Algebra II/Trig
Unit 11 Day 8 – Inverse Trig Functions (notes)
page 1
Introduction to Inverse Functions:
If you have a pup tent with sides of 16 feet each and the middle pole of 8 feet,
find the angle that the tent makes with the ground.
8
1
 1
sin  
sin  
  sin1  
  300
16
2
2
 
What other angles on the unit circle gives you this value when you take the sin?
16 ft
16 ft
8 ft.
θ
θ
1500, 3900, 5100, etc.
Why couldn’t this value be the angle the tent makes with the ground?
the tent makes a triangle so if each angle was 1500, then 1500 + 1500 > 1800 which makes no sense
When you are given the values of sin , cos , or tan  and you want to find the angle  which makes it true, you have to use
inverse trig. functions.
Remember that functions must pass the vertical line test. The graphs for each of the six trigonometric functions that we have
learned so far have passed that vertical line test. Now let us look at the graphs for the inverse trigonometric functions.
Inverse Trig Functions
 we use the inverse trig functions to find the angle when we know the value of the trig ratio
 the symbol for inverse is a -1 exponent, such as sin-1, cos-1 or tan-1
 sometimes the notation arcsin, arccos and arctan are used to indicate the inverse as well
From Algebra II, you should remember that to find the inverse of a function, you switch the x and y values. This also
changes the graph. Look below to see the graph of y = sin x and x = sin y (which is y = sin-1 x)
y = sin-1 x
2
y = sin x
rotate
clockwise/switch
x&y
1
/2

3/2
3/2

2
/2
cut at the two red lines
to make a function
-1
-1
-/2
1
-
-3/2
-2
You can see that the graph of y = sin-1 x does not pass the vertical line test, so, as it is, it is NOT a function. In order to
make the graph of y = sin-1 x a function, we must restrict (cut off) the range of the function to -900 < x < 900 or -/2 < y <
/2. This makes the graph look like this:
y = sin-1 x
/2
-1
-/2
1
Algebra II/Trig
Unit 11 Day 8 – Inverse Trig Functions (notes)
page 2
The inverse cosine graph and inverse tangent graph must also be restricted to make them functions. The restriction on
the inverse cosine function is 0 < y <  and the restriction on the inverse tangent is -/2 < y < /2 (these are not “or equal
to” inequalities because of the asymptotes). The graphs look like the following.
y = cos-1 x
y = tan-1 x

/2
/2
-1
-/2
-1
1
1
To summarize:
inverse trig function
written as
y = sin-1 x
inverse sine
or
restriction
-/2 < y < /2
(QIV and QI)
y = arcsin x
y = cos-1 x
inverse cosine
or
0<y<
(QI and QII)
y = arccos x
y = tan-1 x
inverse tangent
or
-/2 < y < /2
(QIV and QI)
y = arctan x

In a trig. function, the domain represents an angle and the range represents the side ratio.

In an inverse trig function, the domain represents the side ratio and the range represents the angle.

Since the inverse is not a function, there would be infinite answers for each question. Because of this, we have to restrict
the range, which is the value of the angle. We can do this by restricting the domains of the initial function so that the
inverse will be a function.


For sine and tangent, we use -900 to 900 (or  to ). We do not use 2700 to 3600 for Quadrant IV because we want a
2
2
consecutive number interval.


Capital letters are used to distinguish trigonometric functions with restricted domains from those with unrestricted
domains. If it says sin-1(.5), it would have infinite answers, but if it says Sin-1(.5), it would yield only one answer.

The values in the restricted domains are called principal values.
Algebra II/Trig
Unit 11 Day 8 – Inverse Trig Functions (notes)

in radians). Because the ratio is
3
negative, and we know from the restrictions that the angle must be in Quadrant IV or
Quadrant I, that means that the answer must be an angle in Quadrant IV and must be

between -/2 and /2. That makes the answer to this problem - .
3
ex.
Sin-1 -
3
2
The reference angle for this is 600 (
ex.
Cos-1 -
2
2
The reference angle for this is 450 (
ex.
Find sin (Tan-1 ¾) without a calculator.
5

adj = 4
c2 = a2 + b2
c2 = 32 + 42
c2 = 9 + 16
c2 = 25
c=5
opp = 3
page 3

in radians). Because the ratio is
4
negative, and we know from the restrictions that the angle must be in Quadrant I or Quadrant
II, that means that answer must be an angle in QII and must be between 0 and . That makes
3
the answer
.
4
For this problem, we need to use some logical thought. tan-1 ¾ means “the angle  whose
tangent is ¾” and the whole problem means “find the sine of the angle whose tangent is ¾.”
opp
From right triangle trig, we know that tangent is
, so if the tangent is ¾, then the
adj
opposite side is 3 and adjacent side is 4. This is shown on the diagram to the left. Using the
opp
Pythagorean Theorem, the hypotenuse will be 5. This means that the sin of  will be
or
hyp
3
.
5
Algebra II/Trig
Unit 11 Day 8 – Inverse Trig Functions (notes)

3
Example:
Find Cos-1  
 .
 2 

3
Let  = Cos-1  
 .
 2 
Then, Cos   
3
. . . Therefore,
2
page 4
For which angle in the restricted

3
area is the cos  
 =?
 2 

3
o
Cos-1  
 = 150
2


Evaluate the following on your calculator in degree mode.
1. cos-1 (0.7) = ____________
2. sin-1 (0.26) = ____________
3. tan-1 (3.5) = ____________
4. cos-1 (-0.7) = ___________
5. sin-1 (-0.26) = ___________
6. tan-1 (-3.5) = ___________
How does the 2nd row compare with the first row? Why?
Evaluate the following without a calculator.
1
)
2

3
8. Sin-1  

2


9. Tan-1 (0)
10. cot (Tan-1 2)
11. Arctan (-1)
12. Cot-1 1

2
13. cos  Sin 1 

2 

 3
14. Sin 1 

 2 

15. sin  Arc sin


3
16. Arccos  

 2 
 2
17. Arcsin 

 2 

3
18. Arccot  

 3 
7. Cos-1 (
3

2 
Find the exact value of the expression (hint: sketch a right triangle).
4
19. cos (sin-1 5 )
2
20. tan (cos-1 3 )
5

21. tan  Arc sin - 
7

5

22. sin  Tan 1 
12 

23. sin [Arctan (  2 )]
11 

24. csc  Arc sin

15 
