Trig without Tears Part 9:

# Inverse Functions

revised 17 Aug 2015

Copyright © 1997–2015 Stan Brown, BrownMath.com

Trig without Tears Part 9:

revised 17 Aug 2015

Copyright © 1997–2015 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).

Sometimes you have a sine or cosine or tangent and need to find the
associated angle. For instance, this happens whenever you
solve a triangle.
When you **have a sine function value and find the corresponding angle**,
we say you are finding the
**arcsine or inverse sine** of that value, and similarly for the other
functions.

Different books use different notation: *sin* with a superscript
−1, or *arcsin*. I prefer the “arc” forms because the superscript
−1 looks too much like an exponent.

You may be less intimidated by the arcfunctions if you pronounce
them as **the angle whose**. For instance, *arccos* 0.6 becomes
“the angle whose cosine is 0.6”.

What is *arcsin*(0.5)? You probably recognize that 0.5 is ½,
and it must be a sine of one of the special
angles. In fact, equation 15 tells you that sin(30°) = ½. So
you can say that *arcsin*(0.5) = 30° or π/6.

But wait, there’s more! You know from equation 22 that

*sin*(30°) = *sin*(150°) =
*sin*(390°) = *sin*(−210°)

and so on, and therefore they all equal ½. In fact,

*sin*(30° + 360°*k*) =
*sin*(150° + 360°*k*) = 0.5 for all integer
*k*

So which of this infinite number of values is *the* arcsine?

To make an arcsine *function*, we have to **restrict the range**
(the possible answers given by each function)
so that **each number has at most one arcsine**.
(Why do I say “at most one” and not “one and
only one”? The sine of any angle is between −1 and +1
inclusive; therefore only those numbers have arcsines. There is no
angle whose sine is 2, so 2 has no arcsines and *arcsin* 2 is
a meaningless noise.)

The arcsine is defined so that its range is the
interval [−π/2;+π/2], which is the
same as [−90°;+90°].
Some books use a capital letter for the function *Arcsin*
distinguish it from the multi-valued relation *arcsin*.
So we could say

*arcsin*(0.5) = π/6+2*k*π or 5π/6+2*k*π
for all integer *k*

but

*Arcsin*(0.5) = π/6

Why the particular range −π/2 to +π/2?
To start with, it seems tidy that any arcfunction of a positive
number should fall in Quadrant I, [0;+π/2]. So the only real
question is arcfunctions of negative numbers. If we prefer
the numerically smallest values for the arcsine function,
then *Arcsin*(−0.5) = −30° = −π/6
fits that rule, and a negative number’s arcsine (and arctangent,
too) will be in Quadrant IV, [−π/2;0].

What about the arccosine? 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, or [0;π].

You can remember the range of the
arctangent function this
way: the only two choices that make sense are 0 to π and
−π/2 to +π/2. But *tan*(π/2) is undefined, so if you
picked [0;π] for the range of *Arctan* it would have an ugly
hole in it at π/2. Therefore the range is defined to be
(−π/2;+π/2), an open interval but with no points missing
inside.

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) as the range.

Thomas defines the arcsecant and arccosecant 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] | [−π/2; +π/2] |

Arccos |
Arccos x = π/2 − Arcsin x |
[−1; +1] | [0; π] |

Arctan |
inverse of tangent function | all reals | (−π/2; +π/2) |

Arccot |
Arccot x = π/2 − Arctan x |
all reals | (0; π) |

Arcsec |
Arcsec x = Arccos(1/x) |
(−∞; −1], [1; ∞) | [0; π] |

Arccsc |
Arccsc x = Arcsin(1/x) |
(−∞; −1], [1; ∞) | [−π/2; +π/2] |

Remember that the *relations* 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.

**Advice to the reader:** The formulas 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.

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 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, you may need to add some absolute value signs. See Example 3.)

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 is how we needed to set up angle A.

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,
*sqrt*(1+x²). (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/*sqrt*(1+*x*²)

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

*cos*(*Arctan* *x*) = 1/*sqrt*(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
*sqrt*(1−*x*²), and write that
down. Now you need *cos* A, which is the adjacent side over
the hypotenuse, which is *sqrt*(1−*x*²)/1.
Answer:

*cos*(*Arcsin* *x*) =
*sqrt*(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 *sqrt*(1+x²).

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

*cos* A = *x* / *sqrt*(1+*x*²)

*cos*(*Arctan* 1/x) =
*x* / *sqrt*(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 in the previous paragraph,
*x*/*sqrt*(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* | / *sqrt*(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.

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 to the reader:** The formulas 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.

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, 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. 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.

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