Unit 6, lesson 55

The Arccosine Function

Why arccosine reverses the decreasing cosine branch over the upper half of the unit circle, how to find its exact values, and how its domain, range, endpoints and graph follow from that choice.

  • Theory part 109
  • Practice part 110
  • Inverse Trigonometric Functions

The angle between two directions is never negative and never more than a half turn. If you know its cosine, you want back exactly one angle between 00 and π\pi. Arccosine does that job. Its interval differs from the one arcsine uses, and there is a good reason for that.

Key terms

Restricted cosine branch

Definition The restricted cosine branch is the cosine function with its domain cut down to the closed interval [0,π][0, \pi]. On this interval cosine decreases from cos⁡0=1\cos 0 = 1 to cos⁡π=−1\cos \pi = -1 and takes every value in [−1,1][-1, 1] exactly once, so it is one-to-one with range [−1,1][-1, 1]. The interval [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] used for sine would not work, since cosine is even and cos⁡(−π3)=cos⁡π3\cos\left(-\frac{\pi}{3}\right) = \cos \frac{\pi}{3}.

In plain words On the unit circle, angles from 00 to π\pi sweep the upper half. Each xx-coordinate between −1-1 and 11 belongs to exactly one point of that upper arc, so the angle is determined by its cosine.

Example The equation cos⁡x=0\cos x = 0 has the solutions π2+kπ\frac{\pi}{2} + k\pi for every integer kk. Of these, only π2\frac{\pi}{2} lies in [0,π][0, \pi].

Arccosine

Definition For −1≤u≤1-1 \le u \le 1, the arccosine of uu is the unique angle θ\theta in radians with

0≤θ≤πandcos⁡θ=u.0 \le \theta \le \pi \quad \text{and} \quad \cos \theta = u.

It is written θ=arccos⁡u\theta = \arccos u or θ=cos⁡−1u\theta = \cos^{-1} u, where cos⁡−1\cos^{-1} means the inverse function and not the reciprocal 1cos⁡u=sec⁡u\frac{1}{\cos u} = \sec u. The inverse identities are

cos⁡(arccos⁡u)=u for every u∈[−1,1],arccos⁡(cos⁡θ)=θ only for θ∈[0,π].\begin{aligned} &\cos(\arccos u) = u \\[0.4em] &\quad \text{ for every } u \in [-1, 1], \\[0.4em] &\arccos(\cos \theta) = \theta \\[0.4em] &\quad \text{ only for } \theta \in [0, \pi]. \end{aligned}

In plain words The expression arccos⁡u\arccos u asks which angle between 00 and π\pi has cosine uu. There is exactly one.

Example The angle π3\frac{\pi}{3} has cosine 12\frac{1}{2}. A cosine of −12-\frac{1}{2} needs a point on the left side of the upper arc with the same reference angle, so

arccos⁡(−12)=π−π3=2π3.\arccos\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}.

Principal angle

Definition The value arccos⁡u\arccos u is the principal angle with cosine uu. It always lies in [0,π][0, \pi], so it is never negative. It is acute when u>0u > 0, a right angle when u=0u = 0 and obtuse when u<0u < 0. For negative inputs the identity

arccos⁡(−u)=π−arccos⁡u,−1≤u≤1,\begin{gathered} \arccos(-u) = \pi - \arccos u, \\[0.4em] -1 \le u \le 1, \end{gathered}

holds, because cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos \theta and π−θ\pi - \theta stays in [0,π][0, \pi] when θ\theta does. Every other angle with cosine uu is ±arccos⁡u+2kπ\pm \arccos u + 2k\pi for some integer kk.

In plain words A negative input does not give a negative angle. It gives an obtuse angle, on the left half of the upper arc.

Example Since arccos⁡22=π4\arccos \frac{\sqrt{2}}{2} = \frac{\pi}{4}, the identity gives

arccos⁡(−22)=π−π4=3π4.\arccos\left(-\frac{\sqrt{2}}{2}\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.

The answer −π4-\frac{\pi}{4} would be wrong twice over, since it is negative and its cosine is +22+\frac{\sqrt{2}}{2}.

Decreasing function

Definition Arccosine is decreasing on [−1,1][-1, 1], which means that −1≤u1<u2≤1-1 \le u_1 < u_2 \le 1 implies arccos⁡u1>arccos⁡u2\arccos u_1 > \arccos u_2. It inherits this from the restricted cosine branch, which is decreasing. Arccosine is neither even nor odd. Its graph is instead symmetric about the point (0,π2)\left(0, \frac{\pi}{2}\right), which is what the identity arccos⁡(−u)=π−arccos⁡u\arccos(-u) = \pi - \arccos u says.

In plain words The larger the cosine, the smaller the angle. An input of 11 gives the smallest angle and an input of −1-1 gives the largest.

Example Three inputs in increasing order give angles in decreasing order.

arccos⁡(−12)=2π3> arccos⁡0=π2> arccos⁡12=π3.\begin{gathered} \arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \\[0.4em] > \ \arccos 0 = \frac{\pi}{2} \\[0.4em] > \ \arccos \frac{1}{2} = \frac{\pi}{3}. \end{gathered}

The values at −12-\frac{1}{2} and 12\frac{1}{2} are neither equal nor opposite, so arccosine is not even and not odd.

Domain

Definition The domain of arccosine is the closed interval [−1,1][-1, 1]. For a composition arccos⁡(g(x))\arccos(g(x)) the domain is the set of xx in the domain of gg with −1≤g(x)≤1-1 \le g(x) \le 1.

In plain words Arccosine accepts exactly the numbers that can be cosines.

Example The expression arccos⁡(−1.2)\arccos(-1.2) is undefined. For h(x)=arccos⁡(x−2)h(x) = \arccos(x - 2) the condition −1≤x−2≤1-1 \le x - 2 \le 1 gives the domain [1,3][1, 3].

Range

Definition The range of arccosine is the closed interval [0,π][0, \pi], the domain of the restricted cosine branch. Every value in it is attained.

In plain words Arccosine always answers with an angle from 00 to a half turn.

Example Because cos⁡(−π3)=12\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}, we get arccos⁡(cos⁡(−π3))=π3\arccos\left(\cos\left(-\frac{\pi}{3}\right)\right) = \frac{\pi}{3} and not −π3-\frac{\pi}{3}. The same happens at 5π3\frac{5\pi}{3}, since cos⁡5π3=12\cos \frac{5\pi}{3} = \frac{1}{2} as well. The output has to land in [0,π][0, \pi], whatever angle went in.

Endpoint values

Definition The landmark values of arccosine are

arccos⁡(−1)=π,arccos⁡0=π2,arccos⁡1=0.\begin{gathered} \arccos(-1) = \pi, \qquad \arccos 0 = \frac{\pi}{2}, \\[0.4em] \arccos 1 = 0. \end{gathered}

The first and last are attained at the included endpoints of the domain, and π2\frac{\pi}{2} is an interior value. So the graph begins at the solid point (−1,π)(-1, \pi) and ends at the solid point (1,0)(1, 0).

In plain words The ends of the curve are real points of the function. An input of 11 gives angle 00 and an input of −1-1 gives a straight angle.

Example For h(x)=arccos⁡(x−2)h(x) = \arccos(x - 2) on [1,3][1, 3] the landmarks are h(1)=πh(1) = \pi, h(2)=π2h(2) = \frac{\pi}{2} and h(3)=0h(3) = 0, so its range is still [0,π][0, \pi].

Inverse graph

Definition The graph of y=arccos⁡xy = \arccos x is the set of points (b,a)(b, a) for which (a,b)(a, b) lies on the graph of y=cos⁡xy = \cos x with 0≤x≤π0 \le x \le \pi. Swapping the coordinates is what is meant by reflecting across the line y=xy = x, and the picture looks like a mirror image only when both axes use the same scale. The curve falls from (−1,π)(-1, \pi) through (0,π2)\left(0, \frac{\pi}{2}\right) to (1,0)(1, 0). The cosine branch has horizontal tangent lines at θ=0\theta = 0 and θ=π\theta = \pi, so the arccosine graph has vertical tangent lines at its endpoints (−1,π)(-1, \pi) and (1,0)(1, 0) and falls more and more steeply as xx approaches −1-1 or 11.

In plain words Take the falling arch of cosine between 00 and π\pi, swap xx and yy in every point, and the arccosine curve appears, still falling.

Example For the scaled input y=arccos⁡x2y = \arccos \frac{x}{2} the domain is [−2,2][-2, 2], twice as wide. The arccosine graph is stretched horizontally by a factor of 22, so each point (a,b)(a, b) moves to (2a,b)(2a, b).

xx−2-2−1-1001122
arccos⁡x2\arccos \frac{x}{2}π\pi2π3\frac{2\pi}{3}π2\frac{\pi}{2}π3\frac{\pi}{3}00

The range is [0,π][0, \pi], with solid dots at (−2,π)(-2, \pi) and (2,0)(2, 0).

Check yourself

  1. Find the exact value of arccos⁡(−32)\arccos\left(-\frac{\sqrt{3}}{2}\right).
Show the answer

Since arccos⁡32=π6\arccos \frac{\sqrt{3}}{2} = \frac{\pi}{6}, the identity for negative inputs gives

arccos⁡(−32)=π−π6=5π6.\arccos\left(-\frac{\sqrt{3}}{2}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}.
  1. Find the domain, the range and the endpoints of the graph of y=arccos⁡2xy = \arccos 2x.
Show the answer

The condition −1≤2x≤1-1 \le 2x \le 1 gives the domain [−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right]. The graph starts at the solid point (−12,π)\left(-\frac{1}{2}, \pi\right), passes through (0,π2)\left(0, \frac{\pi}{2}\right) and ends at the solid point (12,0)\left(\frac{1}{2}, 0\right). The range is [0,π][0, \pi].

  1. Find arccos⁡(cos⁡4π3)\arccos\left(\cos \frac{4\pi}{3}\right).
Show the answer

The inner value is cos⁡4π3=−12\cos \frac{4\pi}{3} = -\frac{1}{2}, and the angle in [0,π][0, \pi] with that cosine is 2π3\frac{2\pi}{3}. The answer is 2π3\frac{2\pi}{3}, because 4π3\frac{4\pi}{3} lies outside [0,π][0, \pi].

How this lesson connects