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Chapter 2 · Inverse Trigonometric Functions

Miscellaneous · Q2: tan⁡−1(tan⁡7π6)\tan^{-1}\left(\tan\frac{7\pi}{6}\right)

Miscellaneous ExerciseFolding into the principal range · Miscellaneous Exercise
The question

Find the value of the following:

tan⁡−1(tan⁡7π6)\tan^{-1}\left(\tan\frac{7\pi}{6}\right)
Exam answer

The principal range of tan⁡−1\tan^{-1} is (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, and 7π6\frac{7\pi}{6} is not in it.

tan⁡−1(tan⁡7π6)=tan⁡−1(tan⁡(π+π6))=tan⁡−1(tan⁡π6)\begin{aligned} &\tan^{-1}\left(\tan\frac{7\pi}{6}\right) \\ &= \tan^{-1}\left(\tan\left(\pi + \frac{\pi}{6}\right)\right) \\ &= \tan^{-1}\left(\tan\frac{\pi}{6}\right) \end{aligned}

As π6∈(−π2,π2)\frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠,

∴ tan⁡−1(tan⁡7π6)=π6\therefore\ \tan^{-1}\left(\tan\dfrac{7\pi}{6}\right) = \dfrac{\pi}{6}⁠.

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

The one idea this problem is built on

tan⁡−1\tan^{-1} always returns an angle in its principal range (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠, so tan⁡−1(tan⁡x)=x\tan^{-1}(\tan x) = x only when xx is in that interval. The angle 7π6\frac{7\pi}{6} is not. But tangent repeats every π\pi (half a turn, not a full turn), and 7π6−π=π6\frac{7\pi}{6} - \pi = \frac{\pi}{6}⁠, which is in the interval and has the same tangent. So the answer is π6\frac{\pi}{6}⁠.

Given, to find, and the plan
Given
The expression tan⁡−1(tan⁡7π6)\tan^{-1}\left(\tan\frac{7\pi}{6}\right)⁠, where tan⁡−1\tan^{-1} is the principal value branch, with range (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠.
To find
Its value, as an angle in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠.
Plan
(1) Check whether 7π6\frac{7\pi}{6} is in the principal range; it is not. (2) Use the period π\pi of tangent to compute tan⁡7π6\tan\frac{7\pi}{6}⁠, and confirm it from sine and cosine. (3) Find the principal value of tan⁡−1\tan^{-1} of that number and check it is the only possibility.

Basics you need

1The principal range of tan⁡−1\tan^{-1}

tan⁡−1:R→(−π2,π2)\tan^{-1} : \R \to \left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠. The interval is open: ±π2\pm\frac{\pi}{2} are excluded because tan⁡\tan is not defined there. On this interval tangent is strictly increasing, so it takes each real value exactly once. Basics: tan, cot, sec and cosec

2When tan⁡−1(tan⁡x)=x\tan^{-1}(\tan x) = x

Only for x∈(−π2,π2)x \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠, the range of the principal branch. Outside it, first move the angle into that interval without changing its tangent. Lesson: the folding rules

3Tangent has period π\pi

tan⁡(π+x)=tan⁡x\tan(\pi + x) = \tan x⁠. Adding π\pi moves a point on the unit circle to the diametrically opposite point. Both coordinates change sign, so their ratio sin⁡cos⁡\frac{\sin}{\cos} is unchanged. Basics: sin and cos on the unit circle

4Special values

sin⁡π6=12\sin\frac{\pi}{6} = \frac12⁠, cos⁡π6=32\cos\frac{\pi}{6} = \frac{\sqrt3}{2}⁠, tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt3}⁠. Basics: special angles

Solution

  1. Spot the trap: 7π6\frac{7\pi}{6} is outside (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

    Compare with π2=3π6\frac{\pi}{2} = \frac{3\pi}{6}⁠. Since 7>37 > 3⁠,

    7π6>π2,7π6≈3.665,π2≈1.571.\frac{7\pi}{6} > \frac{\pi}{2}, \qquad \frac{7\pi}{6} \approx 3.665,\quad \frac{\pi}{2} \approx 1.571.

    So 7π6\frac{7\pi}{6} is not in the principal range, and the answer is not 7π6\frac{7\pi}{6}⁠.

    Every output of tan⁡−1\tan^{-1} lies strictly between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}⁠. The rule "tan⁡−1\tan^{-1} undoes tan⁡\tan" only works on the piece of the tangent graph used to define tan⁡−1\tan^{-1}⁠.
    Principal range of tan⁡−1\tan^{-1}⁠: (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) (Basic 1). Lesson: principal values

  2. Remove half a turn: tangent has period π\pi

    Split the angle:

    7π6=6π6+π6=π+π6.\frac{7\pi}{6} = \frac{6\pi}{6} + \frac{\pi}{6} = \pi + \frac{\pi}{6}.

    Tangent repeats every π\pi⁠, so

    tan⁡7π6=tan⁡(π+π6)=tan⁡π6=13.\tan\frac{7\pi}{6} = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\frac{\pi}{6} = \frac{1}{\sqrt3}.

    Check this directly from sine and cosine. The angle 7π6\frac{7\pi}{6} is in the third quadrant, where both are negative:

    sin⁡7π6=−12,cos⁡7π6=−32,\sin\frac{7\pi}{6} = -\frac12, \qquad \cos\frac{7\pi}{6} = -\frac{\sqrt3}{2}, tan⁡7π6=−1/2−3/2=13.\tan\frac{7\pi}{6} = \frac{-1/2}{-\sqrt3/2} = \frac{1}{\sqrt3}.
    Turning a further π\pi takes the point PP at angle π6\frac{\pi}{6} to the opposite point QQ at angle 7π6\frac{7\pi}{6}⁠. Both coordinates flip sign, and two minus signs cancel in the ratio, so the tangent is positive 13\frac{1}{\sqrt3}⁠. In the figure, PP⁠, the centre OO and QQ are on one straight line, and the tangent is the slope of that line.
    tan⁡(π+x)=tan⁡x\tan(\pi + x) = \tan x (Basic 3). Cosine needs a full turn 2π2\pi to repeat, tangent only half a turn. Basics: tan and its period

  3. Find the principal value of tan⁡−113\tan^{-1}\frac{1}{\sqrt3}

    The expression is now

    tan⁡−1(tan⁡7π6)=tan⁡−113.\tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \tan^{-1}\frac{1}{\sqrt3}.

    We need yy with −π2<y<π2-\frac{\pi}{2} < y < \frac{\pi}{2} and tan⁡y=13\tan y = \frac{1}{\sqrt3}⁠. The angle y=π6y = \frac{\pi}{6} works: tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt3}⁠, and −π2<π6<π2-\frac{\pi}{2} < \frac{\pi}{6} < \frac{\pi}{2}⁠. Tangent is strictly increasing on this interval, so no other angle in it has the same tangent. Hence

    tan⁡−113=π6.\tan^{-1}\frac{1}{\sqrt3} = \frac{\pi}{6}.
    Both conditions of a principal value are checked: the right tangent and a place in the principal range. The angles 7π6\frac{7\pi}{6} and −5π6-\frac{5\pi}{6} also have tangent 13\frac{1}{\sqrt3}⁠, but they are outside (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠.
    Principal value: the value that lies in the principal branch (Section 2.2, Note 3). Basics: what an inverse needs

  4. Conclude

    tan⁡−1(tan⁡7π6)=π6.\tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \frac{\pi}{6}.

    In folding terms: 7π6−π=π6\frac{7\pi}{6} - \pi = \frac{\pi}{6} lies in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) and has the same tangent, so it is what tan⁡−1\tan^{-1} returns.

    For tangent the fold is always "shift by a whole number of π\pi's until you land in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)". Exactly one shift lands there, because that interval has length π\pi⁠. Compare Miscellaneous Q1, where cosine needed a shift by 2π2\pi⁠.
π/2 −π/2 x = 1 P Q T 1/√3 7π/6 gold arc: (−π/2, π/2), the range of tan⁻¹ Q is at angle 7π/6 = π + π/6 P is at angle π/6, opposite Q P, O and Q lie on one line, so the two angles have the same tangent: the height of T, which is 1/√3
QQ (angle 7π6\frac{7\pi}{6}⁠) and PP (angle π6\frac{\pi}{6}⁠) are opposite ends of one diameter, so the line QOPQOP has a single slope: the tangent of both angles. It meets the line x=1x = 1 at TT⁠, at height tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt3}⁠. Of the two points, only PP is on the gold right half circle, the principal range of tan⁡−1\tan^{-1}⁠.
Result

tan⁡−1(tan⁡7π6)=π6\tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \frac{\pi}{6}⁠. The angle 7π6\frac{7\pi}{6} lies outside (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠; since tangent has period π\pi⁠, the angle 7π6−π=π6\frac{7\pi}{6} - \pi = \frac{\pi}{6} has the same tangent 13\frac{1}{\sqrt3} and lies inside. (Textbook answer: π6\frac{\pi}{6}⁠.)

Check it with numbers

7π6≈3.6651914\frac{7\pi}{6} \approx 3.6651914⁠. In radian mode, tan⁡(3.6651914)≈0.5773503\tan(3.6651914) \approx 0.5773503⁠, and 13≈0.5773503\frac{1}{\sqrt3} \approx 0.5773503⁠. Then tan⁡−1(0.5773503)≈0.5235988=π6\tan^{-1}(0.5773503) \approx 0.5235988 = \frac{\pi}{6}⁠. ✓ The calculator does not return 3.66523.6652⁠, because its tan⁡−1\tan^{-1} key also uses the principal range.

Common mistakes
  1. Answering 7π6\frac{7\pi}{6}⁠. It is outside (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠, so tan⁡−1\tan^{-1} cannot return it.
  2. Thinking the tangent is negative in the third quadrant. Sine and cosine are both negative there, so their ratio is positive. That gives +13+\frac{1}{\sqrt3}⁠, and an answer of −π6-\frac{\pi}{6} is wrong.
  3. Removing 2π2\pi instead of π\pi⁠. 7π6−2π=−5π6\frac{7\pi}{6} - 2\pi = -\frac{5\pi}{6} has the right tangent but is still outside the range. For tangent, shift by π\pi⁠.
  4. Using the cosine range [0,π][0,\pi]⁠. That belongs to cos⁡−1\cos^{-1} and cot⁡−1\cot^{-1} (the latter open). For tan⁡−1\tan^{-1} it is (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠.
What to take away
  1. tan⁡−1(tan⁡x)=x\tan^{-1}(\tan x) = x only for xx in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠.
  2. Tangent has period π\pi⁠, so fold by adding or removing whole multiples of π\pi until the angle lands in the principal range.
  3. Opposite points on the unit circle have the same tangent: the slope of the diameter through them.

Every fact used, and where it comes from

FactUsed inSource
tan⁡−1:R→(−π2,π2)\tan^{-1} : \R \to \left(-\frac{\pi}{2},\frac{\pi}{2}\right) (principal value branch)Steps 1, 3Section 2.2 and its table of principal branches (Lesson: principal values)
Principal value: the value lying in the principal branchStep 3Section 2.2, Note 3
tan⁡(π+x)=tan⁡x\tan(\pi + x) = \tan xStep 2Periodicity of tangent, Class XI Chapter 3 (Trigonometric Functions)
sin⁡7π6=−12\sin\frac{7\pi}{6} = -\frac12⁠, cos⁡7π6=−32\cos\frac{7\pi}{6} = -\frac{\sqrt3}{2}Step 2sin⁡(π+x)=−sin⁡x\sin(\pi + x) = -\sin x⁠, cos⁡(π+x)=−cos⁡x\cos(\pi + x) = -\cos x⁠, Class XI Chapter 3
tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt3}Steps 2, 3Special angles, Class XI Chapter 3
Tangent is strictly increasing on (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)Step 3Graph of tan⁡x\tan x⁠, Fig 2.5 (i), Section 2.2
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