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Chapter 2 · Inverse Trigonometric Functions

Exercise 2.1 · Q9: Principal value of cos⁡−1(−12)\cos^{-1}\left(-\frac{1}{\sqrt2}\right)

Exercise 2.1Principal values · Section 2.2
The question

Find the principal value of cos⁡−1(−12)\cos^{-1}\left(-\dfrac{1}{\sqrt2}\right)⁠.

Exam answer

Let cos⁡−1(−12)=y\cos^{-1}\left(-\frac{1}{\sqrt2}\right) = y⁠. Using cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos\theta⁠,

cos⁡y=−12=−cos⁡π4=cos⁡(π−π4)=cos⁡3π4\begin{aligned} \cos y &= -\frac{1}{\sqrt2} = -\cos\frac{\pi}{4} \\ &= \cos\left(\pi - \frac{\pi}{4}\right) = \cos\frac{3\pi}{4} \end{aligned}

Range of principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi]⁠, and 3π4∈[0,π]\frac{3\pi}{4} \in [0, \pi]⁠.

∴\therefore Principal value of cos⁡−1(−12)\cos^{-1}\left(-\dfrac{1}{\sqrt2}\right) is 3π4\dfrac{3\pi}{4}⁠.

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

The one idea this problem is built on

cos⁡−1(−12)\cos^{-1}\left(-\frac{1}{\sqrt2}\right) is the one angle in [0,π][0,\pi] whose cosine is −12-\frac{1}{\sqrt2}⁠. The reference angle is π4\frac{\pi}{4}⁠, because cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt2}⁠. A negative cosine inside [0,π][0,\pi] means the second quadrant, and the second-quadrant angle with reference angle π4\frac{\pi}{4} is π−π4=3π4\pi - \frac{\pi}{4} = \frac{3\pi}{4}⁠. The answer is not −π4-\frac{\pi}{4}⁠: that angle has a positive cosine and is not even in [0,π][0,\pi]⁠.

Given, to find, and the plan
Given
The number −12≈−0.707-\frac{1}{\sqrt2} \approx -0.707⁠, which lies in the domain [−1,1][-1,1] of cos⁡−1\cos^{-1}⁠.
To find
The principal value y=cos⁡−1(−12)y = \cos^{-1}\left(-\frac{1}{\sqrt2}\right)⁠, the unique y∈[0,π]y\in[0,\pi] with cos⁡y=−12\cos y = -\frac{1}{\sqrt2}⁠.
Plan
(1) State the principal range. (2) Find the reference angle. (3) Use the negative sign to place the answer in [0,π][0,\pi]⁠. (4) Check, and exclude the other angles with the same cosine.

Basics you need

1What a principal value is

cos⁡−1x\cos^{-1} x means the single angle yy in the principal range [0,π][0,\pi] with cos⁡y=x\cos y = x (Note, points 2 and 3, Section 2.2). Basics: what an inverse needs

2cos on the unit circle

cos⁡y\cos y is the xx-coordinate of the point at angle yy on the unit circle. It is positive on the right half and negative on the left half. Basics: sin and cos on the unit circle

3The special angle π4\frac{\pi}{4}

cos⁡π4=sin⁡π4=12\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt2}⁠. Basics: special angles and reference angles

4The allied-angle identity

cos⁡(π−α)=−cos⁡α\cos(\pi - \alpha) = -\cos\alpha for every α\alpha⁠. Reflecting a point in the yy-axis turns angle α\alpha into π−α\pi-\alpha and flips the sign of the xx-coordinate.

5Principal range of cos⁡−1\cos^{-1}

cos⁡−1:[−1,1]→[0,π]\cos^{-1} : [-1,1] \to [0,\pi]⁠. On [0,π][0,\pi] cos decreases from 11 to −1-1⁠, taking every value in [−1,1][-1,1] exactly once. Basics: domain and range of sin and cos

Solution

  1. Name the unknown and its range

    Let y=cos⁡−1(−12)y = \cos^{-1}\left(-\frac{1}{\sqrt2}\right)⁠. By definition of the principal value branch,

    cos⁡y=−12and0≤y≤π.\cos y = -\frac{1}{\sqrt2} \qquad\text{and}\qquad 0 \le y \le \pi.

    The input is allowed: −1≤−12≤1-1 \le -\frac{1}{\sqrt2} \le 1⁠.

    An answer must pass both tests. The range test 0≤y≤π0\le y\le\pi is what makes the answer unique, so it has to be written down before we start, not checked as an afterthought.
    cos⁡−1:[−1,1]→[0,π]\cos^{-1} : [-1,1] \to [0,\pi]⁠, with both ends included. Lesson: principal values

  2. Find the reference angle

    Drop the minus sign: which acute angle α\alpha has cos⁡α=12\cos\alpha = \frac{1}{\sqrt2}⁠? From the special angles, cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt2}⁠, so

    α=π4.\alpha = \frac{\pi}{4}.
    The size of the cosine, 12\frac{1}{\sqrt2}⁠, fixes the reference angle; the sign only decides the quadrant. Splitting the problem this way means we only ever need the table of acute special angles.
    cos⁡π6=32\cos\frac{\pi}{6} = \frac{\sqrt3}{2}⁠, cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt2}⁠, cos⁡π3=12\cos\frac{\pi}{3} = \frac12⁠. Basics: special angles and reference angles

  3. Use the sign to place the answer

    Inside [0,π][0,\pi]⁠, cos⁡y\cos y is positive on [0,π2)\left[0,\frac{\pi}{2}\right)⁠, zero at π2\frac{\pi}{2}⁠, and negative on (π2,π]\left(\frac{\pi}{2},\pi\right]⁠. Our cosine is negative, so yy is in the second quadrant. The second-quadrant angle with reference angle α\alpha is π−α\pi - \alpha⁠, and

    cos⁡(π−π4)=−cos⁡π4=−12.\cos\left(\pi - \frac{\pi}{4}\right) = -\cos\frac{\pi}{4} = -\frac{1}{\sqrt2}.

    Hence

    y=π−π4=3π4.y = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.
    This is the sign rule for cos⁡−1\cos^{-1}⁠: a positive input gives α\alpha⁠, a negative input gives π−α\pi - \alpha⁠. The range [0,π][0,\pi] contains no negative angles at all, so the answer to a "negative input" question is never a negative angle. In short, cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x for every xx in [−1,1][-1,1]⁠.
    cos⁡(π−α)=−cos⁡α\cos(\pi - \alpha) = -\cos\alpha (Basic 4). Basics: sin and cos on the unit circle

  4. Check it, and exclude every other candidate

    Check. 0≤3π4≤π0 \le \frac{3\pi}{4} \le \pi ✓ and cos⁡3π4=−12\cos\frac{3\pi}{4} = -\frac{1}{\sqrt2} ✓ (Step 3).

    Other angles with the same cosine. Cos has period 2π2\pi and cos⁡(−y)=cos⁡y\cos(-y) = \cos y⁠, so all solutions of cos⁡y=−12\cos y = -\frac{1}{\sqrt2} are

    y=±3π4+2nπ,n∈Z.y = \pm\frac{3\pi}{4} + 2n\pi, \qquad n\in\Z.

    The nearest ones to the range are −3π4<0-\frac{3\pi}{4} < 0⁠, which is the same point as 5π4\frac{5\pi}{4}⁠, and 5π4>π\frac{5\pi}{4} > \pi⁠. Both lie outside [0,π][0,\pi]⁠, and every other solution is further out. So 3π4\frac{3\pi}{4} is the only candidate in the principal range.

    O 3π/4 5π/4 0 π x = −1/√2 Principal range of cos⁻¹: [0, π], the closed upper half Blue: y = 3π/4, the answer. Its x-coordinate is −1/√2. Rose: 5π/4 (that is, −3π/4) has the same x-coordinate, but it is not in [0, π]. cos y is the x-coordinate, so cos y = −1/√2 means the point is on the gold line.
    The principal range [0,π][0,\pi] of cos⁡−1\cos^{-1} is the closed upper half of the unit circle (green). The points with cos⁡y=−12\cos y = -\frac{1}{\sqrt2} lie on the vertical line x=−12x = -\frac{1}{\sqrt2} (gold), which meets the circle at 3π4\frac{3\pi}{4} (blue, in the range) and 5π4\frac{5\pi}{4} (rose, outside it).
    Equal cosines mean equal xx-coordinates, so the candidates sit on one vertical line. A vertical line cuts the circle at a point and its mirror image in the xx-axis, one in the upper half and one in the lower half. The upper half is the principal range, so exactly one survives.
    On [0,π][0,\pi] cos is one-one, so cos⁡−1\cos^{-1} is a function there. Lesson: branches
Result cos⁡−1(−12)=3π4.\cos^{-1}\left(-\frac{1}{\sqrt2}\right) = \frac{3\pi}{4}.

It is the only angle in [0,π][0,\pi] whose cosine is −12-\frac{1}{\sqrt2}⁠. (Textbook answer: 3π4\frac{3\pi}{4}⁠.)

Check it with numbers

3π4≈2.3561945\frac{3\pi}{4} \approx 2.3561945⁠, and cos⁡(2.3561945)≈−0.7071068\cos(2.3561945) \approx -0.7071068⁠, while −12≈−0.7071068-\frac{1}{\sqrt2} \approx -0.7071068 ✓. The tempting wrong answer −π4≈−0.7853982-\frac{\pi}{4} \approx -0.7853982 has cos⁡≈+0.7071068\cos \approx +0.7071068⁠: wrong sign, and it is below 00 ✓.

Common mistakes
  1. Answering −π4-\frac{\pi}{4}⁠. The rule "put a minus sign on the reference angle" works for sin⁡−1\sin^{-1} and tan⁡−1\tan^{-1}⁠, not for cos⁡−1\cos^{-1}⁠. Here cos⁡(−π4)=+12\cos\left(-\frac{\pi}{4}\right) = +\frac{1}{\sqrt2}⁠, and −π4-\frac{\pi}{4} is not in [0,π][0,\pi]⁠.
  2. Answering 5π4\frac{5\pi}{4}⁠. Its cosine is correct, but it lies outside the principal range.
  3. Writing π+π4\pi + \frac{\pi}{4} instead of π−π4\pi - \frac{\pi}{4}⁠. The second-quadrant angle with reference angle α\alpha is π−α\pi - \alpha⁠.
What to take away
  1. For cos⁡−1\cos^{-1}⁠: positive input gives α\alpha⁠, negative input gives π−α\pi - \alpha⁠. The answer is always between 00 and π\pi⁠.
  2. Equal cosines lie on one vertical line; the upper half circle keeps one of the two points.
  3. Compare with Q5: cos⁡−1(−12)=π−π3=2π3\cos^{-1}\left(-\frac12\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}⁠, the same pattern.

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, 4Section 2.2 and its table of principal value branches (Lesson: principal values)
Principal value means the value in the principal rangeStep 1Note, points 2 and 3, Section 2.2
cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt2}Step 2Class XI trigonometry (special angles)
cos⁡(π−α)=−cos⁡α\cos(\pi-\alpha) = -\cos\alpha⁠; signs of cos by quadrantStep 3Class XI trigonometry (allied angles)
Cos has period 2π2\pi and cos⁡(−y)=cos⁡y\cos(-y) = \cos yStep 4Class XI trigonometry
All numerical valuesCheckRecomputed in the site's verification script
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