The Inverse Trigonometric
Functions
Let's again review a few things about inverse functions.
• To have an inverse function, a function must be one-to-one (remember if a horizontal line intersections the graph of a function in more than one place it is NOT one-to-one).
•If we have points on a function graph and we trade x and y places we'll have points on the inverse function graph.
1f f x •Functions and their inverses "undo" each other so
•Since x and y trade places, the domain of the function is the range of the inverse and the range of the function is the domain of the inverse
•The graph of a function and its inverse are reflections about the line y = x (a 45° line).
Is y = sin x a one-to-one function?
No! A horizontal line intersects its graph many times.
If we want to find an inverse sine function, we can't have the sine repeat itself so we are only going to look at part of the sine graph.
If we only look at part of the sine graph where the x values go from -/2 to /2 and the y values go from -1 to 1, we could find an inverse function.
x y = sin x
6
0
1
2
1
2
0
2
1
1
x y = sin-1 x
6
2
Remember we are only looking at x values from -/2 to /2 so that we can have a one-to-one function that will have an inverse.
1
2
1
0
2
1
1
6
0
2
6
2
So for y = sin-1 x, we can put in numbers between -1 and 1. What we get out is the angle between -/2 to /2 that has that sine value.
We are going to define the inverse function of sine then to be y = sin-1x. Remember for inverse functions x and y trade places so let's take the values we got for sin x and we'll trade them places for y = sin-1x with our restricted domain.
Let's graph both of these.
x y = sin x
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0
1
2
1
2
0
2
1
1
x y = sin-1 x
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2
1
2
1
0
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1
1
6
0
2
6
2
Here is the graph of sin x between -/2 to /2
2
2
1 1
Here is the graph of sin-1 x from -1 to 1
Notice they are reflections about a 45° line
Since sin x and sin-1 x are inverse functions they "undo" each other.
11 wheresinsin
22 wheresinsin
11
11
xxxxff
xxxxff
6sinsin 1
6
CAUTION!!! You must be careful that the angle is between -/2 to /2 to cancel these.
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Looking at the unit circle, if we choose from -/2 to /2 it would be the right half of the circle.
Notice the y values (sine values) never repeat themselves in this half the circle so the sine function would be 1-to-1 here.
When you see inverse sine then, it means they'll give you the sine value and it's asking which angle on the right half of the unit circle has this sine value.
2
1sin 1
This is asking, where on the right half of the unit circle is the sine value -1/2?
6
It looks like the answer should be 11/6 but remember the range or what you get out must be an angle between -/2 to /2, so we'd use a coterminal angle in this range which is:
Remember when you have sin-1 x it means "What angle between -/2 to /2 has a sine value of x?"
2
2sin 1
This is asking, "What angle from -/2 to /2 has a sine value of square root 2 over 2?"
Let's look at the unit circle to answer this.
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1,0
0,1
1,0
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4
1 3sin
2
This is asking, "What angle from -/2 to /2 has a sine value of negative square root 3 over 2?"
This is 5/3 but need
3
4
3sinsin 1
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1,0
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4
This looks like sine and its inverse should cancel out but then you'd be getting an answer out that was not in the range so you must be careful on these. Let's work the stuff in the parenthesis first and see what happens.
2
2sin 1
Now remember for inverse sign we only use values on the right half of the unit circle. So this is asking, "Where on the right half of the circle is the sine value square root 2 over 2?"
2
1
3cos
3cos
2
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1,0
0,1
1,0
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Let's think about an inverse cosine function. If we chose the right half of the unit circle for cosine values would we have a one-to-one function?
NO! For example:
Let's look at the graph of y = cos x and see where it IS one-to-one
From 0 to
So cos-1 x is the inverse function of cos x but the domain is between -1 to 1 and the range is from 0 to .
11- wherecoscos
0 wherecoscos11
11
xxxxff
xxxxff
So this is asking where on the upper half of the unit circle does the cosine value equal 1/2.
2
1cos 1
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33
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1 3cos
2
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5
tan-1 x is the inverse function of tan x but again we must have restrictions to have tan x a one-to-one function.
1 1
1 1
tan tan where 2 2
tan tan where
f f x x x x
f f x x x x
We'll take tan x from -/2 to /2
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For 1x 1
ysin 1 x provided
sin y x and / 2 y / 2.
Definition: The Inverse Sine Function
Another notation for inverse sine is
arcsin( ) which is read "arc sine of "x x
For 1x 1
ycos 1 x provided
cos y x and 0 y .
Definition: The Inverse Cosine Function
Another notation for inverse cosine is
arccos( ) which is read "arc cosine of "x x
For any real number x,
ytan 1 x provided
tan y x and / 2 y / 2.
Another notation for inverse tangent is
arctan( ) which is read "arc tangent of "x x
Definition: The Inverse Tangent Function
We've talked about inverse functions for sine, cosine and tangent and saw that we must restrict the range so we have a one-to-one function. Here is a summary:
1sin x
Restricted to angles on right half of unit circle listed from -/2 to /2
1cos x
Restricted to angles on upper half of unit circle listed from 0 to
1tan x
Restricted to angles on right half of unit circle listed from -/2 to /2
What about the inverses of the reciprocal trig functions?
1sin x
Restricted to angles on right half of unit circle listed from -/2 to /2
1cos x
Restricted to angles on upper half of unit circle listed from 0 to
1tan x
Restricted to angles on right half of unit circle listed from -/2 to /2
1csc x 1sec x 1cot x
Cosecant and secant have same restrictions as their reciprocal functions but cotangent does not. Cosecant excludes 0 since it would be undefined there and secant excludes /2.
Restricted to angles on upper half of unit circle listed from 0 to
We've seen in this section how to work some of these if the values are exact values on the unit circle and if not how to use the calculator. Now we need to find exact values even if not on the unit circle.
3
1costan 1
The secret is to draw a triangle in the correct quadrant and label the sides you know, figure out the other side by Pythagorean Theorem, and then find the trig function you need.
Inverse cosine is quadrants I and II and since the cosine value is negative, the terminal side must be in quadrant II. drop down line to x axis
This means the cosine of some angle (with reference angle ) is -1/3 so label adjacent and hypotenuse.
-1
322
a
o22
1
22
If you need the exact value of a reciprocal function you can find the "flip" over the value you have and find the inverse of the reciprocal function (being careful that you meet the restrictions).
2sec 1
This is asking what angle (on the upper half of the unit circle) has a cosine value of -1/2 or in other words, cos = -1/2 so what is ?
2
1cos 1
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If you use your calculator to find one of the reciprocal functions you use the same strategy of "flipping" over the given value and then you can use the inverse trig function button for the reciprocal.
5csc 1
5
1sin 1
Now use your calculator and compute this.
20.0
We've talked about inverse functions for sine, cosine and tangent and saw that we must restrict the range so we have a one-to-one function. Here is a summary:
1sin
Restricted to angles on right half of unit circle listed from -/2 to /2
1cos
Restricted to angles on upper half of unit circle listed from 0 to
1tan
Restricted to angles on right half of unit circle listed from -/2 to /2
What about the inverses of the reciprocal trig functions?
1sin
Restricted to angles on right half of unit circle listed from -/2 to /2
1cos
Restricted to angles on upper half of unit circle listed from 0 to
1tan
Restricted to angles on right half of unit circle listed from -/2 to /2
1csc 1sec 1cot
Cosecant and secant have same restrictions as their reciprocal functions but cotangent does not.
Restricted to angles on upper half of unit circle listed from 0 to
For help on using your calculator to compute inverse trig functions, click here.
Also in the video clips I show you how to use the calculator.
We've seen in the last section how to work some of these if the values are exact values on the unit circle and if not how to use the calculator. Now we need to find exact values even if not on the unit circle.
3
1costan 1
The secret is to draw a triangle in the correct quadrant and label the sides you know, figure out the other side by Pythagorean Theorem, and then find the trig function you need.
Inverse cosine is quadrants I and II and since the cosine value is negative, the terminal side must be in quadrant II. drop down line to x axis
This means the cosine of some angle x (with reference angle x’ ) is -1/3 so label adjacent and hypotenuse.
x’
-1
322
x a
o22
1
22
x
If you need the exact value of a reciprocal function you can find the "flip" over the value you have and find the inverse of the reciprocal function (being careful that you meet the restrictions).
2sec 1
This is asking what angle (on the upper half of the unit circle) has a cosine value of -1/2 or in other words, cos = -1/2 so what is x?
2
1cos 1
2
1,
2
3
1,0
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1,0
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33
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If you use your calculator to find one of the reciprocal functions you use the same strategy of "flipping" over the given value and then you can use the inverse trig function button for the reciprocal.
5csc 1
5
1sin 1
Now use your calculator and compute this.
20.0
csc 1 x sin-1 1
x
for x 1
sec 1 x cos-1 1
x
for x 1
cot 1 x 2
tan 1 x
Definition: The Inverse Cosecant, Secant and Cotangent Functions
cot 1 x
tan-1 1
x
for x 0
tan-1 1
x
for x 0
2 for x 0
More on Cotangent
Function Gallery:Inverse Trigonometric Functions