2.2 Graphs and Principal Values
Reflect each restricted trigonometric graph in the line $y=x$ and you have the graph of its inverse. The same picture explains why $\sin^{-1}(\sin\theta)$ is not always $\theta$ — and how to find it when it isn't.
Type $\sin\!\left(\tfrac{2\pi}3\right)$ into a calculator in radian mode, then press $\sin^{-1}$. You might expect to get back $\tfrac{2\pi}3\approx2.0944$. You get $1.0472$, which is $\tfrac\pi3$. The calculator is not wrong: $\tfrac{2\pi}3$ and $\tfrac\pi3$ have the same sine, and $\sin^{-1}$ must answer with the angle in its branch $\left[-\tfrac\pi2,\tfrac\pi2\right]$. Undoing a function only gives back what you started with if you started inside the branch.
This section makes that precise with pictures. First we see that the graph of any inverse function is the mirror image of the original graph in the line $y=x$, and draw all six inverse trigonometric graphs that way. Then we use the graphs to evaluate expressions such as $\sin^{-1}(\sin\theta)$, $\cos^{-1}(\cos\theta)$ and $\tan^{-1}(\tan\theta)$ for any angle $\theta$ — a favourite of Board and JEE papers.
The Graph of an Inverse: Reflection in y = x
By Definition 2.1, $f^{-1}(b)=a$ exactly when $f(a)=b$. In terms of points: $(a,b)$ lies on the graph of $f$ exactly when $(b,a)$ lies on the graph of $f^{-1}$. Swapping the coordinates of a point is a geometric move — it reflects the point in the line $y=x$.
Proof. The first statement is Definition 2.1 written with points. For the second, take $P(a,b)$ and $P'(b,a)$ with $a\ne b$. The mid-point of $PP'$ is $\left(\tfrac{a+b}2,\tfrac{a+b}2\right)$, which lies on $y=x$; the slope of $PP'$ is $\dfrac{a-b}{b-a}=-1$, so $PP'$ is perpendicular to $y=x$ (slope $1$, and $1\times(-1)=-1$). Hence $P'$ is the mirror image of $P$ in $y=x$. (If $a=b$, the point lies on the mirror and is its own image.) Reflecting every point of the graph of $f$ therefore gives every point of the graph of $f^{-1}$. $\square$
The Six Graphs and What They Tell You
Reflecting each principal branch in $y=x$ gives the six graphs below (switch between them in Interactive 1). Three features carry over from the original functions:
- Domain and range swap. The domain of $f^{-1}$ is the range of the restricted $f$, and the range of $f^{-1}$ is the branch.
- Increasing stays increasing, decreasing stays decreasing. If $f$ rises on its branch, so does its mirror image. Sine and tangent rise on their branches, cosine and cotangent fall; secant rises and cosecant falls on each piece of theirs.
- Vertical asymptotes become horizontal ones. $\tan x$ shoots up at $x=\pm\tfrac\pi2$, so $\tan^{-1}x$ levels off towards $y=\pm\tfrac\pi2$ as $x\to\pm\infty$.
| graph | domain | range | behaviour | key points / asymptotes |
|---|---|---|---|---|
| $y=\sin^{-1}x$ | $[-1,1]$ | $\left[-\tfrac\pi2,\tfrac\pi2\right]$ | increasing | $\left(-1,-\tfrac\pi2\right)$, $(0,0)$, $\left(1,\tfrac\pi2\right)$ |
| $y=\cos^{-1}x$ | $[-1,1]$ | $[0,\pi]$ | decreasing | $(-1,\pi)$, $\left(0,\tfrac\pi2\right)$, $(1,0)$ |
| $y=\tan^{-1}x$ | $\mathbb R$ | $\left(-\tfrac\pi2,\tfrac\pi2\right)$ | increasing | $(0,0)$; $y=\pm\tfrac\pi2$ |
| $y=\cot^{-1}x$ | $\mathbb R$ | $(0,\pi)$ | decreasing | $\left(0,\tfrac\pi2\right)$; $y=0$ (as $x\to\infty$), $y=\pi$ (as $x\to-\infty$) |
| $y=\sec^{-1}x$ | $\mathbb R-(-1,1)$ | $[0,\pi]-\left\{\tfrac\pi2\right\}$ | increasing on each piece | $(-1,\pi)$, $(1,0)$; $y=\tfrac\pi2$ |
| $y=\operatorname{cosec}^{-1}x$ | $\mathbb R-(-1,1)$ | $\left[-\tfrac\pi2,\tfrac\pi2\right]-\{0\}$ | decreasing on each piece | $\left(-1,-\tfrac\pi2\right)$, $\left(1,\tfrac\pi2\right)$; $y=0$ |
Evaluating sin⁻¹(sin θ) and Its Cousins
Theorem 2.1 says $\sin^{-1}(\sin\theta)=\theta$ when $\theta$ is in the branch. For any other $\theta$ the value is still defined — $\sin\theta$ is always in $[-1,1]$ — but it is a different angle: the one angle in the branch with the same sine as $\theta$. The Class 11 allied-angle facts tell us which angles share a sine, cosine or tangent:
- same sine as $\theta$: $\pi-\theta$ and $\theta\pm2\pi,\ \theta\pm4\pi,\dots$;
- same cosine as $\theta$: $-\theta$, $2\pi-\theta$ and $\theta\pm2\pi,\dots$;
- same tangent as $\theta$: $\theta\pm\pi,\ \theta\pm2\pi,\dots$.
(a) For every real $\theta$, $\sin^{-1}(\sin\theta)$ is the unique angle $\varphi\in\left[-\tfrac\pi2,\tfrac\pi2\right]$ with $\sin\varphi=\sin\theta$. Explicitly, if $k$ is an integer and $k\pi-\tfrac\pi2\le\theta\le k\pi+\tfrac\pi2$, then $$\sin^{-1}(\sin\theta)=(-1)^k(\theta-k\pi).$$ So it equals $\theta$ on $\left[-\tfrac\pi2,\tfrac\pi2\right]$, $\pi-\theta$ on $\left[\tfrac\pi2,\tfrac{3\pi}2\right]$, $\theta-2\pi$ on $\left[\tfrac{3\pi}2,\tfrac{5\pi}2\right]$, $-\pi-\theta$ on $\left[-\tfrac{3\pi}2,-\tfrac\pi2\right]$, and so on.
(b) For every real $\theta$, $\cos^{-1}(\cos\theta)$ is the unique $\varphi\in[0,\pi]$ with $\cos\varphi=\cos\theta$: if $(2k-1)\pi\le\theta\le(2k+1)\pi$ then $\cos^{-1}(\cos\theta)=|\theta-2k\pi|$. So it equals $\theta$ on $[0,\pi]$, $2\pi-\theta$ on $[\pi,2\pi]$ and $-\theta$ on $[-\pi,0]$.
(c) For every real $\theta$ that is not an odd multiple of $\tfrac\pi2$, $\tan^{-1}(\tan\theta)$ is the unique $\varphi\in\left(-\tfrac\pi2,\tfrac\pi2\right)$ with $\tan\varphi=\tan\theta$: if $k\pi-\tfrac\pi2\lt\theta\lt k\pi+\tfrac\pi2$ then $\tan^{-1}(\tan\theta)=\theta-k\pi$.
Proof of (a). $\sin\theta\in[-1,1]$, so $\sin^{-1}(\sin\theta)$ is defined, and by Definition 2.2 it is the one angle of the branch whose sine is $\sin\theta$. For the formula, put $\varphi=(-1)^k(\theta-k\pi)$. From $k\pi-\tfrac\pi2\le\theta\le k\pi+\tfrac\pi2$ we get $-\tfrac\pi2\le\theta-k\pi\le\tfrac\pi2$, and multiplying by $\pm1$ keeps it in this symmetric interval, so $\varphi$ is in the branch. Also $\sin(\theta-k\pi)=(-1)^k\sin\theta$ (each shift by $\pi$ changes the sign of sine), so $\sin\varphi=(-1)^k(-1)^k\sin\theta=\sin\theta$. By uniqueness, $\varphi=\sin^{-1}(\sin\theta)$. Parts (b) and (c) are proved the same way, using $\cos(\theta-2k\pi)=\cos\theta$, $\cos(-\alpha)=\cos\alpha$ and $\tan(\theta-k\pi)=\tan\theta$. $\square$
- Is $\theta$ already in the branch? Then the answer is $\theta$ (Theorem 2.1). Stop.
- Otherwise move $\theta$ into the branch without changing the value, using only the allied angles listed above: for sine try $\pi-\theta$, then add or subtract $2\pi$; for cosine try $2\pi-\theta$ or $-\theta$, then $\pm2\pi$; for tangent add or subtract $\pi$ as often as needed.
- Check that the new angle is in the branch and has the same sine, cosine or tangent. (Or compute $\sin\theta$ first and find the principal value directly, as in Section 2.1.)
The other order — a trigonometric function of an inverse — is simpler, because a principal value is a known angle. If it is not a standard angle, draw a right triangle for its reference angle and use the branch to fix the sign.
1. “Inverse cancels the function, always.” $\sin^{-1}(\sin\theta)=\theta$ only for $\theta\in\left[-\tfrac\pi2,\tfrac\pi2\right]$; $\cos^{-1}(\cos\theta)=\theta$ only for $\theta\in[0,\pi]$. Writing $\sin^{-1}\!\left(\sin\tfrac{2\pi}3\right)=\tfrac{2\pi}3$ scores zero.
2. The answer has the right value but the wrong branch. For $\cos^{-1}\!\left(\cos\tfrac{7\pi}6\right)$, $-\tfrac{5\pi}6$ has the right cosine, but $\cos^{-1}$ never gives negative angles: the answer is $\tfrac{5\pi}6$. Always check both conditions.
3. Using the sine shift for tangent. $\pi-\theta$ has the same sine as $\theta$ but the opposite tangent. For $\tan^{-1}(\tan\theta)$ only shifts by multiples of $\pi$ are allowed.
4. Dropping the sign in triangle problems. A right triangle gives sizes only. $\tan\!\left(\sin^{-1}\!\left(-\tfrac5{13}\right)\right)$ is $-\tfrac5{12}$, not $\tfrac5{12}$: the angle lies in $\left[-\tfrac\pi2,0\right)$.
5. Plain numbers are radians. In $\sin^{-1}(\sin2)$ the angle is $2$ radians, not $2^\circ$: the answer is $\pi-2$, because $2\gt\tfrac\pi2$.
In JEE Main and JEE Advanced 2026; not in the CBSE Class 12 syllabus. JEE Main regularly uses the addition formula for $\tan^{-1}$, which comes from $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ (Class 11):
$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}\quad(xy\lt1),$$ $$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\dfrac{x-y}{1+xy}\quad(xy\gt-1),$$ and for $|x|\lt1$, $2\tan^{-1}x=\tan^{-1}\dfrac{2x}{1-x^2}=\sin^{-1}\dfrac{2x}{1+x^2}$. The condition matters: if $x,y\gt0$ and $xy\gt1$, then $\tan^{-1}x+\tan^{-1}y$ lies in $\left(\tfrac\pi2,\pi\right)$, outside the range of $\tan^{-1}$, and the right side of the first formula must be increased by $\pi$.
Worked example. Show that $\tan^{-1}\tfrac12+\tan^{-1}\tfrac13=\tfrac\pi4$. Solution. $xy=\tfrac16\lt1$, so the sum is $\tan^{-1}\dfrac{\tfrac12+\tfrac13}{1-\tfrac16}=\tan^{-1}\dfrac{5/6}{5/6}=\tan^{-1}1=\tfrac\pi4$. Check: $0.4636+0.3218=0.7854$ ✓. (With $x=2$, $y=3$: $xy=6\gt1$, and $\tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}(-1)=\tfrac{3\pi}4$.)
Exercises
- 1 For $y=\sin^{-1}x$, state the domain, the range, whether it is increasing or decreasing, and the end points of the graph.
- 2 Do the same for $y=\cot^{-1}x$, and give its horizontal asymptotes.
- 3 Do the same for $y=\sec^{-1}x$.
- 4 Do the same for $y=\operatorname{cosec}^{-1}x$.
- 5 Write the equations of the horizontal asymptotes of $y=\tan^{-1}x$, and say which end of the graph approaches each.
- 6 The point $(0.6,\,0.6435)$ lies (to 4 decimal places) on $y=\sin^{-1}x$. Which point on $y=\sin x$ corresponds to it, and why?
- 7 Find the domain and range of $y=\sin^{-1}\tfrac x3$ and of $y=2\cos^{-1}x+1$.
- 8 Show that the equation $\sin^{-1}x=x$ has exactly one solution. (Hint: if $y=\sin^{-1}x\gt0$, compare $\sin y$ with $y$.)
- 9 Explain using the graphs why $\cos^{-1}x=x$ has exactly one solution, and show that it lies between $0.73$ and $0.74$.
- 1 $\sin^{-1}\!\left(\sin\tfrac{3\pi}4\right)$
- 2 $\sin^{-1}\!\left(\sin\tfrac{7\pi}6\right)$
- 3 $\sin^{-1}\!\left(\sin\!\left(-\tfrac{4\pi}5\right)\right)$
- 4 $\cos^{-1}\!\left(\cos\tfrac{5\pi}4\right)$
- 5 $\cos^{-1}\!\left(\cos\!\left(-\tfrac\pi3\right)\right)$
- 6 $\cos^{-1}\!\left(\cos\tfrac{13\pi}6\right)$
- 7 $\tan^{-1}\!\left(\tan\tfrac{5\pi}6\right)$
- 8 $\tan^{-1}\!\left(\tan\tfrac{7\pi}4\right)$
- 9 $\sec^{-1}\!\left(\sec\tfrac{7\pi}4\right)$
- 10 $\cot^{-1}\!\left(\cot\!\left(-\tfrac\pi3\right)\right)$
- 11 $\operatorname{cosec}^{-1}\!\left(\operatorname{cosec}\tfrac{5\pi}6\right)$
- 12 $\sin^{-1}(\sin3)$
- 13 $\cos^{-1}(\cos4)$
Type an angle in radians using numbers, + − * / ( ) and π (or pi), e.g. -4π/5, 13*pi/6, 3.
- 1 $\sin\!\left(\cos^{-1}\tfrac5{13}\right)$
- 2 $\tan\!\left(\sin^{-1}\!\left(-\tfrac35\right)\right)$
- 3 $\cos\!\left(\tan^{-1}(-2)\right)$
- 4 $\sec\!\left(\cot^{-1}\!\left(-\tfrac34\right)\right)$
- 5 $\sin\!\left(\tfrac\pi2-\sin^{-1}\tfrac13\right)$
- 6 $\cos\!\left(2\cos^{-1}\tfrac14\right)$
- 7 $\sin\!\left(2\tan^{-1}\tfrac12\right)$
- 8 For which $x\in[-2\pi,2\pi]$ is $\cos^{-1}(\cos x)=x$? For which $x\in[-2\pi,2\pi]$ is $\sin^{-1}(\sin x)=\pi-x$?
- 1 Giant wheel. A rider on a wheel of radius $10$ m is at height $h=10\sin\theta$ metres above the centre, where $\theta=\tfrac\pi{30}t$ radians after $t$ seconds, $0\le t\le60$. A student solves $h=5$ with a calculator: $t=\tfrac{30}\pi\sin^{-1}0.5=5$ s. Find all times in $[0,60]$ when the rider is $5$ m above the centre, and all times when the rider is $5$ m below it. Why did the calculator give only one?
- 2 Electric circuits. In an AC circuit with resistance $R=40\ \Omega$ and reactance $X=30\ \Omega$, the phase angle is $\varphi=\tan^{-1}\tfrac XR$. Find $\varphi$ in radians and degrees, and $\varphi$ when the reactance is $X=-30\ \Omega$ (a capacitor).
- 3 Cinema seats. A screen's lower edge is $2$ m and its upper edge $5$ m above a viewer's eye level. Standing $x$ metres from the wall, the viewer sees the screen under the angle $\theta=\tan^{-1}\tfrac5x-\tan^{-1}\tfrac2x$. Compute $\theta$ for $x=2$, $4$ and $6$ m. Which of these positions gives the largest view?
- 4 True or false? Give a reason or a counterexample. (a) $\sin^{-1}(\sin x)=x$ for every real $x$. (b) $\sin\!\left(\sin^{-1}x\right)=x$ for every $x\in[-1,1]$. (c) $\cos^{-1}(\cos x)=|x|$ for every $x\in[-\pi,\pi]$. (d) $\tan^{-1}(\tan x)=x$ wherever $\tan x$ is defined.
How this is examined: the Board asks for values such as $\sin^{-1}\!\left(\sin\tfrac{3\pi}5\right)$ in 1- and 2-mark questions and occasionally for a sketch of an inverse graph; the classic error is to “cancel” and write $\tfrac{3\pi}5$. JEE uses angles in plain radians ($\sin^{-1}(\sin10)$) and piecewise expressions built from Theorem 2.3.
E1. Board · 1 mark The value of $\sin^{-1}\!\left(\sin\tfrac{4\pi}5\right)$ is (A) $\tfrac{4\pi}5$ (B) $\tfrac\pi5$ (C) $-\tfrac\pi5$ (D) $\tfrac{6\pi}5$
E2. Board · 1 mark Assertion–Reason. Assertion (A): $\cos^{-1}\!\left(\cos\tfrac{7\pi}6\right)=\tfrac{7\pi}6$. Reason (R): $\cos^{-1}(\cos\theta)=\theta$ for every $\theta\in[0,\pi]$. (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
E3. Board · 2 marks Find the value of $\sin^{-1}\!\left(\sin\tfrac{3\pi}5\right)+\cos^{-1}\!\left(\cos\tfrac{6\pi}5\right)$.
E4. Board · 3 marks Sketch $y=\sin^{-1}x$ and $y=\cos^{-1}x$ on the same axes. Find their point of intersection, and the values of $x$ for which $\sin^{-1}x\gt\cos^{-1}x$.
E5. JEE Main $\sin^{-1}(\sin10)$ equals (A) $10$ (B) $10-3\pi$ (C) $3\pi-10$ (D) $\pi-10$
E6. JEE Advanced (One or more options correct.) Let $f(x)=\sin^{-1}(\sin x)+\cos^{-1}(\cos x)$ for real $x$. Then (A) $f(x)=\pi$ for all $x\in\left[\tfrac\pi2,\pi\right]$ (B) $f(x)=2x$ for all $x\in\left[0,\tfrac\pi2\right]$ (C) $f(x)=0$ for all $x\in\left[-\tfrac\pi2,0\right]$ (D) $f$ is periodic with period $\pi$
Research question. A phone works out which way up it is from its accelerometer, which measures the two components $(a_x,a_y)$ of gravity in the plane of the screen. The tilt angle $\theta$ satisfies $\tan\theta=\tfrac{a_y}{a_x}$. Why does $\tan^{-1}\tfrac{a_y}{a_x}$ give the wrong angle half the time, and how do programmers fix it?
Why it matters. Screen rotation, drone and robot balance, game controllers and the compass in maps all turn a pair of sensor readings into an angle. Because $\tan^{-1}$ only returns angles in $\left(-\tfrac\pi2,\tfrac\pi2\right)$, a naive formula confuses directions that differ by $\pi$ — a tilt to the left reads as a tilt to the right.
What to do.
- Install a free sensor-logging app and record $(a_x,a_y)$ while you turn the phone slowly through a full circle in a vertical plane (or simulate it with $a_x=9.8\cos\theta$, $a_y=9.8\sin\theta$ in a spreadsheet for $\theta=0^\circ,15^\circ,\dots,360^\circ$).
- Compute $\tan^{-1}\tfrac{a_y}{a_x}$ for each reading and plot it against the true angle. Describe the graph and link it to Theorem 2.3(c).
- Design a rule using the signs of $a_x$ and $a_y$ that corrects the angle to the interval $(-\pi,\pi]$ (hint: add or subtract $\pi$ when $a_x\lt0$), and test it.
- Compare your rule with the spreadsheet function
ATAN2(note its argument order in your software).
Tools: Phone with a sensor app; spreadsheet (ATAN, ATAN2, DEGREES).
Hand in: Your data or simulation, the two graphs (naive and corrected), your correction rule in words and as a formula, and one paragraph explaining the link with principal value branches.
Going further. GPS and mapping software find the bearing from one place to another with the same two-argument idea. Find the bearing from Chennai to Delhi using approximate coordinates and a flat-map approximation.
🧪 Self-check quiz
Q1. The graph of $y=\sin^{-1}x$ is obtained from the graph of $y=\sin x$, $-\tfrac\pi2\le x\le\tfrac\pi2$, by reflecting it in
Q2. $\cos^{-1}\!\left(\cos\tfrac{4\pi}3\right)=$
Q3. The horizontal asymptotes of $y=\tan^{-1}x$ are
Q4. $\sin^{-1}(\sin x)=x$ holds exactly when
Q5. $\tan\!\left(\cos^{-1}\tfrac35\right)=$