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

Miscellaneous · Q1: cos⁡−1(cos⁡13π6)\cos^{-1}\left(\cos\frac{13\pi}{6}\right)

Miscellaneous ExerciseFolding into the principal range · Miscellaneous Exercise
The question

Find the value of the following:

cos⁡−1(cos⁡13π6)\cos^{-1}\left(\cos\frac{13\pi}{6}\right)
Exam answer

The principal range of cos⁡−1\cos^{-1} is [0,π][0, \pi]⁠, and 13π6\frac{13\pi}{6} is not in it.

cos⁡−1(cos⁡13π6)=cos⁡−1(cos⁡(2π+π6))=cos⁡−1(cos⁡π6)\begin{aligned} &\cos^{-1}\left(\cos\frac{13\pi}{6}\right) \\ &= \cos^{-1}\left(\cos\left(2\pi + \frac{\pi}{6}\right)\right) \\ &= \cos^{-1}\left(\cos\frac{\pi}{6}\right) \end{aligned}

As π6∈[0,π]\frac{\pi}{6} \in [0, \pi]⁠,

∴ cos⁡−1(cos⁡13π6)=π6\therefore\ \cos^{-1}\left(\cos\dfrac{13\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

cos⁡−1\cos^{-1} always returns an angle in its principal range [0,π][0,\pi]⁠. So cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x) = x only when xx is already in [0,π][0,\pi]. The angle 13π6\frac{13\pi}{6} is bigger than π\pi⁠, so the answer cannot be 13π6\frac{13\pi}{6}⁠. Instead we find the one angle in [0,π][0,\pi] that has the same cosine. Since 13π6\frac{13\pi}{6} is one full turn plus π6\frac{\pi}{6}⁠, that angle is π6\frac{\pi}{6}⁠.

Given, to find, and the plan
Given
The expression cos⁡−1(cos⁡13π6)\cos^{-1}\left(\cos\frac{13\pi}{6}\right)⁠, where cos⁡−1\cos^{-1} means the principal value branch, with range [0,π][0,\pi]⁠.
To find
Its value, as an angle in [0,π][0,\pi]⁠.
Plan
(1) Check whether 13π6\frac{13\pi}{6} is in [0,π][0,\pi]⁠; it is not. (2) Use the period 2π2\pi of cosine to work out cos⁡13π6\cos\frac{13\pi}{6}⁠. (3) Find the principal value of cos⁡−1\cos^{-1} of that number, and check that it is the only possible answer.

Basics you need

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

cos⁡−1:[−1,1]→[0,π]\cos^{-1} : [-1,1] \to [0,\pi]⁠. Whatever the input, the output is an angle from 00 to π\pi inclusive. On [0,π][0,\pi] cosine is strictly decreasing, so it takes each value in [−1,1][-1,1] exactly once there. Basics: domain and range of sin and cos

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

The book states sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x for xx in the principal range of sin⁡−1\sin^{-1}⁠, and says similar results hold for the other functions "for suitable values of domain". For cosine the suitable values are exactly x∈[0,π]x \in [0,\pi]⁠. Outside that interval, you must first move the angle into [0,π][0,\pi]⁠. Lesson: the folding rules

3Cosine has period 2π2\pi

Adding or removing a full turn brings you back to the same point on the unit circle, so cos⁡(2π+x)=cos⁡x\cos(2\pi + x) = \cos x for every xx⁠. Basics: sin and cos on the unit circle

4A special value

cos⁡π6=cos⁡30∘=32\cos\frac{\pi}{6} = \cos 30^\circ = \frac{\sqrt3}{2}⁠. Basics: special angles

Solution

  1. Spot the trap: 13π6\frac{13\pi}{6} is outside [0,π][0,\pi]

    Compare 13π6\frac{13\pi}{6} with π=6π6\pi = \frac{6\pi}{6}⁠. Since 13>613 > 6⁠,

    13π6>π,13π6≈6.807,π≈3.142.\frac{13\pi}{6} > \pi, \qquad \frac{13\pi}{6} \approx 6.807,\quad \pi \approx 3.142.

    So 13π6\frac{13\pi}{6} is not in the principal range [0,π][0,\pi]⁠, and the answer is not 13π6\frac{13\pi}{6}⁠.

    Whatever number goes into cos⁡−1\cos^{-1}⁠, the output lies in [0,π][0,\pi]⁠. An answer of 13π6\frac{13\pi}{6} is impossible because it is not in that interval. The rule "cos⁡−1\cos^{-1} undoes cos⁡\cos" works only on the piece of the cosine graph that was used to define cos⁡−1\cos^{-1}⁠, which is [0,π][0,\pi]⁠.
    Principal range of cos⁡−1\cos^{-1}⁠: [0,π][0,\pi] (Basic 1). Lesson: principal values

  2. Remove one full turn

    Split the angle into a whole number of turns plus a remainder:

    13π6=12π6+π6=2π+π6.\frac{13\pi}{6} = \frac{12\pi}{6} + \frac{\pi}{6} = 2\pi + \frac{\pi}{6}.

    Cosine repeats every 2π2\pi⁠, so

    cos⁡13π6=cos⁡(2π+π6)=cos⁡π6=32.\cos\frac{13\pi}{6} = \cos\left(2\pi + \frac{\pi}{6}\right) = \cos\frac{\pi}{6} = \frac{\sqrt3}{2}.
    On the unit circle, turning through 13π6\frac{13\pi}{6} means going once all the way round (2π2\pi⁠) and then another π6\frac{\pi}{6}⁠. You stop at exactly the same point as if you had turned through π6\frac{\pi}{6} only. The cosine is the xx-coordinate of that point, so it is the same for both angles. The figure shows this.
    cos⁡(2π+x)=cos⁡x\cos(2\pi + x) = \cos x (Basic 3), and cos⁡π6=32\cos\frac{\pi}{6} = \frac{\sqrt3}{2} (Basic 4).

  3. Find the principal value of cos⁡−132\cos^{-1}\frac{\sqrt3}{2}

    The expression is now

    cos⁡−1(cos⁡13π6)=cos⁡−132.\cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \cos^{-1}\frac{\sqrt3}{2}.

    We need the angle yy with y∈[0,π]y \in [0,\pi] and cos⁡y=32\cos y = \frac{\sqrt3}{2}⁠. The angle y=π6y = \frac{\pi}{6} works: cos⁡π6=32\cos\frac{\pi}{6} = \frac{\sqrt3}{2}⁠, and 0≤π6≤π0 \le \frac{\pi}{6} \le \pi⁠.

    It is the only such angle, because cosine is strictly decreasing on [0,π][0,\pi] and so cannot take the value 32\frac{\sqrt3}{2} twice there. Hence

    cos⁡−132=π6.\cos^{-1}\frac{\sqrt3}{2} = \frac{\pi}{6}.
    The definition of the principal value has two parts: the cosine must be right and the angle must lie in [0,π][0,\pi]⁠. Other angles also have cosine 32\frac{\sqrt3}{2}⁠, for instance −π6-\frac{\pi}{6} and 13π6\frac{13\pi}{6} itself, but they fail the second condition. Checking both parts is what makes π6\frac{\pi}{6} the answer.
    Principal value: the value of the inverse function that lies in the principal branch (Section 2.2, Note 3). Basics: what an inverse needs

  4. Conclude

    Putting Steps 2 and 3 together,

    cos⁡−1(cos⁡13π6)=π6.\cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \frac{\pi}{6}.

    Equivalently: 13π6−2π=π6\frac{13\pi}{6} - 2\pi = \frac{\pi}{6} is the angle in [0,π][0,\pi] with the same cosine as 13π6\frac{13\pi}{6}⁠, so it is what cos⁡−1\cos^{-1} returns.

    This is the general "folding" method. For any xx⁠, shift it by whole turns (and, for cosine, use cos⁡(−x)=cos⁡x\cos(-x)=\cos x if needed) until it lands in [0,π][0,\pi] with the cosine unchanged. The landing point is cos⁡−1(cos⁡x)\cos^{-1}(\cos x)⁠. Here one shift by 2π2\pi was enough.
0 π P √3/2 gold arc: [0, π], the range of cos⁻¹ violet spiral: turning through 13π/6, one full turn (2π) and then π/6 more both turns end at the same point P, so both angles have cosine √3/2 P lies on the gold arc at angle π/6
Turning through 13π6\frac{13\pi}{6} (violet) is one full turn plus π6\frac{\pi}{6}⁠, so it ends at the same point PP as the angle π6\frac{\pi}{6}⁠. The cosine is the xx-coordinate of PP⁠, which is 32\frac{\sqrt3}{2}⁠. The only angle in [0,π][0,\pi] (gold) that ends at PP is π6\frac{\pi}{6}⁠.
Result

cos⁡−1(cos⁡13π6)=π6\cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \frac{\pi}{6}⁠. The angle 13π6\frac{13\pi}{6} lies outside [0,π][0,\pi]⁠; removing one full turn gives π6\frac{\pi}{6}⁠, which has the same cosine 32\frac{\sqrt3}{2} and lies inside [0,π][0,\pi]⁠. (Textbook answer: π6\frac{\pi}{6}⁠.)

Check it with numbers

13π6≈6.8067841\frac{13\pi}{6} \approx 6.8067841⁠. A calculator in radian mode gives cos⁡(6.8067841)≈0.8660254\cos(6.8067841) \approx 0.8660254⁠, which is 32≈0.8660254\frac{\sqrt3}{2} \approx 0.8660254⁠. Then cos⁡−1(0.8660254)≈0.5235988\cos^{-1}(0.8660254) \approx 0.5235988⁠, and π6≈0.5235988\frac{\pi}{6} \approx 0.5235988⁠. ✓ The calculator returns 0.52360.5236⁠, not 6.80686.8068⁠, because its cos⁡−1\cos^{-1} key also uses the principal range [0,π][0,\pi]⁠.

Common mistakes
  1. Writing 13π6\frac{13\pi}{6} as the answer. "Inverse cancels function" holds only inside [0,π][0,\pi]⁠. The output of cos⁡−1\cos^{-1} can never exceed π\pi⁠.
  2. Writing −π6-\frac{\pi}{6}⁠. It has the right cosine, since cos⁡(−π6)=32\cos(-\frac{\pi}{6}) = \frac{\sqrt3}{2}⁠, but it is negative, so it is not in [0,π][0,\pi]⁠. This often comes from computing 2π−13π62\pi - \frac{13\pi}{6} instead of 13π6−2π\frac{13\pi}{6} - 2\pi⁠.
  3. Using the sin⁡−1\sin^{-1} range by mistake. The interval [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] belongs to sin⁡−1\sin^{-1} and tan⁡−1\tan^{-1}⁠. For cos⁡−1\cos^{-1} it is [0,π][0,\pi]⁠.
  4. Degree mode on the calculator. 13π6\frac{13\pi}{6} is in radians; in degrees it is 390∘390^\circ⁠. Mixing the two gives nonsense.
What to take away
  1. cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x) = x only for x∈[0,π]x \in [0,\pi]⁠. Always check the range first.
  2. To fold an angle into [0,π][0,\pi]⁠, remove whole turns of 2π2\pi (and use cos⁡(−x)=cos⁡x\cos(-x)=\cos x if the result is negative), keeping the cosine unchanged.
  3. A principal value must satisfy two conditions: the right function value and a place in the principal range.

Every fact used, and where it comes from

FactUsed inSource
cos⁡−1:[−1,1]→[0,π]\cos^{-1} : [-1,1] \to [0,\pi] (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
cos⁡(2π+x)=cos⁡x\cos(2\pi + x) = \cos xStep 2Periodicity of cosine, Class XI Chapter 3 (Trigonometric Functions)
cos⁡π6=32\cos\frac{\pi}{6} = \frac{\sqrt3}{2}Steps 2, 3Special angles, Class XI Chapter 3
Cosine is strictly decreasing on [0,π][0,\pi]Step 3Graph of cos⁡x\cos x⁠, Fig 2.2 (i), Section 2.2
Folding an angle into the principal rangeStep 4Section 2.3; compare Miscellaneous Example 6 (Example 6), which does the same for sin⁡−1\sin^{-1}
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 →