\ \Longleftrightarrow\ \tan y=x\ \text{ and }\ -\tfrac\pi2\lt y\lt\tfrac\pi2.$$ It is defined for every real $x$; its principal value branch is the open interval $\left(-\tfrac\pi2,\tfrac\pi2\right)$, so $\tan^{-1}x$ is never equal to $\pm\tfrac\pi2$.

Notice the pattern. Where the branch includes negative angles (sine, tangent), a negative input gives a negative angle. The cosine branch $[0,\pi]$ has no negative angles, so a negative input gives an angle in the second quadrant, $\left(\tfrac\pi2,\pi\right]$, where cosine is negative. The Class 11 allied-angle facts $\cos(\pi-\alpha)=-\cos\alpha$ and $\tan(-\alpha)=-\tan\alpha$ do the work.

✏️ Example 2 — Inverse cosine and inverse tangent
Find (a) $\cos^{-1}\tfrac{\sqrt3}2$  (b) $\cos^{-1}\!\left(-\tfrac12\right)$  (c) $\cos^{-1}(-1)$  (d) $\tan^{-1}1$  (e) $\tan^{-1}\!\left(-\sqrt3\right)$.
(a) $\cos\tfrac\pi6=\tfrac{\sqrt3}2$ and $\tfrac\pi6\in[0,\pi]$: $\cos^{-1}\tfrac{\sqrt3}2=\tfrac\pi6$.
(b) The acute angle with cosine $\tfrac12$ is $\tfrac\pi3$. We need cosine $-\tfrac12$ inside $[0,\pi]$: $\cos\!\left(\pi-\tfrac\pi3\right)=-\cos\tfrac\pi3=-\tfrac12$, so $\cos^{-1}\!\left(-\tfrac12\right)=\tfrac{2\pi}3$.
(c) $\cos\pi=-1$ and $\pi\in[0,\pi]$: $\cos^{-1}(-1)=\pi$.
(d) $\tan\tfrac\pi4=1$ and $\tfrac\pi4\in\left(-\tfrac\pi2,\tfrac\pi2\right)$: $\tan^{-1}1=\tfrac\pi4$.
(e) $\tan\tfrac\pi3=\sqrt3$, so $\tan\!\left(-\tfrac\pi3\right)=-\sqrt3$, and $-\tfrac\pi3$ is in the branch: $\tan^{-1}\!\left(-\sqrt3\right)=-\tfrac\pi3$.
Check In (b), the tempting answer $-\tfrac\pi3$ does have cosine $\tfrac12$ — not $-\tfrac12$ — and is not in $[0,\pi]$ anyway. Always ask both questions: right value? right branch?

Cosecant, Secant and Cotangent; Principal Values

The reciprocal functions are inverted on (almost) the same branches as their partners, so that the six inverses fit together neatly.

📘 Definition 2.5 — Inverse cosecant, secant and cotangent

$\operatorname{cosec}^{-1}:\mathbb R-(-1,1)\to\left[-\tfrac\pi2,\tfrac\pi2\right]-\{0\}$,   $y=\operatorname{cosec}^{-1}x\iff\operatorname{cosec}y=x$ and $y\in\left[-\tfrac\pi2,\tfrac\pi2\right]-\{0\}$;

$\sec^{-1}:\mathbb R-(-1,1)\to[0,\pi]-\left\{\tfrac\pi2\right\}$,   $y=\sec^{-1}x\iff\sec y=x$ and $y\in[0,\pi]-\left\{\tfrac\pi2\right\}$;

$\cot^{-1}:\mathbb R\to(0,\pi)$,   $y=\cot^{-1}x\iff\cot y=x$ and $0\lt y\lt\pi$.

These are the inverse cosecant, inverse secant and inverse cotangent functions; the target intervals are their principal value branches.

Here are all six side by side. Learn this table: almost every question in this chapter starts from it.

functiondomainprincipal value branch (range)
$\sin^{-1}x$$[-1,1]$$\left[-\tfrac\pi2,\tfrac\pi2\right]$
$\cos^{-1}x$$[-1,1]$$[0,\pi]$
$\tan^{-1}x$$\mathbb R$$\left(-\tfrac\pi2,\tfrac\pi2\right)$
$\operatorname{cosec}^{-1}x$$\mathbb R-(-1,1)$$\left[-\tfrac\pi2,\tfrac\pi2\right]-\{0\}$
$\sec^{-1}x$$\mathbb R-(-1,1)$$[0,\pi]-\left\{\tfrac\pi2\right\}$
$\cot^{-1}x$$\mathbb R$$(0,\pi)$

A memory aid: the three “co-less” functions sin, tan, cosec use branches centred at $0$ (first and fourth quadrants); cos, sec, cot use branches starting at $0$ (first and second quadrants).

📘 Definition 2.6 — Principal value
The value of an inverse trigonometric function at a point of its domain — which by definition lies in the principal value branch — is called the principal value of that function at the point. For example, the principal value of $\cos^{-1}\!\left(-\tfrac12\right)$ is $\tfrac{2\pi}3$, although infinitely many other angles (such as $-\tfrac{2\pi}3$ and $\tfrac{4\pi}3$) have cosine $-\tfrac12$.

Because each inverse undoes a one-one, onto restriction, Definition 2.1 gives two “cancellation” rules — but each holds only on the right set.

📐 Theorem 2.1 — Cancelling a function and its inverse

(a) $\sin\!\left(\sin^{-1}x\right)=x$ for every $x\in[-1,1]$, and $\sin^{-1}(\sin y)=y$ for every $y\in\left[-\tfrac\pi2,\tfrac\pi2\right]$.

(b) $\cos\!\left(\cos^{-1}x\right)=x$ for every $x\in[-1,1]$, and $\cos^{-1}(\cos y)=y$ for every $y\in[0,\pi]$.

(c) $\tan\!\left(\tan^{-1}x\right)=x$ for every $x\in\mathbb R$, and $\tan^{-1}(\tan y)=y$ for every $y\in\left(-\tfrac\pi2,\tfrac\pi2\right)$.

The same holds for $\operatorname{cosec}^{-1}$, $\sec^{-1}$, $\cot^{-1}$: $f\!\left(f^{-1}(x)\right)=x$ for every $x$ in the domain of $f^{-1}$, and $f^{-1}(f(y))=y$ for every $y$ in the principal value branch.

Proof of (a). Let $x\in[-1,1]$ and put $y=\sin^{-1}x$. By Definition 2.2, $\sin y=x$, i.e. $\sin\!\left(\sin^{-1}x\right)=x$. Next let $y\in\left[-\tfrac\pi2,\tfrac\pi2\right]$ and put $x=\sin y$, which lies in $[-1,1]$. Then $y$ is an angle in the branch whose sine is $x$; the branch contains only one such angle, so $y=\sin^{-1}x=\sin^{-1}(\sin y)$. The other parts are proved in the same way from Definitions 2.3–2.5. $\square$

The hypotheses matter: $\sin^{-1}(\sin\pi)=\sin^{-1}0=0\ne\pi$, because $\pi$ is outside the branch, and $\sin\!\left(\sin^{-1}2\right)$ is meaningless because $2$ is outside the domain. Section 2.2 shows exactly what $\sin^{-1}(\sin y)$ is when $y$ is outside the branch.

🧭 Problem-solving strategy — finding a principal value
  1. Check the domain. $\sin^{-1}$, $\cos^{-1}$ need $|x|\le1$; $\operatorname{cosec}^{-1}$, $\sec^{-1}$ need $|x|\ge1$; $\tan^{-1}$, $\cot^{-1}$ accept every $x$.
  2. Find the reference angle $\alpha\in\left[0,\tfrac\pi2\right]$ whose sine, cosine, … equals $|x|$ (a standard angle: $0,\tfrac\pi6,\tfrac\pi4,\tfrac\pi3,\tfrac\pi2$, or a calculator value).
  3. If $x\ge0$, the answer is $\alpha$ for every one of the six functions.
  4. If $x\lt0$: for $\sin^{-1}$, $\tan^{-1}$, $\operatorname{cosec}^{-1}$ (branches centred at $0$) the answer is $-\alpha$; for $\cos^{-1}$, $\sec^{-1}$, $\cot^{-1}$ (branches in $[0,\pi]$) it is $\pi-\alpha$.
  5. Check that the answer is in the branch and has the right value.
✏️ Example 3 — The reciprocal inverses
Find (a) $\operatorname{cosec}^{-1}2$  (b) $\operatorname{cosec}^{-1}\!\left(-\sqrt2\right)$  (c) $\sec^{-1}(-2)$  (d) $\sec^{-1}\tfrac{2}{\sqrt3}$  (e) $\cot^{-1}\sqrt3$  (f) $\cot^{-1}(-1)$.
(a) $\operatorname{cosec}y=2\iff\sin y=\tfrac12$; in the branch, $y=\tfrac\pi6$. So $\operatorname{cosec}^{-1}2=\tfrac\pi6$.
(b) $|x|=\sqrt2$ gives $\alpha=\tfrac\pi4$ ($\operatorname{cosec}\tfrac\pi4=\sqrt2$). The branch is centred at $0$ and $x\lt0$: $\operatorname{cosec}^{-1}\!\left(-\sqrt2\right)=-\tfrac\pi4$.
(c) $\sec\alpha=2$ gives $\alpha=\tfrac\pi3$. The branch is in $[0,\pi]$ and $x\lt0$: $\sec^{-1}(-2)=\pi-\tfrac\pi3=\tfrac{2\pi}3$. Check: $\cos\tfrac{2\pi}3=-\tfrac12$, so $\sec\tfrac{2\pi}3=-2$ ✓.
(d) $\sec y=\tfrac2{\sqrt3}\iff\cos y=\tfrac{\sqrt3}2$: $y=\tfrac\pi6$.
(e) $\cot\tfrac\pi6=\dfrac{\cos(\pi/6)}{\sin(\pi/6)}=\dfrac{\sqrt3/2}{1/2}=\sqrt3$, and $\tfrac\pi6\in(0,\pi)$: $\cot^{-1}\sqrt3=\tfrac\pi6$.
(f) $\alpha=\tfrac\pi4$, branch $(0,\pi)$, $x\lt0$: $\cot^{-1}(-1)=\pi-\tfrac\pi4=\tfrac{3\pi}4$. Not $-\tfrac\pi4$, which is outside $(0,\pi)$ — the most common slip with $\cot^{-1}$.
Interactive 2 — Principal-Value Explorer for All Six

Choose a function. The graph shows the trigonometric function $y=f(\theta)$ over $[-2\pi,2\pi]$; the shaded strip is the principal value branch and the thick part of the curve is the branch that gets inverted. Move $x$ (or press Play to visit the standard values one by one): the dashed line $y=x$ meets the curve at every angle with $f(\theta)=x$ (hollow dots), but only one of them — the red dot — lies in the strip. Its $\theta$ is the principal value $f^{-1}(x)$.

$x$—
principal value—
in degrees—
other angles in view—
✏️ Example 4 — Expressions built from principal values
Find the value of (a) $\tan^{-1}1+\cos^{-1}\!\left(-\tfrac12\right)+\sin^{-1}\!\left(-\tfrac12\right)$  (b) $\cos^{-1}\tfrac12+2\sin^{-1}\tfrac12$  (c) $\sin\!\left(\tfrac\pi3-\sin^{-1}\!\left(-\tfrac12\right)\right)$.
(a) Each term separately: $\tan^{-1}1=\tfrac\pi4$; $\cos^{-1}\!\left(-\tfrac12\right)=\tfrac{2\pi}3$ (Example 2); $\sin^{-1}\!\left(-\tfrac12\right)=-\tfrac\pi6$. Sum: $\tfrac\pi4+\tfrac{2\pi}3-\tfrac\pi6=\tfrac{3\pi+8\pi-2\pi}{12}=\tfrac{9\pi}{12}=\tfrac{3\pi}4$.
(b) $\cos^{-1}\tfrac12=\tfrac\pi3$ and $\sin^{-1}\tfrac12=\tfrac\pi6$, so the value is $\tfrac\pi3+2\cdot\tfrac\pi6=\tfrac{2\pi}3$.
(c) $\sin^{-1}\!\left(-\tfrac12\right)=-\tfrac\pi6$, so the bracket is $\tfrac\pi3+\tfrac\pi6=\tfrac\pi2$ and the value is $\sin\tfrac\pi2=1$.
Check (a) numerically: $0.7854+2.0944-0.5236=2.3562=\tfrac{3\pi}4$ ✓.
✏️ Example 5 — Domains and ranges of related functions
(a) Find the domain of $f(x)=\sin^{-1}(2x-1)$.  (b) Find the domain of $g(x)=\cos^{-1}\!\left(x^2-3\right)$.  (c) Find the range of $h(x)=3\cos^{-1}x-\tfrac\pi2$.
(a) $\sin^{-1}u$ needs $-1\le u\le1$: $-1\le2x-1\le1\iff0\le2x\le2\iff0\le x\le1$. Domain $[0,1]$.
(b) $-1\le x^2-3\le1\iff2\le x^2\le4\iff\sqrt2\le|x|\le2$. Domain $\left[-2,-\sqrt2\right]\cup\left[\sqrt2,2\right]$. (Remember both signs when you take square roots.)
(c) The domain is $[-1,1]$ and $0\le\cos^{-1}x\le\pi$, with every value in $[0,\pi]$ taken. So $0\le3\cos^{-1}x\le3\pi$ and $-\tfrac\pi2\le h(x)\le\tfrac{5\pi}2$. Range $\left[-\tfrac\pi2,\tfrac{5\pi}2\right]$.
Check (b): $x=1.8$ gives $x^2-3=0.24\in[-1,1]$ ✓; $x=1$ gives $-2$, outside ✓ (and $1\notin$ the domain).
✏️ Example 6 — Angles from measurements
(a) A $5$ m ladder leans against a wall with its foot $1.5$ m from the wall. Find the angle it makes with the ground.  (b) A $12$ m pole casts a $9$ m shadow on level ground. Find the angle of elevation of the Sun. Give both answers in radians (4 decimal places) and degrees.
(a) The ground, the wall and the ladder form a right triangle with hypotenuse $5$ m and side adjacent to the angle $\theta$ equal to $1.5$ m. So $\cos\theta=\tfrac{1.5}5=0.3$. The angle is acute, i.e. in $[0,\pi]$, so $\theta=\cos^{-1}0.3\approx1.2661$ rad $\approx72.54^\circ$.
(b) $\tan\theta=\tfrac{12}9=\tfrac43$ and $0\lt\theta\lt\tfrac\pi2$, so $\theta=\tan^{-1}\tfrac43\approx0.9273$ rad $\approx53.13^\circ$.
Why it is legitimate In both problems the angle is known to be acute, and the acute angles lie inside every principal branch. So the calculator's principal value is the physical angle. When the physical angle may be obtuse or negative, you must check the branch (Section 2.2).
Check (a): $5\cos(1.2661)=1.5000$ m ✓. (b): $9\tan(0.9273)=12.000$ m ✓.
⚠️ Common mistakes when studying alone

1. Giving an angle outside the branch. $\cos^{-1}\!\left(-\tfrac12\right)=-\tfrac\pi3$ and $\cot^{-1}(-1)=-\tfrac\pi4$ are wrong — negative angles never come out of $\cos^{-1}$, $\sec^{-1}$ or $\cot^{-1}$. Likewise $\sin^{-1}\!\left(-\tfrac12\right)=\tfrac{11\pi}6$ is wrong: it has the right sine but is outside $\left[-\tfrac\pi2,\tfrac\pi2\right]$.

2. Reading $\sin^{-1}x$ as $\dfrac1{\sin x}$. See the warning after Definition 2.2.

3. Ignoring the domain. $\sin^{-1}2$, $\cos^{-1}(-1.5)$ and $\sec^{-1}\tfrac12$ do not exist. A calculator shows “Math ERROR”; in an exam, say “not defined” and give the reason.

4. Degrees and radians mixed. Principal values are angles in radians unless degrees are asked for: write $\tfrac\pi6$, not $30$. If a calculator is used, set it to radian mode.

5. Open and closed ends. $\tan^{-1}x$ is never $\pm\tfrac\pi2$ and $\cot^{-1}x$ is never $0$ or $\pi$; $\sec^{-1}x$ is never $\tfrac\pi2$ and $\operatorname{cosec}^{-1}x$ is never $0$. These missing values matter when you state a range.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026; not in the CBSE Class 12 syllabus. JEE uses a few elementary properties that follow straight from the definitions; with them, many principal values come out in one line.

(i) $\sin^{-1}(-x)=-\sin^{-1}x$, $\tan^{-1}(-x)=-\tan^{-1}x$, $\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}x$;   $\cos^{-1}(-x)=\pi-\cos^{-1}x$, $\sec^{-1}(-x)=\pi-\sec^{-1}x$, $\cot^{-1}(-x)=\pi-\cot^{-1}x$ (each for $x$ in the domain).
(ii) $\operatorname{cosec}^{-1}x=\sin^{-1}\tfrac1x$ and $\sec^{-1}x=\cos^{-1}\tfrac1x$ for $|x|\ge1$; $\cot^{-1}x=\tan^{-1}\tfrac1x$ for $x\gt0$.
(iii) $\sin^{-1}x+\cos^{-1}x=\tfrac\pi2$ for $x\in[-1,1]$; $\tan^{-1}x+\cot^{-1}x=\tfrac\pi2$ for $x\in\mathbb R$.

Why (iii). Let $y=\sin^{-1}x\in\left[-\tfrac\pi2,\tfrac\pi2\right]$. Then $\tfrac\pi2-y\in[0,\pi]$ and $\cos\!\left(\tfrac\pi2-y\right)=\sin y=x$, so $\tfrac\pi2-y=\cos^{-1}x$.

Worked example. Solve $\sin^{-1}x=2\cos^{-1}x$. Solution. By (iii), $\tfrac\pi2-\cos^{-1}x=2\cos^{-1}x$, so $\cos^{-1}x=\tfrac\pi6$ and $x=\cos\tfrac\pi6=\tfrac{\sqrt3}2$. Check: $\sin^{-1}\tfrac{\sqrt3}2=\tfrac\pi3=2\cdot\tfrac\pi6$ ✓.

Exercises

I
Find the principal value.
  • 1 $\sin^{-1}\tfrac{\sqrt3}2$
  • 2 $\sin^{-1}(-1)$
  • 3 $\cos^{-1}\!\left(-\tfrac{\sqrt3}2\right)$
  • 4 $\cos^{-1}\tfrac1{\sqrt2}$
  • 5 $\tan^{-1}\sqrt3$
  • 6 $\tan^{-1}\!\left(-\tfrac1{\sqrt3}\right)$
  • 7 $\operatorname{cosec}^{-1}(-2)$
  • 8 $\sec^{-1}\sqrt2$
  • 9 $\sec^{-1}\!\left(-\tfrac2{\sqrt3}\right)$
  • 10 $\cot^{-1}\!\left(-\sqrt3\right)$
  • 11 $\operatorname{cosec}^{-1}\tfrac2{\sqrt3}$
  • 12 $\cot^{-1}0$
Check: principal-value calculator
—

Type numbers with + − * / ( ), √ (or sqrt) and π (or pi), e.g. -1/√3, 2/sqrt(3).

1 $\sin\tfrac\pi3=\tfrac{\sqrt3}2$, $\tfrac\pi3\in\left[-\tfrac\pi2,\tfrac\pi2\right]$: $\tfrac\pi3$.
2 $\sin\!\left(-\tfrac\pi2\right)=-1$: $-\tfrac\pi2$.
3 Reference angle $\tfrac\pi6$; branch $[0,\pi]$, $x\lt0$: $\pi-\tfrac\pi6=\tfrac{5\pi}6$.
4 $\cos\tfrac\pi4=\tfrac1{\sqrt2}$: $\tfrac\pi4$.
5 $\tan\tfrac\pi3=\sqrt3$: $\tfrac\pi3$.
6 Reference angle $\tfrac\pi6$; branch centred at $0$, $x\lt0$: $-\tfrac\pi6$.
7 $\operatorname{cosec}\tfrac\pi6=2$; centred branch, $x\lt0$: $-\tfrac\pi6$.
8 $\cos\tfrac\pi4=\tfrac1{\sqrt2}$, so $\sec\tfrac\pi4=\sqrt2$: $\tfrac\pi4$.
9 $\sec\tfrac\pi6=\tfrac2{\sqrt3}$; branch in $[0,\pi]$, $x\lt0$: $\pi-\tfrac\pi6=\tfrac{5\pi}6$. Check: $\cos\tfrac{5\pi}6=-\tfrac{\sqrt3}2$ ✓.
10 $\cot\tfrac\pi6=\sqrt3$; branch $(0,\pi)$, $x\lt0$: $\tfrac{5\pi}6$.
11 $\sin\tfrac\pi3=\tfrac{\sqrt3}2$, so $\operatorname{cosec}\tfrac\pi3=\tfrac2{\sqrt3}$: $\tfrac\pi3$.
12 $\cot\tfrac\pi2=\dfrac{\cos(\pi/2)}{\sin(\pi/2)}=0$ and $\tfrac\pi2\in(0,\pi)$: $\tfrac\pi2$.
II
Evaluate.
  • 1 $\tan^{-1}\sqrt3-\sec^{-1}(-2)$
  • 2 $\sin^{-1}\!\left(-\tfrac12\right)+\cos^{-1}\!\left(-\tfrac{\sqrt3}2\right)$
  • 3 $\cot^{-1}(-1)+\operatorname{cosec}^{-1}\sqrt2$
  • 4 $2\cos^{-1}0-3\tan^{-1}1+\sin^{-1}1$
  • 5 $\tan^{-1}1+\cos^{-1}(-1)+\sin^{-1}\!\left(-\tfrac{\sqrt3}2\right)$
  • 6 $\cos\!\left(2\sin^{-1}\tfrac12\right)$
  • 7 $\sin\!\left(\sin^{-1}0.3\right)+\cos\!\left(\cos^{-1}(-0.8)\right)$
  • 8 Is $\sin^{-1}\!\left(\sin^{-1}1\right)$ defined? Is $\sin^{-1}\!\left(\sin^{-1}\tfrac12\right)$ defined? Explain.
1 $\tfrac\pi3-\tfrac{2\pi}3=-\tfrac\pi3$.
2 $-\tfrac\pi6+\tfrac{5\pi}6=\tfrac{4\pi}6=\tfrac{2\pi}3$.
3 $\tfrac{3\pi}4+\tfrac\pi4=\pi$.
4 $2\cdot\tfrac\pi2-3\cdot\tfrac\pi4+\tfrac\pi2=\pi-\tfrac{3\pi}4+\tfrac\pi2=\tfrac{3\pi}4$.
5 $\tfrac\pi4+\pi-\tfrac\pi3=\tfrac{3\pi+12\pi-4\pi}{12}=\tfrac{11\pi}{12}$.
6 $2\sin^{-1}\tfrac12=\tfrac\pi3$, and $\cos\tfrac\pi3=\tfrac12$.
7 By Theorem 2.1 (both inputs lie in $[-1,1]$): $0.3+(-0.8)=-0.5$.
8 $\sin^{-1}1=\tfrac\pi2\approx1.571\gt1$, outside the domain of $\sin^{-1}$: not defined. But $\sin^{-1}\tfrac12=\tfrac\pi6\approx0.5236\le1$, so $\sin^{-1}\!\left(\tfrac\pi6\right)$ is defined (about $0.5511$).
III
Domains, ranges and equations.
  • 1 Find the domain of $\sin^{-1}(3x)$.
  • 2 Find the domain of $\cos^{-1}\!\left(\tfrac x2-1\right)$.
  • 3 Find the domain of $\sin^{-1}\!\left(x^2\right)$.
  • 4 Find the domain of $\sec^{-1}(2x+1)$.
  • 5 Find the range of $2\sin^{-1}x+\pi$.
  • 6 Find the range of $3\tan^{-1}x$.
  • 7 Solve $\cos^{-1}(2x-1)=\tfrac\pi3$.
  • 8 Solve $\tan^{-1}(x+1)=-\tfrac\pi4$.
  • 9 Explain why neither $\cos^{-1}x=-\tfrac\pi3$ nor $\sin^{-1}x=2$ has a solution.
  • 10 Without a calculator, decide which is larger: $\sin^{-1}0.9$ or $\cos^{-1}0.9$. (Hint: both equal $\tfrac\pi4$ at $x=\tfrac1{\sqrt2}\approx0.707$.)
1 $-1\le3x\le1$: $\left[-\tfrac13,\tfrac13\right]$.
2 $-1\le\tfrac x2-1\le1\iff0\le\tfrac x2\le2\iff0\le x\le4$: $[0,4]$.
3 $x^2\ge0\gt-1$ always, so we need only $x^2\le1$: $[-1,1]$.
4 Need $|2x+1|\ge1$: $2x+1\ge1$ ($x\ge0$) or $2x+1\le-1$ ($x\le-1$). Domain $(-\infty,-1]\cup[0,\infty)$.
5 $-\tfrac\pi2\le\sin^{-1}x\le\tfrac\pi2$, so $-\pi\le2\sin^{-1}x\le\pi$ and the range is $[0,2\pi]$.
6 $-\tfrac\pi2\lt\tan^{-1}x\lt\tfrac\pi2$ with every value taken: range $\left(-\tfrac{3\pi}2,\tfrac{3\pi}2\right)$ (open ends).
7 $\tfrac\pi3\in[0,\pi]$, so the equation means $2x-1=\cos\tfrac\pi3=\tfrac12$: $x=\tfrac34$. Check: $\cos^{-1}\tfrac12=\tfrac\pi3$ ✓.
8 $-\tfrac\pi4\in\left(-\tfrac\pi2,\tfrac\pi2\right)$, so $x+1=\tan\!\left(-\tfrac\pi4\right)=-1$: $x=-2$.
9 Every value of $\cos^{-1}$ lies in $[0,\pi]$, which does not contain $-\tfrac\pi3$. Every value of $\sin^{-1}$ lies in $\left[-\tfrac\pi2,\tfrac\pi2\right]$, and $2\gt\tfrac\pi2\approx1.571$.
10 $\sin$ increases on $\left[-\tfrac\pi2,\tfrac\pi2\right]$, so $\sin^{-1}$ increases; $\cos$ decreases on $[0,\pi]$, so $\cos^{-1}$ decreases. As $0.9\gt\tfrac1{\sqrt2}$: $\sin^{-1}0.9\gt\tfrac\pi4\gt\cos^{-1}0.9$. (Values: $1.1198$ and $0.4510$.)
IV
Angles in the real world. Give answers in radians (4 decimal places) and in degrees.
  • 1 Accessibility. A ramp $8$ m long rises $0.5$ m. Find its angle with the ground. Accessibility guidelines in India and many other countries ask for a gradient no steeper than $1$ in $12$, i.e. $\tan\theta\le\tfrac1{12}$. Does this ramp meet them?
  • 2 Shadows. A $30$ m tower casts a shadow $30\sqrt3$ m long. Find the angle of elevation of the Sun exactly.
  • 3 Safety. A $6$ m ladder stands with its foot $2$ m from a wall. Find the angle it makes with the ground. (Safety advice for ladders suggests about $75^\circ$.)
  • 4 Roads. A hill road rises $1$ m for every $20$ m measured horizontally. Find the angle of the road with the horizontal.
  • 5 Rocket tracking. A tracking station $25$ km from a launch pad sees a rocket climbing vertically. Find the angle of elevation when the rocket is $10$ km high, and the height at which the angle is exactly $\tfrac\pi4$.
1 $\sin\theta=\tfrac{0.5}8=0.0625$, $\theta=\sin^{-1}0.0625\approx0.0625$ rad $\approx3.58^\circ$. The horizontal run is $\sqrt{64-0.25}\approx7.984$ m, so $\tan\theta\approx0.0626\lt\tfrac1{12}\approx0.0833$: yes, it meets the guideline (the limit angle is $\tan^{-1}\tfrac1{12}\approx4.76^\circ$).
2 $\tan\theta=\tfrac{30}{30\sqrt3}=\tfrac1{\sqrt3}$, so $\theta=\tan^{-1}\tfrac1{\sqrt3}=\tfrac\pi6$ ($30^\circ$, $0.5236$ rad).
3 $\cos\theta=\tfrac26=\tfrac13$, $\theta=\cos^{-1}\tfrac13\approx1.2310$ rad $\approx70.53^\circ$ — a little flatter than the advised $75^\circ$.
4 $\tan\theta=\tfrac1{20}=0.05$, $\theta=\tan^{-1}0.05\approx0.0500$ rad $\approx2.86^\circ$.
5 $\tan\theta=\tfrac{10}{25}=0.4$, $\theta=\tan^{-1}0.4\approx0.3805$ rad $\approx21.80^\circ$. The angle is $\tfrac\pi4$ when $\tfrac h{25}=\tan\tfrac\pi4=1$: $h=25$ km.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for principal values (1-mark MCQs and Assertion–Reason) and for the value of a short expression built from them (2 marks); marks are lost by giving an angle outside the branch. JEE asks for the domain of a function built from $\sin^{-1}$ or $\cos^{-1}$ and counts its integer points.

E1. Board · 1 mark The principal value of $\sec^{-1}\!\left(-\sqrt2\right)$ is  (A) $-\tfrac\pi4$  (B) $\tfrac{3\pi}4$  (C) $\tfrac{5\pi}4$  (D) $\tfrac\pi4$

Idea $\sec y=-\sqrt2\iff\cos y=-\tfrac1{\sqrt2}$, with $y\in[0,\pi]-\left\{\tfrac\pi2\right\}$. The reference angle is $\tfrac\pi4$ and $x\lt0$, so $y=\pi-\tfrac\pi4=\tfrac{3\pi}4$.
Answer (B). Trap (A): $-\tfrac\pi4$ has secant $+\sqrt2$ and is not in the branch. (C) has the right secant but is outside $[0,\pi]$; (D) has secant $+\sqrt2$.

E2. Board · 1 mark Assertion–Reason. Assertion (A): the principal value of $\tan^{-1}(-1)$ is $\tfrac{3\pi}4$. Reason (R): the principal value branch of $\tan^{-1}$ is $\left(-\tfrac\pi2,\tfrac\pi2\right)$.  (A) Both A and R are true and R is the correct explanation of A  (B) Both A and R are true but R is not the correct explanation of A  (C) A is true but R is false  (D) A is false but R is true

Idea R is Definition 2.4, so R is true. $\tfrac{3\pi}4$ does satisfy $\tan\tfrac{3\pi}4=-1$, but it is not in $\left(-\tfrac\pi2,\tfrac\pi2\right)$; the principal value is $-\tfrac\pi4$. So A is false.
Answer (D). ($\tfrac{3\pi}4$ is the principal value of $\cot^{-1}(-1)$ — a common mix-up.)

E3. Board · 2 marks Find the value of $\cot^{-1}\!\left(-\tfrac1{\sqrt3}\right)+\operatorname{cosec}^{-1}(-2)+\sec^{-1}2$.

Term 1 $\cot\tfrac\pi3=\tfrac1{\sqrt3}$; branch $(0,\pi)$ and $x\lt0$, so $\cot^{-1}\!\left(-\tfrac1{\sqrt3}\right)=\pi-\tfrac\pi3=\tfrac{2\pi}3$. (½ mark)
Terms 2, 3 $\operatorname{cosec}^{-1}(-2)=-\tfrac\pi6$ (centred branch) and $\sec^{-1}2=\tfrac\pi3$. (1 mark)
Answer $\tfrac{2\pi}3-\tfrac\pi6+\tfrac\pi3=\tfrac{4\pi-\pi+2\pi}6=\tfrac{5\pi}6$. (½ mark)

E4. Board · Case study A kite is flown on a taut string $50$ m long, held at ground level. When the kite is $h$ m above the ground, the string makes an angle $\theta=\sin^{-1}\tfrac h{50}$ with the ground. (i) Find $\theta$ when $h=25$ and when $h=25\sqrt2$. [2 marks] (ii) For which values of $h$ is the formula defined as a function, and what is the range of $\theta$ for the physically possible heights $0\le h\le50$? [1 mark] (iii) The angle of the string with the vertical is $\cos^{-1}\tfrac h{50}$. Find it when $h=25$. [1 mark]

(i) $h=25$: $\theta=\sin^{-1}\tfrac12=\tfrac\pi6$ ($30^\circ$). $h=25\sqrt2$: $\theta=\sin^{-1}\tfrac1{\sqrt2}=\tfrac\pi4$ ($45^\circ$).
(ii) $\sin^{-1}u$ needs $-1\le u\le1$, so $-50\le h\le50$. For $0\le h\le50$, $\tfrac h{50}\in[0,1]$ and $\theta$ takes every value in $\left[0,\tfrac\pi2\right]$.
(iii) $\cos^{-1}\tfrac12=\tfrac\pi3$ ($60^\circ$) — and indeed $\tfrac\pi6+\tfrac\pi3=\tfrac\pi2$, as the two angles of a right triangle must be.

E5. JEE Main The domain of $f(x)=\cos^{-1}\!\left(\dfrac{x^2-5}4\right)$ is  (A) $[1,3]$  (B) $[-3,3]$  (C) $[-3,-1]\cup[1,3]$  (D) $[-1,1]$

Method $-1\le\dfrac{x^2-5}4\le1\iff-4\le x^2-5\le4\iff1\le x^2\le9\iff1\le|x|\le3$.
Answer (C). Trap (A) forgets the negative square roots; (B) forgets the lower bound $x^2\ge1$ (at $x=0$ the input is $-\tfrac54$). ⏱ Fast route: test $x=0$ (input $-1.25$, rejected) and $x=-2$ (input $-0.25$, accepted) — only (C) survives.

E6. JEE Advanced (Numerical answer.) Find the number of integers in the domain of $g(x)=\sin^{-1}\!\left(\log_2\dfrac x2\right)$.

Method We need $x\gt0$ (for the logarithm) and $-1\le\log_2\tfrac x2\le1$. Since $\log_2$ is increasing, this means $2^{-1}\le\tfrac x2\le2^1$, i.e. $1\le x\le4$.
Answer The domain is $[1,4]$, containing the integers $1,2,3,4$: the answer is $\mathbf 4$. Check $x=4$: $\log_22=1$ and $\sin^{-1}1=\tfrac\pi2$ ✓.
🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. Are the ramps around your school or neighbourhood gentle enough for a wheelchair user? Use inverse trigonometric functions to turn tape-measure readings into angles and compare them with the guideline.

Why it matters. Accessibility guidelines in India (and in most countries) recommend a ramp gradient of at most $1$ in $12$ — a rise of $1$ unit for every $12$ units of horizontal run. Too steep a ramp is tiring or unsafe for wheelchair users, parents with prams and elderly people. Engineers state the limit as a ratio, a percentage or an angle; $\tan^{-1}$ converts between them.

What to do.

  1. Find three to five ramps. For each, measure the vertical rise $h$ and either the horizontal run $d$ or the sloping length $\ell$ (with permission and safely).
  2. Compute the angle in two ways where possible: $\theta=\tan^{-1}\tfrac hd$ and $\theta=\sin^{-1}\tfrac h\ell$. Explain why both are principal values that equal the physical angle.
  3. Compare each angle with the limit $\tan^{-1}\tfrac1{12}$. Rank the ramps from gentlest to steepest.
  4. If your phone has a free clinometer/level app, measure each angle directly and compare with your calculation. Estimate the measurement error.

Tools: Tape measure; spreadsheet (DEGREES(ATAN(h/d)), DEGREES(ASIN(h/l))); optional phone clinometer.

Hand in: A table of $h$, $d$ or $\ell$, the computed angles in radians and degrees, the phone readings, and a short recommendation for any ramp that is too steep (how long would it need to be for the same rise?).

Going further. Railways and roads use the same idea: a road gradient of $10\%$ means $\tan\theta=0.1$. Find the angle, and compare it with the steepest gradient you can find on a hill road in India.

Instructor note: $\tan^{-1}\tfrac1{12}\approx0.0831$ rad $\approx4.76^\circ$; $\tan^{-1}0.1\approx0.0997$ rad $\approx5.71^\circ$. For small angles $\sin^{-1}\tfrac h\ell$ and $\tan^{-1}\tfrac hd$ differ by well under $0.1^\circ$, so measurement error dominates. A ramp with rise $0.75$ m needs a horizontal run of at least $9$ m.

🧪 Self-check quiz

0/5 correct

Q1. The principal value of $\sin^{-1}\!\left(-\tfrac12\right)$ is

Q2. The principal value branch of $\cos^{-1}$ is

Q3. For $-1\le x\le1$, $\sin^{-1}x$ means

Q4. The domain of $\sec^{-1}x$ is

Q5. $\cos^{-1}(-1)+\tan^{-1}1$ equals

← Chapter 2 — Inverse Trigonometric Functions2.2 Graphs and Principal Values →