Class 12 Maths, Worked in Full Past papersGet updates
Chapter 2 · Inverse Trigonometric Functions

Example 2: Principal value of cot⁡−1(−13)\cot^{-1}\left(\frac{-1}{\sqrt3}\right)

ExamplesPrincipal values · Section 2.2
The question

Find the principal value of cot⁡−1(−13)\cot^{-1}\left(\dfrac{-1}{\sqrt3}\right)⁠.

Exam answer

Let cot⁡−1(−13)=y\cot^{-1}\left(-\frac{1}{\sqrt3}\right) = y⁠. Using cot⁡(π−θ)=−cot⁡θ\cot(\pi - \theta) = -\cot\theta⁠,

cot⁡y=−13=−cot⁡π3=cot⁡(π−π3)=cot⁡2π3\begin{aligned} \cot y &= -\frac{1}{\sqrt3} = -\cot\frac{\pi}{3} \\ &= \cot\left(\pi - \frac{\pi}{3}\right) = \cot\frac{2\pi}{3} \end{aligned}

Range of principal value branch of cot⁡−1\cot^{-1} is (0,π)(0, \pi)⁠, and 2π3∈(0,π)\frac{2\pi}{3} \in (0, \pi)⁠.

∴\therefore Principal value of cot⁡−1(−13)\cot^{-1}\left(-\dfrac{1}{\sqrt3}\right) is 2π3\dfrac{2\pi}{3}⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The principal range of cot⁡−1\cot^{-1} is (0,π)(0, \pi)⁠, the upper half of the unit circle, not (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) like tan⁡−1\tan^{-1}⁠. A negative cotangent therefore cannot come from a negative angle. It must come from an angle between π2\frac{\pi}{2} and π\pi (the second quadrant). The reference angle with cot⁡=13\cot = \frac{1}{\sqrt3} is π3\frac{\pi}{3}⁠, and the identity cot⁡(π−θ)=−cot⁡θ\cot(\pi-\theta) = -\cot\theta moves it to π−π3=2π3\pi - \frac{\pi}{3} = \frac{2\pi}{3}⁠.

Given, to find, and the plan
Given
The real number −13-\frac{1}{\sqrt3}⁠. (Every real number is in the domain R\R of cot⁡−1\cot^{-1}⁠.)
To find
The unique yy in (0,π)(0,\pi) with cot⁡y=−13\cot y = -\frac{1}{\sqrt3}⁠.
Plan
(1) Set y=cot⁡−1(−13)y = \cot^{-1}\left(-\frac{1}{\sqrt3}\right) and write the two conditions on yy⁠. (2) Find the reference angle for the size 13\frac{1}{\sqrt3}⁠. (3) Use cot⁡(π−θ)=−cot⁡θ\cot(\pi-\theta) = -\cot\theta to get the sign right. (4) Check the angle is in (0,π)(0,\pi) and that no other angle there works.

Basics you need

1Cotangent on the unit circle

If the angle θ\theta takes you to the point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta)⁠, then cot⁡θ=cos⁡θsin⁡θ=xy\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{x}{y}⁠. It is defined whenever sin⁡θ≠0\sin\theta \ne 0⁠, that is, θ≠nπ\theta \ne n\pi⁠. Basics: tan, cot, sec and cosec

2The special angle π3\frac{\pi}{3}

cos⁡π3=12\cos\frac{\pi}{3} = \frac12 and sin⁡π3=32\sin\frac{\pi}{3} = \frac{\sqrt3}{2}⁠, so cot⁡π3=1/23/2=13\cot\frac{\pi}{3} = \frac{1/2}{\sqrt3/2} = \frac{1}{\sqrt3}⁠. Basics: special angles and reference angles

3Supplementary angles

The angle π−θ\pi - \theta is the mirror image of θ\theta in the yy-axis: same height, opposite xx⁠. So sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta) = \sin\theta⁠, cos⁡(π−θ)=−cos⁡θ\cos(\pi-\theta) = -\cos\theta⁠, and dividing, cot⁡(π−θ)=−cot⁡θ\cot(\pi-\theta) = -\cot\theta⁠.

4The principal value branch of cot⁡−1\cot^{-1}

cot⁡−1:R→(0,π)\cot^{-1} : \R \to (0,\pi)⁠. The endpoints are left out because cot⁡0\cot 0 and cot⁡π\cot\pi are not defined. Basics: what an inverse needs

Solution

  1. Name the answer and write down what it must satisfy

    Let y=cot⁡−1(−13)y = \cot^{-1}\left(-\frac{1}{\sqrt3}\right)⁠. By the definition of the principal value branch,

    cot⁡y=−13and0<y<π.\cot y = -\frac{1}{\sqrt3} \qquad\text{and}\qquad 0 < y < \pi.
    As with every inverse trigonometric value, the equation alone has infinitely many solutions (cot repeats every π\pi⁠), and the range condition picks out one. Writing both down at the start stops us from grabbing the first angle that happens to fit the equation.
    The ranges differ from function to function: [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] for sin⁡−1\sin^{-1}⁠, [0,π][0,\pi] for cos⁡−1\cos^{-1}⁠, (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) for tan⁡−1\tan^{-1} and (0,π)(0,\pi) for cot⁡−1\cot^{-1}⁠. Lesson: principal values

  2. Find the reference angle: which angle has cot⁡=+13\cot = +\frac{1}{\sqrt3}⁠?

    Ignore the minus sign for a moment. From Basic 2,

    cot⁡π3=cos⁡(π/3)sin⁡(π/3)=1/23/2=13.\cot\frac{\pi}{3} = \frac{\cos(\pi/3)}{\sin(\pi/3)} = \frac{1/2}{\sqrt3/2} = \frac{1}{\sqrt3}.
    Working with the size first and the sign second splits one hard question into two easy ones. The size 13\frac{1}{\sqrt3} tells us the reference angle is π3\frac{\pi}{3}⁠; the sign then tells us which quadrant the answer is in.
    A reference angle is the acute angle between the radius and the xx-axis. Angles with the same reference angle have trig values of the same size, differing only in sign. Basics: reference angles

  3. Fix the sign with cot⁡(π−θ)=−cot⁡θ\cot(\pi-\theta) = -\cot\theta

    We need a negative cotangent. Since cot⁡y=cos⁡ysin⁡y\cot y = \frac{\cos y}{\sin y} and sin⁡y>0\sin y > 0 for every yy in (0,π)(0,\pi)⁠, the sign of cot⁡y\cot y there is the sign of cos⁡y\cos y⁠. So we need cos⁡y<0\cos y < 0⁠, which on (0,π)(0,\pi) means π2<y<π\frac{\pi}{2} < y < \pi⁠. The supplementary-angle identity (Basic 3) supplies such an angle:

    cot⁡y=−13=−cot⁡π3=cot⁡(π−π3)=cot⁡2π3.\begin{aligned} \cot y &= -\frac{1}{\sqrt3} = -\cot\frac{\pi}{3} \\ &= \cot\left(\pi - \frac{\pi}{3}\right) = \cot\frac{2\pi}{3}. \end{aligned}

    So y=2π3y = \frac{2\pi}{3} satisfies the equation. Directly: cos⁡2π3=−12\cos\frac{2\pi}{3} = -\frac12 and sin⁡2π3=32\sin\frac{2\pi}{3} = \frac{\sqrt3}{2}⁠, so cot⁡2π3=−1/23/2=−13\cot\frac{2\pi}{3} = \frac{-1/2}{\sqrt3/2} = -\frac{1}{\sqrt3}⁠.

    This is exactly the textbook's line of working. The identity cot⁡(π−θ)=−cot⁡θ\cot(\pi-\theta)=-\cot\theta is used because it turns a negative cotangent into a positive angle inside (0,π)(0,\pi)⁠. The other identity people reach for, cot⁡(−θ)=−cot⁡θ\cot(-\theta) = -\cot\theta⁠, would give −π3-\frac{\pi}{3}⁠, which is outside (0,π)(0,\pi) and so cannot be the principal value.
    Proof of cot⁡(π−θ)=−cot⁡θ\cot(\pi-\theta) = -\cot\theta⁠: cot⁡(π−θ)=cos⁡(π−θ)sin⁡(π−θ)\cot(\pi-\theta) = \frac{\cos(\pi-\theta)}{\sin(\pi-\theta)} =−cos⁡θsin⁡θ=−cot⁡θ= \frac{-\cos\theta}{\sin\theta} = -\cot\theta⁠, valid whenever sin⁡θ≠0\sin\theta\ne0⁠.

  4. Check the range and that the answer is the only one

    2π3=120∘\frac{2\pi}{3} = 120^\circ lies in (0,π)(0,\pi)⁠, so both conditions of Step 1 hold. It is the only such angle: on (0,π)(0,\pi) cotangent is strictly decreasing. As yy runs from just above 0 to just below π\pi⁠, the point (cos⁡y,sin⁡y)(\cos y, \sin y) moves over the top of the circle from right to left, and cos⁡ysin⁡y\frac{\cos y}{\sin y} falls from very large positive values, through 00 at y=π2y=\frac{\pi}{2}⁠, to very large negative values. Each real number is passed exactly once. Hence

    cot⁡−1(−13)=2π3.\cot^{-1}\left(-\frac{1}{\sqrt3}\right) = \frac{2\pi}{3}.
    x y O 2π/3 π/3 A B C 0 π thick blue arc: angles in (0, π), the range of cot⁻¹ A: angle 2π/3, point (−1/2, √3/2), cot = x/y = −1/√3 B: angle −π/3, also cot −1/√3, outside (0, π) C: reference angle π/3, cot = +1/√3 gold line through O: all points with cot −1/√3
    Points with the same cotangent xy\frac{x}{y} lie on the same line through OO⁠. The gold line meets the circle at AA (angle 2π3\frac{2\pi}{3}⁠) and BB (angle −π3-\frac{\pi}{3}⁠); only AA is on the thick upper half, the range (0,π)(0,\pi) of cot⁡−1\cot^{-1}⁠.
    The picture explains why cot⁡\cot is one-one on (0,π)(0,\pi)⁠: every line through the origin (other than the xx-axis) crosses the open upper half of the circle exactly once. The line for cot⁡=−13\cot = -\frac{1}{\sqrt3} crosses it at AA⁠, and its other crossing BB is on the lower half, outside the range.
    π−π3=3π3−π3=2π3\pi - \frac{\pi}{3} = \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3}⁠. Lesson: graphs of the inverse functions
Result

The principal value of cot⁡−1(−13)\cot^{-1}\left(-\frac{1}{\sqrt3}\right) is 2π3\frac{2\pi}{3}. This agrees with the textbook's solution.

Check it with numbers

2π3≈2.094395\frac{2\pi}{3}\approx 2.094395⁠. Then cos⁡(2.094395)≈−0.500000\cos(2.094395)\approx -0.500000 and sin⁡(2.094395)≈0.866025\sin(2.094395)\approx 0.866025⁠, so cot⁡(2.094395)≈−0.5000000.866025\cot(2.094395) \approx \frac{-0.500000}{0.866025}⁠, which is ≈−0.577350\approx -0.577350⁠. And −13≈−0.577350-\frac{1}{\sqrt3}\approx -0.577350⁠. They agree, and 0<2.094395<π≈3.1415930 < 2.094395 < \pi \approx 3.141593⁠.

Calculators have no cot⁡−1\cot^{-1} key. If you use cot⁡−1a=tan⁡−11a\cot^{-1}a = \tan^{-1}\frac{1}{a}⁠, you get tan⁡−1(−3)≈−1.047198=−π3\tan^{-1}(-\sqrt3) \approx -1.047198 = -\frac{\pi}{3}⁠, which is outside (0,π)(0,\pi)⁠. That shortcut is only valid for a>0a > 0⁠; for a<0a<0 you must add π\pi⁠: −1.047198+3.141593=2.094395-1.047198 + 3.141593 = 2.094395⁠.

Common mistakes
  1. Answering −π3-\frac{\pi}{3}⁠. It does satisfy cot⁡y=−13\cot y = -\frac{1}{\sqrt3}⁠, but the range of cot⁡−1\cot^{-1} is (0,π)(0,\pi)⁠, and −π3-\frac{\pi}{3} is negative. This is the most common error, caused by treating cot⁡−1\cot^{-1} as if it had the range of tan⁡−1\tan^{-1}⁠.
  2. Using cot⁡(−θ)=−cot⁡θ\cot(-\theta) = -\cot\theta⁠. True, but it produces a negative angle. For cot⁡−1\cot^{-1} and cos⁡−1\cos^{-1} (ranges in [0,π][0,\pi]⁠), negative inputs are handled with π−θ\pi - \theta⁠.
  3. Confusing cot⁡π3\cot\frac{\pi}{3} with cot⁡π6\cot\frac{\pi}{6}⁠. cot⁡π6=3\cot\frac{\pi}{6} = \sqrt3 and cot⁡π3=13\cot\frac{\pi}{3} = \frac{1}{\sqrt3}⁠. Check with cos⁡sin⁡\frac{\cos}{\sin} if unsure.
What to take away
  1. For cot⁡−1\cot^{-1}⁠, a negative input gives an answer in the second quadrant, between π2\frac{\pi}{2} and π\pi⁠: cot⁡−1(−a)=π−cot⁡−1a\cot^{-1}(-a) = \pi - \cot^{-1}a⁠. Here π−π3=2π3\pi - \frac{\pi}{3} = \frac{2\pi}{3}⁠. The same holds for cos⁡−1\cos^{-1}⁠, except that cos⁡−1(−1)=π\cos^{-1}(-1) = \pi lies on the axis.
  2. Work out the size from a reference angle, then use the principal range to decide the quadrant.

Every fact used, and where it comes from

FactUsed inSource
cot⁡−1:R→(0,π)\cot^{-1} : \R\to(0,\pi)Steps 1, 4Section 2.2, table of principal value branches (Lesson: principal values)
cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}⁠; cot⁡π3=13\cot\frac{\pi}{3} = \frac{1}{\sqrt3}Steps 2, 3Class XI definitions and special angles
cot⁡(π−θ)=−cot⁡θ\cot(\pi-\theta) = -\cot\thetaStep 3Class XI allied angles (proved in Step 3)
Cot is strictly decreasing on (0,π)(0,\pi)Step 4Unit circle; Section 2.2 (cot restricted to (0,π)(0,\pi) is a bijection onto R\R⁠)
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →