Unit 6, lesson 54

The Arcsine Function

What arcsine returns, why it always picks its angle from one fixed closed interval, and how to find its exact values, domain, range and graph.

  • Theory part 107
  • Practice part 108
  • Inverse Trigonometric Functions

A ramp rises 11 meter for every 22 meters along its surface, and you want its angle. You know the sine of the angle is 12\frac{1}{2}, and you need to go backwards from a sine to an angle. The trouble is that infinitely many angles have sine 12\frac{1}{2}. Arcsine settles this by always picking one of them, from one fixed interval, and calculus relies on that choice every time it differentiates or integrates with arcsine.

Key terms

Restricted sine branch

Definition The restricted sine branch is the sine function with its domain cut down to the closed interval [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. On this interval sine is increasing, from sin⁡(−π2)=−1\sin\left(-\frac{\pi}{2}\right) = -1 to sin⁡π2=1\sin \frac{\pi}{2} = 1, and it takes every value in [−1,1][-1, 1] exactly once. So the restricted branch is one-to-one with range [−1,1][-1, 1].

In plain words The full sine graph fails the horizontal line test badly. The piece over [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] passes it and still reaches every height from −1-1 to 11.

Example On the whole real line the equation sin⁡x=12\sin x = \frac{1}{2} has infinitely many solutions, among them π6\frac{\pi}{6}, 5π6\frac{5\pi}{6} and −7π6-\frac{7\pi}{6}. Only π6\frac{\pi}{6} lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so the restricted branch reaches the height 12\frac{1}{2} just once.

Inverse function

Definition Because the restricted sine branch is one-to-one, it has an inverse function, written arcsin⁡\arcsin or sin⁡−1\sin^{-1}. Here sin⁡−1\sin^{-1} always means this inverse function and never the reciprocal. The reciprocal of sine is cosecant,

1sin⁡x=(sin⁡x)−1=csc⁡x,sin⁡x≠0.\begin{gathered} \frac{1}{\sin x} = (\sin x)^{-1} = \csc x, \\[0.4em] \sin x \ne 0. \end{gathered}

In plain words Arcsine undoes sine, but only the restricted piece of it. The −1-1 in sin⁡−1\sin^{-1} means undo, not one over.

Example At the input 11 the two readings give very different numbers.

sin⁡−1(1)=π2≈1.571,(sin⁡1)−1=csc⁡1≈1.188.\begin{gathered} \sin^{-1}(1) = \frac{\pi}{2} \approx 1.571, \\[0.4em] (\sin 1)^{-1} = \csc 1 \approx 1.188. \end{gathered}

Arcsine

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

−π2≤θ≤π2andsin⁡θ=u.-\frac{\pi}{2} \le \theta \le \frac{\pi}{2} \quad \text{and} \quad \sin \theta = u.

This angle is written θ=arcsin⁡u\theta = \arcsin u. The two inverse identities hold with these domains.

sin⁡(arcsin⁡u)=u for every u∈[−1,1],arcsin⁡(sin⁡θ)=θ only for θ∈[−π2,π2].\begin{aligned} &\sin(\arcsin u) = u \\[0.4em] &\quad \text{ for every } u \in [-1, 1], \\[0.4em] &\arcsin(\sin \theta) = \theta \\[0.4em] &\quad \text{ only for } \theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]. \end{aligned}

In plain words The expression arcsin⁡u\arcsin u asks which angle between −π2-\frac{\pi}{2} and π2\frac{\pi}{2} has sine equal to uu. There is always exactly one.

Example Since sin⁡5π6=12\sin \frac{5\pi}{6} = \frac{1}{2} and the angle in the allowed interval with sine 12\frac{1}{2} is π6\frac{\pi}{6},

arcsin⁡(sin⁡5π6)=arcsin⁡12=π6,\arcsin\left(\sin \frac{5\pi}{6}\right) = \arcsin \frac{1}{2} = \frac{\pi}{6},

which is not 5π6\frac{5\pi}{6}. The angle 5π6\frac{5\pi}{6} lies outside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so the second identity does not apply to it.

Principal angle

Definition The value arcsin⁡u\arcsin u is called the principal angle with sine uu. It is a signed angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so its terminal side lies in quadrant I or IV or on the yy-axis or the positive xx-axis. For u<0u < 0 it is negative, measured clockwise from the positive xx-axis. Every angle with sine uu is given by the periodic rule θ=arcsin⁡u+2kπ\theta = \arcsin u + 2k\pi or θ=π−arcsin⁡u+2kπ\theta = \pi - \arcsin u + 2k\pi with k∈Zk \in \mathbb{Z}.

In plain words Among all angles with the right sine, arcsine picks the one on the right half of the unit circle, using a negative angle when the sine is negative.

Example To find arcsin⁡(−22)\arcsin\left(-\frac{\sqrt{2}}{2}\right), note that sin⁡π4=22\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} and that a negative sine in the allowed interval means quadrant IV. So the value is −π4-\frac{\pi}{4}. The angles 5π4\frac{5\pi}{4} and 7π4\frac{7\pi}{4} also have this sine, but neither lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Domain

Definition The domain of arcsine is the closed interval [−1,1][-1, 1], the range of the restricted sine branch. For a composition arcsin⁡(g(x))\arcsin(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 Arcsine only accepts numbers that a sine can actually equal. Anything bigger than 11 or smaller than −1-1 is not allowed.

Example The expression arcsin⁡2\arcsin 2 is undefined, since no angle has sine 22. For f(x)=arcsin⁡(2x−1)f(x) = \arcsin(2x - 1) we need

−1≤2x−1≤1⟺0≤2x≤2⟺0≤x≤1,\begin{gathered} -1 \le 2x - 1 \le 1 \\[0.4em] \Longleftrightarrow \quad 0 \le 2x \le 2 \\[0.4em] \Longleftrightarrow \quad 0 \le x \le 1, \end{gathered}

so the domain of ff is [0,1][0, 1].

Range

Definition The range of arcsine is the closed interval [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], the domain of the restricted sine branch. Arcsine is increasing on [−1,1][-1, 1] and odd, which means

arcsin⁡(−u)=−arcsin⁡ufor every u∈[−1,1].\begin{gathered} \arcsin(-u) = -\arcsin u \\[0.4em] \text{for every } u \in [-1, 1]. \end{gathered}

In plain words Every answer arcsine gives is an angle between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}. A larger input always gives a larger angle, and flipping the sign of the input flips the sign of the angle.

Example For g(x)=2arcsin⁡xg(x) = 2\arcsin x the outputs are twice those of arcsine, so the range of gg is [−π,π][-\pi, \pi]. Oddness gives arcsin⁡(−12)=−arcsin⁡12=−π6\arcsin\left(-\frac{1}{2}\right) = -\arcsin \frac{1}{2} = -\frac{\pi}{6}, not 11π6\frac{11\pi}{6}.

Arcsine graph

Definition The graph of y=arcsin⁡xy = \arcsin x consists of the points (b,a)(b, a) for which (a,b)(a, b) lies on the graph of the restricted sine branch. With equal scales on both axes it is the reflection of that branch across the line y=xy = x. It runs from (−1,−π2)\left(-1, -\frac{\pi}{2}\right) through (0,0)(0, 0) to (1,π2)\left(1, \frac{\pi}{2}\right), rises throughout and is symmetric about the origin. The sine branch has horizontal tangent lines at its ends θ=±π2\theta = \pm\frac{\pi}{2}, so after the swap the arcsine graph has vertical tangent lines at its endpoints (±1,±π2)\left(\pm 1, \pm\frac{\pi}{2}\right). It rises more and more steeply as xx approaches 11 or −1-1.

In plain words Take the piece of the sine wave between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}, swap the two coordinates of every point, and you have the arcsine curve lying on its side.

Example Swapping coordinates of points on the sine branch gives these points on the arcsine graph.

xx−1-1−22-\frac{\sqrt{2}}{2}−12-\frac{1}{2}0012\frac{1}{2}22\frac{\sqrt{2}}{2}11
arcsin⁡x\arcsin x−π2-\frac{\pi}{2}−π4-\frac{\pi}{4}−π6-\frac{\pi}{6}00π6\frac{\pi}{6}π4\frac{\pi}{4}π2\frac{\pi}{2}

Included endpoints

Definition Both endpoints of the domain belong to it and give real values of arcsine.

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

So the domain [−1,1][-1, 1] and the range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] are both closed intervals, and the endpoints of the graph are drawn as solid dots. Beyond x=±1x = \pm 1 there is no graph at all.

In plain words The curve stops at its two ends, and it really reaches them. The end points are part of the function, not missing values.

Example For f(x)=arcsin⁡(2x−1)f(x) = \arcsin(2x - 1) on its domain [0,1][0, 1] the landmarks are

f(0)=−π2,f(12)=0,f(1)=π2.\begin{gathered} f(0) = -\frac{\pi}{2}, \qquad f\left(\frac{1}{2}\right) = 0, \\[0.4em] f(1) = \frac{\pi}{2}. \end{gathered}

Its graph is the arcsine curve squeezed horizontally by a factor of 22 and moved right by 12\frac{1}{2}, so that it runs from (0,−π2)\left(0, -\frac{\pi}{2}\right) to (1,π2)\left(1, \frac{\pi}{2}\right), with solid dots at both ends and range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Check yourself

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

The angle π3\frac{\pi}{3} has sine 32\frac{\sqrt{3}}{2}, and arcsine is odd, so the value is −π3-\frac{\pi}{3}. It lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], as it must.

  1. Give the domain, the range and the values at the two endpoints and at the center for g(x)=2arcsin⁡x3g(x) = 2\arcsin\frac{x}{3}.
Show the answer

The input condition −1≤x3≤1-1 \le \frac{x}{3} \le 1 gives the domain [−3,3][-3, 3]. The landmarks are g(−3)=−πg(-3) = -\pi, g(0)=0g(0) = 0 and g(3)=πg(3) = \pi, so the range is [−π,π][-\pi, \pi], with both endpoints attained.

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

The inner value is sin⁡4π3=−32\sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}, so the answer is arcsin⁡(−32)=−π3\arcsin\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}. It is not 4π3\frac{4\pi}{3}, because that angle lies outside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

How this lesson connects