Precalculus · F.BF.B.4D · F.TF.B.6

Inverse trig functions

A sine key turns an angle into a number. Can a key turn the number back into the angle, when infinitely many angles share every sine?

You need

Periodic functionsPrecalculus Inverse functionsGrades 10 to 11

Unlocks

Composing trig and inverse trig functionsPrecalculus Solving basic trig equationsPrecalculus The complex plane, modulus and argumentPrecalculus Derivatives of inverse functionsCalculus 1 Derivatives of inverse trigonometric functionsCalculus 1 +2 more

Where you are 8 in Trigonometric functions and their graphs

Trigonometry7

Trigonometric functions and their graphs8

▴ 2 above Graphing tangent and the reciprocal functions Periodic functions

Inverse trig functions

Writing an equation from a wave Composing trig and inverse trig functions ▾ Show all 8 (1 more)

Why this exists

Mirjana runs a spotlight from a theater balcony. A rangefinder on the light reads how far the actor is along the beam, and she knows how far the light sits above the actor. Their ratio is the sine of the tilt, so she knows \sin\theta and wants \theta. That is sine run backward: not “what is the sine of this angle” but “which angle has this sine”.

Running a function backward is what an inverse function does: if f(a) = b, then f^{-1}(b) = a. That only works when no two inputs share an output, and sine is periodic, so every value it takes once it takes again in every cycle. The sine of \displaystyle -\frac{\pi}{3}, of \displaystyle \frac{4\pi}{3} and of \displaystyle \frac{5\pi}{3} is \displaystyle -\frac{\sqrt{3}}{2} every time, and infinitely many more angles join them. A backward key would not know which to give.

The cure is the one from restricting a domain to get an inverse: keep one piece of the graph on which the function is one to one, and invert only that piece. This lesson makes that choice for sine, cosine and tangent, once, so that every book and every calculator gives the same answer. The inverse ratio keys of right triangle trigonometry are these same functions; there they only ever met acute angles, where nothing could go wrong.

Definition

Part 1: the sine slice and arcsine

On the interval \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], sine climbs steadily from -1 to 1 and takes every value in [-1, 1] exactly once. The restricted sine is \sin x with its domain cut down to that interval.

The arcsine of a number x in [-1, 1] is the one angle in that slice whose sine is x:

\begin{aligned} y = \arcsin x \quad &\text{means} \quad \sin y = x \\ &\text{and} \quad -\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2} \end{aligned}

Its domain is [-1, 1], the range of sine, and its range is \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], the slice. The output is an angle in radians, which is a real number. Arcsine is also written \sin^{-1} x. That -1 marks an inverse function, not a power: \sin^{-1} x is not \displaystyle \frac{1}{\sin x}, which is \csc x. This lesson writes \arcsin x.

Part 2: arccosine and arctangent

Cosine needs a different slice, because on \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] it repeats itself: \displaystyle \cos\left(-\frac{\pi}{3}\right) = \cos\frac{\pi}{3}. On [0, \pi] cosine falls steadily from 1 to -1 and takes every value in [-1, 1] once. Tangent is cut to the open interval \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), the stretch between two neighboring asymptotes of its graph, where it climbs through every real number once.

\begin{aligned} y = \arccos x \quad &\text{means} \quad \cos y = x \\ &\text{and} \quad 0 \le y \le \pi \\ y = \arctan x \quad &\text{means} \quad \tan y = x \\ &\text{and} \quad -\tfrac{\pi}{2} < y < \tfrac{\pi}{2} \end{aligned}

Function	Domain	Range
\arcsin x	[-1, 1]	\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\arccos x	[-1, 1]	[0, \pi]
\arctan x	all reals	\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Each range is the slice that was kept. The ends of the tangent slice are left out because tangent has no value there.

Intuition

Figure 1The line \displaystyle y = -\frac{\sqrt{3}}{2} meets the sine wave again and again, but it meets the thick slice on \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] exactly once.

A horizontal line at any height between -1 and 1 crosses the sine graph infinitely often. Cut the graph down to one rising stretch and the line crosses once, so that stretch has an inverse. The slices are chosen with care. Each contains 0 and every acute angle, so arcsine agrees with the triangle keys on acute angles. Each is as long as it can be while staying one to one, so the inverse still reaches every value.

On the unit circle, the sine slice is the right half of the circle, from the bottom (0, -1) up to the top (0, 1); every height appears there once. The cosine slice is the upper half, from (1, 0) over to (-1, 0); every horizontal coordinate appears there once. The tangent slice is the right half without its two ends, where every slope of the ray appears once.

For Mirjana the sine slice is exactly right: a light tilts anywhere from straight down, \displaystyle -\frac{\pi}{2}, to straight up, \displaystyle \frac{\pi}{2}, and arcsine hands back one tilt with its sign.

Since arcsine undoes the restricted sine, a point (a, b) on one graph is the point (b, a) on the other, and swapping coordinates reflects a graph across the line y = x.

Figure 2The arcsine graph is the restricted sine graph reflected across y = x. Its domain is [-1, 1] and its range is \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Results

Undoing. \sin(\arcsin x) = x for every x in [-1, 1]. In the other order, \arcsin(\sin y) = y only when y lies in \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. Likewise \cos(\arccos x) = x on [-1, 1] and \arccos(\cos y) = y only on [0, \pi]; \tan(\arctan x) = x for every x and \arctan(\tan y) = y only on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

Symmetry. For every x in each domain,

\begin{aligned} \arcsin(-x) &= -\arcsin x \\ \arctan(-x) &= -\arctan x \\ \arccos(-x) &= \pi - \arccos x \end{aligned}

Graphs. Each graph is the restricted graph reflected across y = x. Arcsine rises from \displaystyle \left(-1, -\frac{\pi}{2}\right) to \displaystyle \left(1, \frac{\pi}{2}\right); arccosine falls from (-1, \pi) to (1, 0); arctangent rises across the whole real line between the horizontal asymptotes \displaystyle y = -\frac{\pi}{2} and \displaystyle y = \frac{\pi}{2}.

The other angles. The angles with sine x are exactly \arcsin x + 2\pi k and \pi - \arcsin x + 2\pi k, with k any integer.

Why it is true

Each slice is one to one and reaches every value. As y runs from \displaystyle -\frac{\pi}{2} to \displaystyle \frac{\pi}{2}, the terminal point moves counterclockwise along the right half of the unit circle from (0, -1) to (0, 1). Its height rises the whole way and moves without jumps, so it passes each height in [-1, 1] once. On [0, \pi] the point moves along the upper half from (1, 0) to (-1, 0), and its horizontal coordinate falls the whole way. On \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) the ray turns from pointing straight down to straight up through the right half plane, and its slope, the tangent, rises through every real number.

Symmetry. Let y = \arcsin x. Then -y is also in the slice and \sin(-y) = -\sin y = -x, because sine is odd, so \arcsin(-x) = -y. Tangent is odd too, and the same argument works. For cosine, let y = \arccos x, in [0, \pi]. Then \pi - y is also in [0, \pi], and \cos(\pi - y) = -\cos y = -x, so \arccos(-x) = \pi - y.

Graphs. (a, b) lies on the restricted sine exactly when a is in the slice and \sin a = b, which is exactly when \arcsin b = a, which says (b, a) lies on the arcsine graph.

Other angles. On the unit circle the height x is reached at two points that mirror each other across the vertical axis, at the angles \arcsin x and \pi - \arcsin x (one point when x = \pm 1). Every full turn 2\pi k returns to the same two points.

Worked examples

Example 1

Part 1: two exact arcsines

Find \displaystyle \arcsin\left(-\frac{\sqrt{3}}{2}\right) and \arcsin 1.

Solution

The first asks for the angle in \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] whose sine is \displaystyle -\frac{\sqrt{3}}{2}. We know \displaystyle \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}, and sine is odd, so \displaystyle \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}. The angle \displaystyle -\frac{\pi}{3} lies in the slice, so \displaystyle \arcsin\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}. The angles \displaystyle \frac{4\pi}{3} and \displaystyle \frac{5\pi}{3} have the same sine, but they lie outside the slice, so neither is the arcsine. Figure 1 shows all four crossings and the one that counts.

For the second, \displaystyle \sin\frac{\pi}{2} = 1 and \displaystyle \frac{\pi}{2} is the right end of the slice, which is included. So \displaystyle \arcsin 1 = \frac{\pi}{2}.

Example 2

Part 1: undoing, and where it stops

Find (a) \sin(\arcsin 0.3), (b) \displaystyle \arcsin\left(\sin\frac{\pi}{5}\right), (c) \displaystyle \arcsin\left(\sin\frac{7\pi}{8}\right).

Solution

(a) By definition \arcsin 0.3 is an angle whose sine is 0.3, so its sine is 0.3. No calculator is needed: \sin(\arcsin 0.3) = 0.3.

(b) \displaystyle \frac{\pi}{5} is less than \displaystyle \frac{\pi}{2}, so it lies in the slice, and the slice angle with sine \displaystyle \sin\frac{\pi}{5} is \displaystyle \frac{\pi}{5} itself. The answer is \displaystyle \frac{\pi}{5}.

(c) Now \displaystyle \frac{7\pi}{8} > \frac{\pi}{2}, outside the slice, so the answer cannot be \displaystyle \frac{7\pi}{8}. Since \sin(\pi - t) = \sin t, \displaystyle \sin\frac{7\pi}{8} = \sin\frac{\pi}{8}, and \displaystyle \frac{\pi}{8} is in the slice. So \displaystyle \arcsin\left(\sin\frac{7\pi}{8}\right) = \frac{\pi}{8}. Check: \displaystyle \sin\frac{7\pi}{8} \approx 0.3827, and a calculator gives \displaystyle \arcsin 0.3827 \approx 0.3927 \approx \frac{\pi}{8}.

Example 3

Part 1: Mirjana’s tilt

Figure 3Example 3. The beam is 34 feet long and drops 20 feet; the tilt \theta is measured from the horizontal, negative when the light points down.

Mirjana’s spotlight is 20 feet above the actor’s head, and the rangefinder reads 34 feet along the beam. The tilt \theta is measured from the horizontal, positive upward and negative downward. Find \theta. Then find the tilt when the actor stands directly below the light.

Solution

The beam, the horizontal through the light and the vertical through the actor make a right triangle. Going along the beam the height changes by -20 feet over 34 feet, so

\sin\theta = \frac{-20}{34} = -\frac{10}{17}

A tilt runs from straight down, \displaystyle -\frac{\pi}{2}, to straight up, \displaystyle \frac{\pi}{2}: exactly the arcsine slice, so there is one answer.

\theta = \arcsin\left(-\frac{10}{17}\right) \approx -0.629 \text{ radians}

That is about 36.0^\circ below the horizontal. The sign carries the direction, which the acute angle from a triangle key would not.

Directly below, the beam is 20 feet long and straight down: \displaystyle \sin\theta = -\frac{20}{20} = -1 and \displaystyle \theta = \arcsin(-1) = -\frac{\pi}{2}.

Example 4

Part 2: an arccosine of a negative number

Find \arccos(-0.4) and \arcsin(-0.4) to four decimal places, explain their signs, and check the first with \arccos(-x) = \pi - \arccos x.

Solution

With the calculator in radian mode, \arccos(-0.4) \approx 1.9823 and \arcsin(-0.4) \approx -0.4115. The cosine slice [0, \pi] holds no negative angles, so an arccosine is never negative; a negative input sends it past \displaystyle \frac{\pi}{2}, into the second quadrant, where the terminal point on the upper half circle has a negative horizontal coordinate. The sine slice reaches below zero, so a negative input gives an angle in the fourth quadrant.

Check: \arccos 0.4 \approx 1.1593 and \pi - 1.1593 \approx 1.9823.

Example 5

Part 2: Ivana’s pan

Ivana’s spotlight hangs above the front edge of the stage, at its center. An actor stands 4 feet upstage of the edge and 12 feet left of center. The pan angle \varphi is measured from pointing straight upstage, positive to the right and negative to the left. Find \varphi.

Solution

Seen from above, sideways distance over depth is the tangent of the pan: \displaystyle \tan\varphi = \frac{-12}{4} = -3. A light aimed onto the stage pans strictly between \displaystyle -\frac{\pi}{2} and \displaystyle \frac{\pi}{2} (pointing exactly sideways would run along the edge), which is the tangent slice. So

\varphi = \arctan(-3) \approx -1.2490 \text{ radians}

about 71.6^\circ to the left. Check: \tan(-1.2490) \approx -3.00.

Example 6

Part 2: the graphs of arccosine and arctangent

Sketch y = \arccos x and y = \arctan x from the restricted graphs. Give key points, domain, range and any asymptotes, and find \arctan 10 and \arctan 100.

Solution

The restricted cosine passes through (0, 1), \displaystyle \left(\frac{\pi}{2}, 0\right) and (\pi, -1). Swapping coordinates gives (1, 0), \displaystyle \left(0, \frac{\pi}{2}\right) and (-1, \pi) on the arccosine graph. A falling graph reflected across y = x still falls, so arccosine falls from (-1, \pi) to (1, 0), with domain [-1, 1] and range [0, \pi].

Figure 4Example 6. Arccosine falls from (-1, \pi) to (1, 0); it is the restricted cosine reflected across y = x.

The restricted tangent passes through \displaystyle \left(-\frac{\pi}{4}, -1\right), (0, 0) and \displaystyle \left(\frac{\pi}{4}, 1\right) and has vertical asymptotes \displaystyle x = \pm\frac{\pi}{2}. Swapping coordinates turns them into horizontal asymptotes \displaystyle y = \pm\frac{\pi}{2}, and gives \displaystyle (-1, -\frac{\pi}{4}), (0, 0), \displaystyle \left(1, \frac{\pi}{4}\right). The domain is all real numbers and the range is \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

Figure 5Example 6. Arctangent takes every real input and stays strictly between the horizontal asymptotes \displaystyle y = -\frac{\pi}{2} and \displaystyle y = \frac{\pi}{2}. The dot is (-3, \arctan(-3)) from Example 5.

\arctan 10 \approx 1.4711 and \arctan 100 \approx 1.5608, against \displaystyle \frac{\pi}{2} \approx 1.5708: the graph keeps climbing but never reaches its asymptote.

Example 7

Part 2: when the angle lies outside the slice

Find \displaystyle \arccos\left(\cos\frac{5\pi}{4}\right) and \displaystyle \arctan\left(\tan\frac{5\pi}{6}\right).

Solution

\displaystyle \frac{5\pi}{4} is past \pi, outside [0, \pi]. Cosine is even and has period 2\pi, so \displaystyle \cos\frac{5\pi}{4} = \cos\left(2\pi - \frac{5\pi}{4}\right) = \cos\frac{3\pi}{4}, and \displaystyle \frac{3\pi}{4} is in the slice. So \displaystyle \arccos\left(\cos\frac{5\pi}{4}\right) = \frac{3\pi}{4}. Check: both cosines equal \displaystyle -\frac{\sqrt{2}}{2}.

Tangent has period \pi, so \displaystyle \tan\frac{5\pi}{6} = \tan\left(\frac{5\pi}{6} - \pi\right) = \tan\left(-\frac{\pi}{6}\right), and \displaystyle -\frac{\pi}{6} is in \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right). So \displaystyle \arctan\left(\tan\frac{5\pi}{6}\right) = -\frac{\pi}{6}. Check: both tangents equal \displaystyle -\frac{\sqrt{3}}{3}.

Boundaries

Domains. Arcsine and arccosine accept only [-1, 1]. No real angle has sine 2, so \arcsin 2 is undefined. Arctangent accepts every real number.
Order matters. \sin(\arcsin x) = x on the whole domain, but \arcsin(\sin y) = y only on the slice; outside it the answer is the slice angle with the same sine (Examples 2 and 7).
One angle, not all. An inverse trig function returns one angle, the slice angle. Every other angle with that sine, cosine or tangent comes from the families in the Results.
The slices are a convention. Sine is also one to one on \displaystyle \left[\frac{\pi}{2}, \frac{3\pi}{2}\right], and inverting that piece gives a valid inverse, but it is not \arcsin. The name fixes the slice.
Units. The output is in radians. A calculator in degree mode gives the same angles in degrees, with slices [-90^\circ, 90^\circ], [0^\circ, 180^\circ] and (-90^\circ, 90^\circ).
Notation. \sin^{-1} x means arcsine, while \sin^2 x means (\sin x)^2. The exponent -1 is the one exception.

Where this goes next

Solving trig equations starts here: the inverse gives one angle and the families give the rest. Composing a trig function with an inverse, such as \cos(\arcsin x), uses a triangle built from the inner angle. In the complex plane, the angle of a point comes from arctangent, adjusted for the quadrant. In calculus, the slope of an inverse and the derivatives of these three functions turn out to have no trig left in them.

Exercises

Part 1: exact arcsines

Find the exact value.

\displaystyle \arcsin\frac{1}{2}
\displaystyle \arcsin\left(-\frac{\sqrt{2}}{2}\right)
\arcsin 0
\displaystyle \arcsin\frac{3}{2}, or explain why it does not exist.

Part 1: with a calculator

Give radians to four decimal places.

\arcsin 0.8
\arcsin(-0.25)

Part 1: undoing

Find the exact value.

\displaystyle \arcsin\left(\sin\frac{2\pi}{3}\right)
\displaystyle \arcsin\left(\sin\left(-\frac{7\pi}{6}\right)\right)
\arcsin(\sin 2)
\sin\left(\arcsin(-0.7)\right)

Part 1: the spotlight

A spotlight is 18 feet above an actor’s head, and the beam to the head is 30 feet long. Find the tilt in radians to three decimal places, and in degrees.
A floor light points up at an actor on a lift whose head is 9 feet above the light, along a 24 foot beam. Find the tilt in radians to three decimal places, and in degrees.
Hrvoje sets a spotlight to a tilt of -0.5 radians, and the beam to the actor’s head is 26 feet. How far above the head is the light, to the nearest tenth of a foot?

Part 2: exact arccosines and arctangents

Find the exact value.

\displaystyle \arccos\left(-\frac{1}{2}\right)
\displaystyle \arccos\frac{\sqrt{3}}{2}
\displaystyle \arctan\sqrt{3}
\arctan(-1)

Part 2: with a calculator

\arccos(-0.9), to four decimal places.
\arctan 0.5, to four decimal places.
With Ivana’s setup from Example 5, an actor stands 20 feet upstage of the edge and 8 feet right of center. Find the pan angle in radians to three decimal places, and in degrees.

Part 2: undoing

\displaystyle \arccos\left(\cos\frac{7\pi}{4}\right)
\displaystyle \arctan\left(\tan\frac{4\pi}{3}\right)

Reasoning

Explain why [0, \pi] would not work as the slice for sine, and why \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] would not work for cosine. Give two angles for each.
Prove that \displaystyle \arcsin x + \arccos x = \frac{\pi}{2} for every x in [-1, 1].
Explain why the graph of y = \arctan x has horizontal asymptotes, while the graph of y = \arcsin x stops at two endpoints.

Answers

\displaystyle \frac{\pi}{6}. \displaystyle \sin\frac{\pi}{6} = \frac{1}{2} and \displaystyle \frac{\pi}{6} is in the slice.
\displaystyle -\frac{\pi}{4}. \displaystyle \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}, and arcsine is odd.
0. \sin 0 = 0 and 0 is in the slice.
It does not exist. Every sine lies in [-1, 1], and \displaystyle \frac{3}{2} > 1, so no angle has sine \displaystyle \frac{3}{2}; the domain of arcsine is [-1, 1].
\arcsin 0.8 \approx 0.9273.
\arcsin(-0.25) \approx -0.2527, a fourth quadrant angle.
\displaystyle \frac{\pi}{3}. \displaystyle \frac{2\pi}{3} is outside the slice, and \displaystyle \sin\frac{2\pi}{3} = \sin\left(\pi - \frac{2\pi}{3}\right) = \sin\frac{\pi}{3}.
\displaystyle \frac{\pi}{6}. \displaystyle \sin\left(-\frac{7\pi}{6}\right) = \frac{1}{2} (the angle ends in the second quadrant, at height \displaystyle \frac{1}{2}), and the slice angle with sine \displaystyle \frac{1}{2} is \displaystyle \frac{\pi}{6}.
\pi - 2 \approx 1.1416. \displaystyle 2 > \frac{\pi}{2} is outside the slice; \sin 2 = \sin(\pi - 2) and \pi - 2 is in the slice.
-0.7. -0.7 is in [-1, 1], and sine undoes arcsine on the whole domain.
\displaystyle \sin\theta = -\frac{18}{30} = -0.6, so \theta = \arcsin(-0.6) \approx -0.644 radians, about 36.9^\circ below the horizontal.
\displaystyle \sin\theta = \frac{9}{24} = 0.375, so \theta = \arcsin 0.375 \approx 0.384 radians, about 22.0^\circ above the horizontal.
The height change is 26\sin(-0.5) \approx -12.47 feet, so the light is about 12.5 feet above the head. This runs sine forward; no inverse is needed.
\displaystyle \frac{2\pi}{3}. \displaystyle \cos\frac{2\pi}{3} = -\frac{1}{2} and \displaystyle \frac{2\pi}{3} is in [0, \pi]; also \displaystyle \pi - \arccos\frac{1}{2} = \pi - \frac{\pi}{3}.
\displaystyle \frac{\pi}{6}. \displaystyle \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}.
\displaystyle \frac{\pi}{3}. \displaystyle \tan\frac{\pi}{3} = \sqrt{3}.
\displaystyle -\frac{\pi}{4}. \displaystyle \tan\frac{\pi}{4} = 1, and arctangent is odd.
\arccos(-0.9) \approx 2.6906, a second quadrant angle; \pi - \arccos 0.9 \approx 3.1416 - 0.4510 agrees.
\arctan 0.5 \approx 0.4636.
\displaystyle \tan\varphi = \frac{8}{20} = 0.4, so \varphi = \arctan 0.4 \approx 0.381 radians, about 21.8^\circ to the right.
\displaystyle \frac{\pi}{4}. \displaystyle \cos\frac{7\pi}{4} = \cos\left(2\pi - \frac{7\pi}{4}\right) = \cos\frac{\pi}{4}, and \displaystyle \frac{\pi}{4} is in [0, \pi].
\displaystyle \frac{\pi}{3}. Tangent has period \pi, so \displaystyle \tan\frac{4\pi}{3} = \tan\frac{\pi}{3}, and \displaystyle \frac{\pi}{3} is in the slice.
On [0, \pi] sine repeats: \displaystyle \sin\frac{\pi}{6} = \sin\frac{5\pi}{6} = \frac{1}{2}, so it is not one to one (and it never goes negative, so it would miss half of [-1, 1]). On \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] cosine repeats: \displaystyle \cos\left(-\frac{\pi}{3}\right) = \cos\frac{\pi}{3} = \frac{1}{2}, and it never goes negative there either.
Let y = \arcsin x, so \sin y = x and \displaystyle -\frac{\pi}{2} \le y \le \frac{\pi}{2}. Then \displaystyle 0 \le \frac{\pi}{2} - y \le \pi, which is the cosine slice, and \displaystyle \cos\left(\frac{\pi}{2} - y\right) = \sin y = x. So \displaystyle \arccos x = \frac{\pi}{2} - y, and adding y to both sides gives \displaystyle \arcsin x + \arccos x = \frac{\pi}{2}.
Each graph is its restricted graph reflected across y = x. The tangent slice is open and tangent grows without bound near its ends, so the restricted graph has vertical asymptotes \displaystyle x = \pm\frac{\pi}{2}; reflecting turns them into horizontal asymptotes \displaystyle y = \pm\frac{\pi}{2}, and every real input is allowed. The sine slice is closed and sine only reaches \pm 1, so the restricted graph ends at two points, \displaystyle \left(\pm\frac{\pi}{2}, \pm 1\right), and the arcsine graph ends at their reflections, \displaystyle \left(\pm 1, \pm\frac{\pi}{2}\right).