Unit 5, lesson 59

Inverse Sine and Cosine Functions

How restricting sine and cosine to one branch makes them invertible, which angle arcsine and arccosine return, and why they only accept inputs that a sine or cosine can actually produce.

  • Theory part 117
  • Practice part 118
  • Inverse Trigonometric Functions

A ramp rises 1 meter over every 2 meters of its sloping surface. What angle does it make with the ground? You know that the sine of the angle is 12\frac{1}{2}, and you want the angle itself, which is 30∘30^\circ or π6\frac{\pi}{6}. But infinitely many angles have sine 12\frac{1}{2}, so going back from a sine to an angle needs a rule that picks exactly one. That rule is the principal branch, and it is what your calculator uses.

Key terms

Restricted sine and cosine

Definition Sine is not one-to-one on the real numbers, since for example sin⁡π6=sin⁡5π6=12\sin \frac{\pi}{6} = \sin \frac{5\pi}{6} = \frac{1}{2}. The restricted sine function is sine with domain [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. On this interval sine is increasing and takes every value in [−1,1][-1, 1] exactly once. The restricted cosine function is cosine with domain [0,π][0, \pi], where cosine is decreasing and takes every value in [−1,1][-1, 1] exactly once. Both restricted functions are one-to-one, so both have inverse functions, as in inverse functions and domain restriction.

In plain words Keep only one stretch of each wave that runs through every height from −1-1 to 11 exactly once. That stretch passes the horizontal line test.

Example On the restricted sine the values only go up.

xx−π2-\frac{\pi}{2}−π6-\frac{\pi}{6}00π6\frac{\pi}{6}π2\frac{\pi}{2}
sin⁡x\sin x−1-1−12-\frac{1}{2}0012\frac{1}{2}11

On the restricted cosine they only go down.

xx00π3\frac{\pi}{3}π2\frac{\pi}{2}2π3\frac{2\pi}{3}π\pi
cos⁡x\cos x1112\frac{1}{2}00−12-\frac{1}{2}−1-1

Of the two angles π6\frac{\pi}{6} and 5π6\frac{5\pi}{6} with sine 12\frac{1}{2}, only π6\frac{\pi}{6} lies in the restricted domain of sine.

Principal branch

Definition When a function is not one-to-one, a branch is its restriction to a set on which it is one-to-one, usually an interval. This is a different use of the word from the branches of the tangent, cotangent, secant and cosecant graphs, which are the pieces of a graph between two consecutive vertical asymptotes. The principal branch is the agreed standard choice of such a restriction, one on which the function still takes every value of its range. For sine the principal branch is the restriction to [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], and for cosine it is the restriction to [0,π][0, \pi]. The inverse of the principal branch is called the principal branch of the inverse function, and its outputs are called principal values. Both intervals contain every angle between 00 and π2\frac{\pi}{2}, so when the sine or cosine of an acute angle in a right triangle is known, the inverse returns that acute angle.

In plain words Many angles share the same sine. The principal branch is the agreement about which of them counts as the answer, so that everyone gets the same one.

Example Sine is also one-to-one on [π2,3π2]\left[\frac{\pi}{2}, \frac{3\pi}{2}\right], where it falls from 11 to −1-1. Inverting that branch would send 12\frac{1}{2} to 5π6\frac{5\pi}{6}, the only number in that interval with sine 12\frac{1}{2}. The principal branch sends 12\frac{1}{2} to π6\frac{\pi}{6} instead. Either choice gives a valid inverse function, but the two agree only at the input 11, so one standard choice is fixed and used everywhere.

Arcsine

Definition For xx in [−1,1][-1, 1], the arcsine of xx, written arcsin⁡x\arcsin x or sin⁡−1x\sin^{-1} x, is the unique number yy in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] with sin⁡y=x\sin y = x. In symbols

y=arcsin⁡x⟺sin⁡y=x  and  −π2≤y≤π2.\begin{gathered} y = \arcsin x \\[0.4em] \Longleftrightarrow \\[0.4em] \sin y = x \ \text{ and } \ -\frac{\pi}{2} \le y \le \frac{\pi}{2}. \end{gathered}

Arcsine is the inverse of the restricted sine, so

sin⁡(arcsin⁡x)=xfor −1≤x≤1,arcsin⁡(sin⁡y)=yfor −π2≤y≤π2.\begin{gathered} \sin(\arcsin x) = x \\[0.4em] \text{for } -1 \le x \le 1, \\[0.8em] \arcsin(\sin y) = y \\[0.4em] \text{for } -\frac{\pi}{2} \le y \le \frac{\pi}{2}. \end{gathered}

It is increasing and odd, arcsin⁡(−x)=−arcsin⁡x\arcsin(-x) = -\arcsin x for every xx in [−1,1][-1, 1].

In plain words The arcsine of xx is the angle between −π2-\frac{\pi}{2} and π2\frac{\pi}{2} whose sine is xx.

Example Since sin⁡(−π4)=−22\sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} and −π4-\frac{\pi}{4} lies in the principal branch, arcsin⁡(−22)=−π4\arcsin\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}. The second cancellation rule needs care,

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

not 5π6\frac{5\pi}{6}, because 5π6\frac{5\pi}{6} lies outside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Arccosine

Definition For xx in [−1,1][-1, 1], the arccosine of xx, written arccos⁡x\arccos x or cos⁡−1x\cos^{-1} x, is the unique number yy in [0,π][0, \pi] with cos⁡y=x\cos y = x. In symbols

y=arccos⁡x⟺cos⁡y=x  and  0≤y≤π.\begin{gathered} y = \arccos x \\[0.4em] \Longleftrightarrow \\[0.4em] \cos y = x \ \text{ and } \ 0 \le y \le \pi. \end{gathered}

Arccosine is the inverse of the restricted cosine, so

cos⁡(arccos⁡x)=xfor −1≤x≤1,arccos⁡(cos⁡y)=yfor 0≤y≤π.\begin{gathered} \cos(\arccos x) = x \\[0.4em] \text{for } -1 \le x \le 1, \\[0.8em] \arccos(\cos y) = y \\[0.4em] \text{for } 0 \le y \le \pi. \end{gathered}

It is decreasing, and since cos⁡(π−y)=−cos⁡y\cos(\pi - y) = -\cos y, it satisfies arccos⁡(−x)=π−arccos⁡x\arccos(-x) = \pi - \arccos x for every xx in [−1,1][-1, 1]. It is neither even nor odd.

In plain words The arccosine of xx is the angle between 00 and π\pi whose cosine is xx. A negative input gives an angle greater than π2\frac{\pi}{2} and at most π\pi, never a negative angle.

Example The value of arccos⁡(−12)\arccos\left(-\frac{1}{2}\right) is 2π3\frac{2\pi}{3}, because cos⁡2π3=−12\cos \frac{2\pi}{3} = -\frac{1}{2} and 2π3\frac{2\pi}{3} lies in [0,π][0, \pi]. The angles 4π3\frac{4\pi}{3} and −2π3-\frac{2\pi}{3} also have cosine −12-\frac{1}{2}, but they lie outside [0,π][0, \pi]. The symmetry rule agrees, since π−arccos⁡12=π−π3=2π3\pi - \arccos \frac{1}{2} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}. In the same way

arccos⁡(cos⁡(−π3))=arccos⁡12=π3.\begin{aligned} \arccos\left(\cos\left(-\frac{\pi}{3}\right)\right) &= \arccos\frac{1}{2} \\[0.8em] &= \frac{\pi}{3}. \end{aligned}

Domain of arcsine and arccosine

Definition The domain of both arcsin⁡\arcsin and arccos⁡\arccos is the closed interval [−1,1][-1, 1], the range of sine and of cosine. For a real number xx with ∣x∣>1\lvert x \rvert > 1 no real angle has sine or cosine equal to xx, so arcsin⁡x\arcsin x and arccos⁡x\arccos x are not defined as real numbers. A composite expression such as arcsin⁡(u(x))\arcsin\bigl(u(x)\bigr) is defined exactly for the xx in the domain of uu with −1≤u(x)≤1-1 \le u(x) \le 1.

In plain words You can only ask which angle has sine xx when some angle does. Sine and cosine never leave [−1,1][-1, 1], so their inverses accept nothing else.

Example The expression arcsin⁡2\arcsin 2 is undefined. For the domain of f(x)=arccos⁡(2x−3)f(x) = \arccos(2x - 3), require

−1≤2x−3≤1,which gives1≤x≤2.\begin{gathered} -1 \le 2x - 3 \le 1, \\[0.4em] \text{which gives} \quad 1 \le x \le 2. \end{gathered}

So the domain is [1,2][1, 2]. At the two ends and the middle, f(1)=arccos⁡(−1)=πf(1) = \arccos(-1) = \pi, f(32)=arccos⁡0=π2f\left(\frac{3}{2}\right) = \arccos 0 = \frac{\pi}{2} and f(2)=arccos⁡1=0f(2) = \arccos 1 = 0.

Principal output ranges

Definition The range of arcsin⁡\arcsin is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and the range of arccos⁡\arccos is [0,π][0, \pi]. The outputs are real numbers, read as angles in radians, and in degrees the ranges are [−90∘,90∘][-90^\circ, 90^\circ] and [0∘,180∘][0^\circ, 180^\circ]. So the terminal side of an arcsine value lies in Quadrant I or IV, on the positive xx-axis or on the yy-axis, and the terminal side of an arccosine value lies in Quadrant I or II, on the xx-axis or on the positive yy-axis. The two outputs are linked. For every xx in [−1,1][-1, 1]

arcsin⁡x+arccos⁡x=π2.\arcsin x + \arccos x = \frac{\pi}{2}.

To see this, let y=arcsin⁡xy = \arcsin x. Then π2−y\frac{\pi}{2} - y lies in [0,π][0, \pi]. Cosine is even and cos⁡u=sin⁡(u+π2)\cos u = \sin\left(u + \frac{\pi}{2}\right) for every real uu, so

cos⁡(π2−y)=cos⁡(y−π2)=sin⁡y=x,\begin{aligned} \cos\left(\frac{\pi}{2} - y\right) &= \cos\left(y - \frac{\pi}{2}\right) \\[0.8em] &= \sin y = x, \end{aligned}

and therefore π2−y=arccos⁡x\frac{\pi}{2} - y = \arccos x.

In plain words The inverse sine key on a calculator never returns an angle above 90∘90^\circ or below −90∘-90^\circ, and the inverse cosine key never returns a negative angle. If the angle you need lies elsewhere, you find it from the principal value yourself.

Example For x=−0.6x = -0.6,

arcsin⁡(−0.6)≈−0.6435,arccos⁡(−0.6)≈2.2143,\begin{gathered} \arcsin(-0.6) \approx -0.6435, \\[0.4em] \arccos(-0.6) \approx 2.2143, \end{gathered}

about −36.87∘-36.87^\circ in Quadrant IV and 126.87∘126.87^\circ in Quadrant II. Their sum is about 1.57081.5708, which is π2\frac{\pi}{2}. If a problem needs the angle between π\pi and 3π2\frac{3\pi}{2} with sine −0.6-0.6, which lies in Quadrant III, use sin⁡(π−y)=sin⁡y\sin(\pi - y) = \sin y to get π−arcsin⁡(−0.6)≈3.7851\pi - \arcsin(-0.6) \approx 3.7851, about 216.87∘216.87^\circ.

Inverse versus reciprocal

Definition The notation sin⁡−1x\sin^{-1} x means arcsin⁡x\arcsin x, the inverse function. It does not mean the reciprocal 1sin⁡x\frac{1}{\sin x}, which is csc⁡x\csc x and is written (sin⁡x)−1(\sin x)^{-1}. Likewise cos⁡−1x=arccos⁡x\cos^{-1} x = \arccos x, while (cos⁡x)−1=sec⁡x(\cos x)^{-1} = \sec x. This clashes with the habit of writing sin⁡2x\sin^2 x for (sin⁡x)2(\sin x)^2, which is one reason many books prefer the names arcsin⁡\arcsin and arccos⁡\arccos.

In plain words The −1-1 means undo the sine, not one over the sine. The reciprocal flips a value upside down, while the inverse goes from a value back to an angle.

Example The two readings give very different numbers,

sin⁡−112=π6≈0.5236,(sin⁡12)−1=csc⁡12≈2.0858.\begin{gathered} \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \approx 0.5236, \\[0.8em] \left(\sin\frac{1}{2}\right)^{-1} = \csc\frac{1}{2} \approx 2.0858. \end{gathered}

The expression sin⁡−12\sin^{-1} 2 is undefined, while (sin⁡2)−1=csc⁡2≈1.0998(\sin 2)^{-1} = \csc 2 \approx 1.0998 is a perfectly good number.

Check yourself

  1. Evaluate arcsin⁡(−12)\arcsin\left(-\frac{1}{2}\right), arccos⁡(−32)\arccos\left(-\frac{\sqrt{3}}{2}\right) and arccos⁡0\arccos 0.
Show the answer

The first is −π6-\frac{\pi}{6}, since sin⁡(−π6)=−12\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} and −π6-\frac{\pi}{6} lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. The second is 5π6\frac{5\pi}{6}, since cos⁡5π6=−32\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} and 5π6\frac{5\pi}{6} lies in [0,π][0, \pi]. The third is π2\frac{\pi}{2}, since cos⁡π2=0\cos \frac{\pi}{2} = 0 and π2\frac{\pi}{2} lies in [0,π][0, \pi].

  1. Evaluate arccos⁡(cos⁡5π3)\arccos\left(\cos \frac{5\pi}{3}\right).
Show the answer

The inner value is cos⁡5π3=12\cos \frac{5\pi}{3} = \frac{1}{2}, so

arccos⁡(cos⁡5π3)=arccos⁡12=π3.\arccos\left(\cos \frac{5\pi}{3}\right) = \arccos \frac{1}{2} = \frac{\pi}{3}.

The answer is not 5π3\frac{5\pi}{3}, because that angle lies outside [0,π][0, \pi].

  1. Find the domain and the range of g(x)=arcsin⁡(x3−1)g(x) = \arcsin\left(\frac{x}{3} - 1\right).
Show the answer

The input must satisfy −1≤x3−1≤1-1 \le \frac{x}{3} - 1 \le 1, so 0≤x≤60 \le x \le 6 and the domain is [0,6][0, 6]. As xx runs over [0,6][0, 6], the input x3−1\frac{x}{3} - 1 runs over all of [−1,1][-1, 1], so the range is the full range of arcsine, [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. For example g(0)=−π2g(0) = -\frac{\pi}{2}, g(3)=0g(3) = 0 and g(6)=π2g(6) = \frac{\pi}{2}.

How this lesson connects

Where this leads in Calculus