Algebra III
Lesson 32
Inverse Functions – Four Quadrant
Signs – Inverse Trigonometric
Functions
Inverse Functions
Every one-to-one functions has an inverse function.
A function takes “input” and gives an “output”.
An inverse function takes an “output” and gives what the
“input” must have been. It works the function backwards.
Steps to determine the inverse function from a starting function.
Starting function: f(x) = 2x – 1
1) Replace the function notation with y
f(x) = 2x – 1
Î
y = 2x – 1
2) Solve for x
2x = y + 1
x=½y+½
3) Swap the x and the y around
x=½y+½
Î
y=½x+½
So, for the function f(x) = 2x – 1,
the inverse function is f(x) = ½ x + ½
Example 32.1
Find the inverse function of y = 2x + 3.
1)
Done
2)
2x = y – 3
y 3
x= −
2 2
Graphically, it looks like this.
3)
x 3
y= −
2 2
original function
y=x
inverse function
Example 32.2
Find the inverse function of y = -2x + 5.
1)
Done
2)
-2x = y – 5
y 5
x=− +
2 2
Graphically, it looks like this.
3)
x 5
y=− +
2 2
original function
y=x
inverse function
Four Quadrant Signs
2nd Quadrant
1st Quadrant
sin Æ +
cos Æ tan Æ -
sin Æ +
cos Æ +
tan Æ +
sin Æ cos Æ tan Æ +
sin Æ cos Æ +
tan Æ -
3rd Quadrant
4th Quadrant
Inverse Trigonometric Functions
Trig functions take an angle and give a number.
Inverse trig functions take a number and give the angle
that goes with it.
Each trig function has its own inverse function.
sin
Î
arcsin or sin-1
cos
Î
arccos or cos-1
tan
Î
arctan or tan-1
Note: the calculator does not give all possible answers from the
inverse trig functions, only the first answer.
Recall the unit circle: for what angle does sin equal 1?
90°, is that all?
How about 450°, 810°, or 1170°, etc.
Each successive full revolution.
The unit circle can be used to solve the common
angles for sin & cos. Tan takes a little work to
figure.
Limits on arcsin, arccos, & arctan
Try to find two touching
quadrants that cover all possible
values (+ & -) for the trig
function, always including the
first quadrant.
The limits on arcsin are
from -90° to +90°.
The limits on arccos are
from 0° to 180°.
The limits on arctan are
from -90° to +90°.
2nd Quadrant
1st Quadrant
sin Æ +
cos Æ tan Æ -
sin Æ +
cos Æ +
tan Æ +
sin Æ cos Æ tan Æ +
sin Æ cos Æ +
tan Æ -
3rd Quadrant
4th Quadrant
Example 32.3
So, Arcsin
45°
2
=
2

3 1
−

 2 , 2


(-1,0)
θ
5°
4
=
θ=1
50°
θ=3

3 1

−
 2 ,− 2 


0°
θ=3
30°
°
θ=
3
(0,-1)
15
°
00°
 1
3

 − ,−

 2
2


θ=3
5°
2
2
θ=
θ=270°

2
2
−

−
,
 2

2


 3 1


 2 , 2


θ=0°
θ=360°
θ=180°
10
θ=2
 2 2


 2 , 2 


0°
θ=6
20°
Sin is x or y?
θ=
13
5°
1 3
 ,

2 2 


(0,1)
θ=1
Recall the unit circle.

2 2

−
 2 , 2 


 1 3

− ,
 2 2 


θ=90°
2
2
θ=
24
0°
Evaluate: Arcsin
1
3
 ,−

2

2


(1,0)
 3 1


 2 ,− 2 


 2
2


−
,
 2

2


But tan = sin/cos or y/x
(-1,0)
θ =1
 3 1
−

 2 ,− 2 



2
2
−

,
−
 2

2


5°
2
2
θ=
 1
3
 − ,− 
 2 2 


1
1
1
3
y
3
1
2
=
−
=
−
•
=
=−
=−
3
3
3
3
x
3
3
3
2
−
0°
θ=6
θ=90°
θ=3
30°
10°
θ=2
θ=
3
(0,-1)
15
°
00°
-30°
θ=0°
θ=360°
θ=3
Try one of the angles.
0°
θ=3
50°
θ=270°
The 3 only comes from
-30° or -60°
θ
 2 2


 2 , 2 


5°
4
=
θ=180°
Reduce options by
elimination.
Tan is negative from
0° to -90°
20°
Tan isn’t on the unit circle.
 3 1
−

 2 , 2


θ=
13
5°
1 3
 , 
2 2 


(0,1)
θ=1
 1

3
Evaluate: Arctan  −
 3

Recall the unit circle.

2 2
−

 2 , 2 


θ=
24
0°
Example 32.4
 1 3
− , 
 2 2 


1
3
 ,− 
2 2 


 3 1
 , 
 2 2


(1,0)
 3 1
 ,− 
 2 2


 2
2


,
−
 2

2


So, -30° is the answer.
Example 32.5

 3 
cos  arctan  −  
 5 

Evaluate:
Since 5 isn’t on the unit circle anywhere, we’ll use a
different trick. We’ll make a triangle.
Remember that tan = y/x, and it is – in the 4th quadrant.
So, -3/5 means go 5 to the right and 3 down.
Since cos = a/h, we need the hypotenuse of this
triangle.
h2 = 32 + 52
= 9 + 25
= 34
5
θ
h
3
h = 34
adj
cosθ =
hyp
5
=
34
5 34
=
34
Example 32.6

 2 
sin  arccos   
 3 

Evaluate:
Since cos = 2/3 isn’t one of the unit circle angles,
we’ll make a triangle.
Remember that cos = a/h, and it is + in the 1st quadrant.
So, 2/3 means go 2 to the right and with a hypotenuse of 3.
3
opp
Since sin = o/h, we need the opposite of this
triangle.
32 = 22 + o2
o2 = 9 - 4 = 5
θ
2
o= 5
opp
sin θ =
hyp
5
=
3
Practice
a) Find the inverse function of y = 3x + 4.
1)
2)
3)
Done
3x = y – 4
x=
y 4
−
3 3
y=
x 4
−
3 3
b) Evaluate. Do not use a calculator. We’ll make a triangle.

 1 
2) sin arccos − 
 2 

cos is -1/2 at 120°
sin 120° =
3
2
h2θ==15402 + 52
°
0°
θ=6
 2 2


 2 , 2 


θ
= 16 + 25
0°
= 41
θ=3
 3 1
 , 
 2 2


h = 41
(-1,0)
θ=180°
cosθ =
 3 1
−

 2 ,− 2 


5
41
=
5 θ=0°
41 (1,0)
= θ=360°
41
θ=3
30°
10°
θ=2
5°
2
2
θ=
θ=
3
h
θ
15
°
5
 1
3
 − ,− 
 2 2 


(0,-1)
 3 1
 ,− 
 2 2


4
00°

2
2
−

,
−
 2

2


adj
hyp
θ=3
Not on unit circle.
5°
θ=270°

 4 
cos
arctan
 
3)

 5 

 3 1
−

 2 , 2


Since cos = a/h, we need the
θ=
5°
4
hypotenuse
of
this
triangle.
13
=
20°
3
at 60°.
2
 , 


go (0,1)
5 to the  2 2 
θ=1
Sin is
1 3
− , 
 2 2 
So, 4/5 means



2 2
−

 2 , 2 right and 4 up.


θ=90°
Recall unit circle.
Remember that tan = y/x, and it is
+ in
the 1st quadrant.
1 3

θ=
24
0°
 3


arcsin
1)
 2 


1
3
 ,− 
2 2 


 2
2


,
−
 2

2


c) The graph of the function f(x) = x2 is shown on the left below. The
graph on the right is the same graph translated two units to the left.
Write the equation of the graph on the right.
Since left is an x direction we’ll substitute there. Left is which sign?
f(x) = (x + 2)2
1
d) Sketch the graph of the function f ( x) = 2 +  
 4
Recognize a translation of a simpler graph.
What is the translation?
2 up.
What is the simpler graph?
1
f ( x) =  
 4
Sketch the simpler graph.
(0,1)
(-1,4)
Now move it.
x
x