Trig without Tears Part 9:

# Inverse Functions

Copyright © 1997–2019 Stan Brown, BrownMath.com

**Summary:**
The inverse trig functions (also called **arcfunctions**)
are similar to any other inverse functions: they go
**from the function value back to the angle (or number)**. Their
**ranges are restricted**, by definition, because an inverse
function must not give multiple answers. Once you understand the
inverse functions, you can simplify **composite functions** like
sin(Arctan x).

An inverse is the math equivalent of an undo. For
example, if you have an angle *A* = 40°, you can find
sin *A* ≈ 0.64. But you have to go the other direction
when you’re solving a
triangle. For instance, you might get
sin *B* = 0.82 and have to find the angle *B*.
You’re not asking what’s the sine of some angle, but
rather, “What angle has a sine equal to 0.82?”

A shorter way to ask that question is, “what’s arcsin(0.82)?” This uses the name “arcsin” for the inverse of the sine function — not going from angle to its sine, but from the sine to the angle.

Is arcsin a function? Well, look at the graph
of *y* = 0.82
against *y* = sin *x*. Where they cross is
arcsin(0.82), and obviously there are many possible answers. So
although sin is a function, arcsin is not. Your
calculator will give you an answer of around 55°, but that’s
just one out of infinitely many. You know from equation 22 that
sin(180° − *x*) = sin *x*, and
since 180° − 55° = 125°,
sin 125° = sin 55. But all the trig functions are
periodic, repeating every
360° forever, so infinitely many angles (or numbers) have a sine
of 0.82:

arcsin(0.82) ≈ (55 + 360*k*)°
and (125 + 360*k*)°, where *k* is any
integer.

arcsin(0.82) ≈ (0.96 + 2π*k*)
and (2.18 + 2π*k*), where *k* is any
integer.

So arcsin(0.82) is all the numbers (or angles) whose sine
is 0.82. But if we want an inverse sine *function*, we have to
have a single answer. That single answer is called the principal
value, and is written Arcsin(0.82):

Arcsin(0.82) ≈ 55°, or 55π/180 = 0.96 in radian measure.

A portion of

Arcsin(*x*), with the capital letter, is the
principal value of arcsin(*x*).
Lower-case arcsin(*x*) is all possible numbers or
angles whose sine is *x*.
The same convention applies to the other five functions.
(See Notation, below, for other ways of
writing the inverse relation and the inverse function.)

Now all we need is a rule for picking the principal values of all the inverse trig functions. We want a continuous interval, with no gaps, and we want that interval to include the range from 0° to 90° (0 to π/2). It turns out that the six functions can’t all have the same range.

For **Arcsin**, the only possibility that meets those
requirements is that Arcsin *x* must return a number in the
interval [−π/2, +π/2], which is the
same as [−90°, +90°].

The inverse tangent, **Arctan**, is almost the
same. But since there’s no value for tan(±90°)
or tan(±π/2), the range of Arctan is the
open interval (−π/2, +π/2) or
(−90°, +90°).

What about **Arccos**? The cosine is positive in both Quadrant
I and Quadrant IV, so the arccosine of a negative number must fall in
Quadrant II or Quadrant III.
Thomas (Calculus and Analytic
Geometry, 4th edition) resolves this in a neat way.
Remember from equation 2 that

cos *A* = sin(π/2 − *A*)

It makes a nice symmetry to write

Arccos *x* = π/2 − Arcsin *x*

And that is how Thomas defines the inverse cosine function. Since the range of Arcsin is the closed interval [−π/2, +π/2], the range of Arccos is π/2 minus that, [0, π] or [0°, 180°].

Once the range for Arctan is defined, there’s really only
one sensible way to define **Arccot**:

cot *x* = tan(π/2 − *x*)
⇒
Arccot *x* = π/2 − Arctan *x*

which gives the single open interval (0, π) or (0°, 180°) as the range.

Thomas defines the **Arcsec** and
**Arccsc** functions using the
reciprocal relationships from equation 5:

sec *x* = 1/(cos *x*)
⇒
Arcsec *x* = Arccos(1/*x*)

csc *x* = 1/(sin *x*)
⇒
Arccsc *x* = Arcsin(1/*x*)

This means that Arcsec and Arccsc have the same ranges as Arccos and Arcsin, respectively.

Here are the **domains (inputs) and ranges (outputs)** of all
six inverse trig functions:

function | derived from | domain | range |
---|---|---|---|

Arcsin | inverse of sine function | [−1, +1] | Q IV, I: [−π/2, +π/2] |

Arccos | Arccos x = π/2 − Arcsin x |
[−1, +1] | Q I, II: [0, π] |

Arctan | inverse of tangent function | all reals | Q IV, I: (−π/2, +π/2) |

Arccot | Arccot x = π/2 − Arctan x |
all reals | Q I, II: (0, π) |

Arcsec | Arcsec x = Arccos(1/x) |
(−∞, −1] and [1, ∞) | Q I, II: [0, π] |

Arccsc | Arccsc x = Arcsin(1/x) |
(−∞, −1] and [1, ∞) | Q IV, I: [−π/2, +π/2] |

Remember that the inverse *relations* arcsin
etc. are many-valued, not
limited to the above ranges of the *functions*. If you see the
capital A in the function name, you know you’re talking about the
function; otherwise you have to depend on context.

In this book, the many-valued inverse *relation* is
arcsin with lower-case a, and the inverse *function*
is Arcsin with capital A. That’s a common choice in
the U.S., but it’s not the only choice.

- Some books, including Thomas, use
sin
^{−1}for the function. That’s actually a logical choice, since^{−1}is the standard way to designate an inverse function. But the problem is that it makes you want to think of sin^{−1}(*x*) as 1/sin(*x*), even though they’re not the same at all. - Many calculators use sin
^{−1}for the function name, though some use lower-case arcsin. - Excel, and many programming languages, use asin for the function.
- Outside the U.S., lower-case arcsin often means the principal value, and if you want the many-valued relation you have to use a periphrasis.

Because there are so many conventions, authors generally explain which notation they’re using, so watch for that.

**Advice tothe reader:**
The methods in this section aren’t
really very useful in trigonometry itself, but are used in integral calculus
and some physics or engineering courses.
You may wish to skip them, especially on a first reading. On the other
hand, they

Sometimes you have to evaluate expressions like

cos(Arctan *x*)

That looks scary, but actually it’s a piece of cake.
You can simplify **any trig function of any inverse trig function**
in two easy steps, using this method:

**Think of the inner arcfunction as an angle**. Draw a right triangle and label that angle and the two relevant sides.**Use the Pythagorean Theorem**to find the third side of the triangle, then write down the value of the outer function according to its definition.- If the answer contains variables raised to
**odd powers, including the first power**, you may need to add some absolute value signs. See Example 3.

Because this textbook helps you,

please click to donate!Because this textbook helps you,

please donate at

BrownMath.com/donate.

please click to donate!Because this textbook helps you,

please donate at

BrownMath.com/donate.

It may be helpful to read the expression out in words: “the cosine of
Arctan *x*.” Doesn’t help much? Well, remember what
Arctan *x* is. It’s the (principal) angle whose tangent is *x*. So
what you have to find reads as “the cosine of the angle whose tangent
is *x*.” And that suggests your plan of attack:
**first identify that angle**, then find its cosine.

Let’s give a name to that “angle whose”. Call it *A*:

*A* = Arctan *x*

from which you know that

tan *A* = *x*

Now all you have to do is find cos *A*, and that’s easy
if you draw a little picture.

Start by drawing a right triangle, and mark one acute angle as *A*.

Using the definition of *A*, write down the lengths of two
sides of the triangle. Since tan *A* = *x*, and
the definition of tangent is opposite side over adjacent side, the
**simplest choice** is to label the opposite
side *x* and the adjacent side 1. Then, by definition, tan *A* =
*x*/1 = *x*, which we needed, because
*A* = Arctan *x*.

The next step is to find the third side. Here you know the two legs,
so you use the theorem of Pythagoras to find the hypotenuse,
√1+*x*². (For some problems, you’ll know one leg and the
hypotenuse, and you’ll use the theorem to find the other leg.)

Once you have all three sides’ lengths, you can write
down the value of any function of *A*. In this case you need
cos *A*, which is adjacent side over hypotenuse:

cos *A* = 1/√(1+*x*²)

But cos *A* = cos(Arctan *x*).
Therefore

cos(Arctan *x*) = 1/√(1+*x*²)

and there’s your answer.

Read this as “the cosine of the angle *A* whose sine is *x*”.
Draw your triangle, and label angle *A*. (Please take a minute and make
the drawing.) You know from equation 1 that

sin *A* = *x* = opposite/hypotenuse

and therefore you label the opposite side *x* and the
hypotenuse 1.

Next, solve for the third side, which is
√(1−*x*²), and write that
down. Now you need cos *A*, which is the adjacent side over
the hypotenuse, which is √(1−*x*²)/1.
Answer:

cos(Arcsin *x*) =
√(1−*x*²)

There you go: quick and painless.

This looks similar to Example 1, but as you’ll see there’s an additional wrinkle. (Thanks to Brian Scott, who raised the issue, albeit inadvertently, in his article “Re: Expression” on 10 Dec 2000 in alt.algebra.help.)

Begin in the regular way by drawing your triangle.
Since *A* =
Arctan(1/*x*), or tan *A* = 1/*x*, you
make 1 the length of the opposite side and *x* the length of the
adjacent side. The hypotenuse is then √(1+*x*²).

Now you can write down cos *A*, which is adjacent over
hypotenuse:

cos *A* = *x*/√(1+*x*²)

cos(Arctan 1/*x*) =
*x*/√(1+*x*²)

But this example has a problem that does not occur in the earlier examples.

Suppose *x* is negative, say −√3. Then
Arctan(−1/√3) = −π/6,
and cos(−π/6) = +(√3)/2.
But the answer above,
*x*/√(1+*x*²), yields
−√3/√(1+3) = −(√3)/2,
which has the wrong sign.

What went wrong? The trouble is that Arctan yields
values in (−π/2, +π/2), which is Quadrants IV and I.
But the cosine is always positive on that interval. Therefore
cos(Arctan *x*) always yields a positive
result. Remember also from equation 22 that
cos(−*A*) = cos *A*.
To ensure this, use the absolute value sign, and the true final answer
is

cos(Arctan(1/*x*)) =
| *x* |/√(1+*x*²)

Why doesn’t every example have this problem? The earlier
examples involved only the square of a variable, which is naturally
nonnegative. Only here, where we have an odd power, does it matter.
Yes, that applies to the first power, even though the exponent
^{1} isn’t written.

Summary: **When your answer contains an odd power** (1, 3,
5, etc.) of a variable, you must **add a third step** to the
process: carefully examine the signs and adjust your answer so that it
has the correct sign for both positive and negative values of the
variable.

**Advice tothe reader:**
The methods in this section are
for the really hard-core trig fan. They aren’t
really very useful in trigonometry itself, but are used in integral calculus
and some physics or engineering courses.
You may wish to skip them, especially on a first reading.

After the previous section, you may be wondering about the inside-out versions, taking the arcfunction of a function. Some of these expressions can be solved algebraically, on a restricted domain at least, but some cannot. (I am grateful to David Cantrell for help with analysis of these problems in general and Example 6 in particular.)

We can say at once that there will be
**no pure algebraic equivalent to an arcfunction of a trig function**.
This means there will be no nice
neat procedure as there was for functions of
arcfunctions

Why? The six trig functions are all
periodic, and therefore any function of any of them must also be
periodic. But no algebraic functions are periodic, except trivial ones
like f(*x*) = 2, and therefore no
function of a trig function can be represented by purely algebraic
operations. As we will see, some can be represented if we add
non-algebraic functions like mod and floor.

This is the angle whose cosine is sin *u*. To come
up with a simpler form,
set *x* equal to the desired expression, and solve the equation
by taking cosine of both sides:

*x* = Arccos(sin *u*)

cos *x* = sin *u*

This could be solved if we could somehow transform it to
sin(something) = sin *u*
or
cos *x* = cos(something else).
In fact, we can use equation 2 to do that.
It tells us that

sin *u* =
cos(π/2 − *u*)

and combining that with the above we have

cos *x* =
cos(π/2 − *u*)

Now if *x* is in Quadrant I, which is the
interval [0, π/2], then *u* will
be in Quadrant I also, and we can write

*x* = π/2−*u*

and therefore

Arccos(sin *u*) =
π/2−*u* for *u* in Quadrant I

But this solution does not work for all quadrants. For instance, try a number from Quadrant II:

Arccos(sin(5π/6)) = Arccos(½) = π/6

but

π/2 − 5π/6 = −π/3

Obviously π/2−*u* isn’t a general solution for
Arccos(sin *u*). Try graphing
Arccos(sin *x*) and π/2−*x* and you’ll see the problem:
one is a sawtooth and the other is a straight line.

Sparing you the gory details, π/2−*u* is right only in
Quadrants IV and I. We have to “decorate” it rather a lot to make it match
Arccos(sin *u*) in the other quadrants, and also to
account for the repetition of values every 2π.
The first modification is not too hard:
On the interval [−π/2, +3π/2],
the absolute-value expression
| π/2−*u* | matches the sawtooth graph of
Arccos(sin *u*).

The repetition every 2π is harder to reflect, but this manages it:

Arccos(sin *u*) =
| π/2 − *u* +
2π*floor[(*u*+π/2)/2π] |

where “floor” means the greatest integer less than or equal to. Messy, eh? (Note also that “floor” is not an algebraic function.)

It could be made a bit shorter with mod (which is also not algebraic):

Arccos(sin *u*) =
| π − mod(*u*+π/2, 2π) |

where mod(*a*, *b*) is the nonnegative
remainder when *a*is divided by *b*.

This one, the angle whose secant is cos *u*, has a
very odd solution. Try the solution method from
Example 4 and you get

*x* = Arcsec(cos *u*)

sec *x* = cos *u*

But sec *x* = 1/cos *x*, and
therefore

1/(cos *x*) = cos *u*

Now think about that equation. The cosine’s values are all between −1 and +1. So the only way one cosine can be the reciprocal of another is if they’re both equal to 1 or both equal to −1; no other solutions exist.

First case: If
cos *u* = 1, then *u* is an even multiple of
π, or in other words a
multiple of 2π. But Arcsec 1 = 0, and therefore

Arcsec(cos *u*) = 0 when
*u* = 2kπ

Second case: If
cos *u* = −1, then *u* is an odd
multiple of π. But Arcsec(−1) = π, and therefore

Arcsec(cos *u*) = π when
*u* = (2k+1)π

If *u* is not a multiple of π, cos *u* will be
less than 1 and greater than −1. The Arcsec function
is not defined for such values, and therefore

Arcsec(cos *u*) does not exist when
*u* is not a multiple of π

The graph of Arcsec(cos *u*) is rather curious:
single points at the ends of an infinite sawtooth: ..., (−3π, π),
(−2π, 0), (−π, π), (0, 0), (π, π), (2π, 0), (3π, π), ...

Proceeding in the regular way, we have

*x* = Arctan(sin *u*)

tan *x* = sin *u*

The most likely approach is the one from Example 4:
try to transform the above into
tan(*x*) = tan(something) or
*sin*(something else) = sin *u*.

If there is any trig identity or combination that can be used to do
that, it is unknown to me. I suspect strongly that
Arctan(sin *u*) can’t be converted to an algebraic
expression, even with the use of mod or floor, but I can’t prove it.

To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.

**Recommendation**: Work them on paper —
it’s harder to fool yourself about whether you really
understand a problem completely.

You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.

1
The possible output values of Arcsin *x* include
±π/2, but the possible output values of Arctan *x* do
not. Why can Arctan *x* never equal −π/2 or
π/2?

2
Find sec(Arcsin *x*). Remember to make a sketch to help you.
Pick a value, like *x* = −0.7, as a test case to
check your answer.

3
Find sin(Arccos 1/*x*). Remember to make a sketch to help you.
Pick a value, like *x* = 1.3, as a test case to
check your answer.

**30 Oct 2016**: Added practice problems. Updated the mathematical notation, particularly the use of italics and spaces, to conform to the standard. I used Jukka Korpela’s comprehensive Writing Mathematical Expressions (2014, Suomen E-painos Oy), ISBN 978-952-6613-25-3.**28 Oct 2016**:- Added What’s an Inverse?; previously I just dived in without properly laying the groundwork.
- Rewrote Principal Values.
- Added Notation. Previously this was mentioned, but just in a short paragraph, buried in another section.

- (intervening changes suppressed)
**15 Dec 2000**: New document.

next: 10/Complex Numbers

Because this textbook helps you,

please click to donate!Because this textbook helps you,

please donate at

BrownMath.com/donate.

please click to donate!Because this textbook helps you,

please donate at

BrownMath.com/donate.

Updates and new info: https://BrownMath.com/twt/