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Math 2/Unit 7/Lesson 1/ TOOLKIT Trigonometric Functions (Investigation 1)
In this investigation, you explored the sine, cosine, and tangent, three members of a new family of functions called
trigonometric functions.
Standard Position of an
angle
Positive angles move
counterclockwise from the
positive x-axis
Negative angles move
clockwise from the positive
x-axis
3 basic trigonometric
functions
tangent of
sine of
cosine of
=tan
= sin
= cos
= = slope =
=
=
Where r =
Example: Find the tangent,
sine and cosine for theta
given the point (12, 5)
r=
tan
=
sin
=
cos
=
Remember that sine and
cosine are never greater than Remember tangent, sine, and cosine are just a way to compare
1 because r > y and r > x
2 sides of a right triangle
Remember these values
2
 .707
2
2
cos 45 
 .707
2
.707
tan 45 
1
.707
sin 45 
Example: Find the tangent,
sine and cosine given a line
y =2x , x > 0 and the fact
that tangent is slope
Example: Find the tangent,
sine and cosine given a line
y =9x/7, x > 0and the fact
that tangent is slope
Example: Find the sine and
cosine given that the
4
tan  
5
Example: Find the tangent
and cosine given that the
5
sin  
9
x  r2  y2
y  r 2  x2
r  x2  y2
As your angle gets bigger from zero to ninety degrees:
The value of sine gets bigger because y gets bigger
The value of cosine gets smaller because x gets smaller
The value of tangent gets bigger because y gets bigger as x gets smaller(tan
tan
=
so r =
tan
=
so r =
4

5
y
sin   
r
tan  
and then sin
and then sin
x
so r  4 2  5 2  41
y
4
x
5
and cos   
r
41
41
5 y
 so x  9 2  5 2  56
9 r
x
56
y
5
and cos   
tan   
r
9
x
56
sin  
=
and cos
=
=
and cos
=
= )
Math 2/Unit 7/Lesson 1/Summarize the Math: Trigonometric Functions (Investigation 2)
The trigonometric functions sine, cosine, and tangent are useful in calculating lengths in situations modeled with
right triangles. Refer to the right triangle below in summarizing your thinking about how to use trigonometric
functions in the situations described.
Record the right triangle
definitions of the sine, cosine,
and tangent functions.
When you focus on angle A:
Side a is opposite
Side b is adjacent
Side c is hypotenuse
a oppoosite

b adjacent
oppoosite
a
sine of A  sinA  
c hypotenuse
adjacent
b
cosine of A  cosA  
c hypotenuse
tangent of A  tanA 
When you focus on angle B:
Side b is opposite
Side a is adjacent
Side c is hypotenuse
The opposite side is always
the same as letter of the
angle
Angle A is opposite a
Angle B is opposite b
Just remember
SOHCAHTOA
b oppoosite

a adjacent
b oppoosite
sine of B  sinB  
c hypotenuse
adjacent
a
cosine of B  cosB  
c hypotenuse
tangent of B  tanB 
Sine
Opposite
Hypotenuse
Cosine
Adjacent
Hypotenuse
Tangent
Opposite
Adjacent
Find angle B, side AB and
side BC given that A  21
and side AC = 22
B  90   21  69 
Side c is the hypotenuse and side a is the opposite since we were given A.
To find side a use tangent because you are given the adjacent side AC and you are
looking for the opposite side.
oppoosite
adjacent
a
tan 21 
22

22tan 21  a M ultiplyboth sides by 22
8.4  a
tanA 
To find side c you can use cosine because you were given the adjacent side AC and
you are looking for the hypotenuse
adjacent
hypotenuse
22
cos 21 
c

ccos 21  22 M ultiplyboth sides by c
22
c
cos 21
c  23.6
cosA 
To find c you could also do Pythagorean theorem
c  22 2  8.4 2  23.6
A  90   45  45
Find angle A, side AC and
side BC given that B  45 
and side AB = 17
Side a is the adjacent and side b is the opposite since we were given B.
To find side a use cosine because you are given the hypotenuse AB and you are
looking for the adjacent side BC.
adjacent
hypotenuse t
a
cos 45  
17

17cos45  a M ultiplyboth sides by 17
12.02  a
cosB 
To find side b you can use sine because you were given the hypotenuse AB and you
are looking for the opposite side of B
opposite
hypotenuse
b
sin 45  
17

17sin45  b M ultiplyboth sides by 17
b  12.02
sin B 
To find b you could also do Pythagorean theorem or even the tangent
b  17 2  12.02 2  12.02
opposite
adjacent
b
tan 45  
12.02
12.02 tan 45   b M ultiplyboth sides by 12.02
b  12.02
tan B 
If you would rather you can switch angle A and B. This will make the opposite and
adjacent always stay in the same place, no matter what angle you are given
opposite
adjacent
a
tan 55  
20

20 tan 55  a M ultiplyboth sides by 20
a  28.562 meters
tan A 
If 55  is the angle of
elevation, Find how tall the
bat is?
If you wanted to, you could find all the other information in that triangle.
B  90   55  35
There are three ways you could find c
c  28.562 2  20 2  34.86 meters
OR
opposite
hypotenuse
28.562
sin 55  
c
28.562
 c M ultiplyboth sides by c and divide by sin 55 

sin 55
c  34.8657 meters
sin A 
adjacent
hypotenuse t
20
cos 55  
c
20
 c M ultiplyboth sides by c and divide by sin 55 
sin 55 
c  34.8657 meters
cosA 
Summarizing the Algebra
If the variable is on the bottom of the fraction, the answer is a division problem.
cos 55  
20
c
20
c
sin 55 
M ultiplyboth sides by c and divide by sin 55 
If the variable is on the bottom of the fraction, the answer is a multiplication problem
a
20

20 tan 55  a
tan 55  
M ultiplyboth sides by 20
One more example:
Just find c
opposite
hypotenuse
7
sin 40  
c
7
 c M ultiplyboth sides by c and divide by sin 40 

sin 40
c  10.9
sin A 
Math 2/Unit 7/Lesson 1 TOOLKIT: Trigonometric Functions (Investigation 3)
The sine, cosine, and tangent functions are useful in finding the measures of acute angles of a right triangle using
the lengths of two sides. Refer to the right triangle below to summarize your thinking about such situations.
To find an angle use the
inverse trigonometric
functions on your calculator
Find angle A given the hypotenuse AB = 28 and the adjacent side a = 11.5
cos 1  / tan 1  / sin 1 
You find these on your
calculator by doing
2nd sin 
2nd cos 
2nd tan 
Since you know the adjacent and hypotenuse use the cosine function
11.5
28
11.5
A  cos 1
28
cos A 
Find angle A given the hypotenuse AB = 28 and the opposite side BC = 11.5
Since you know the opposite and the hypotenuse use the sine function
11.5
28
11.5
A  sin 1
28
sin A 
Find angle A given the adjacent side AC = 28 and the opposite side BC = 11.5
Since you know the opposite and the adjacent use the tangent function
11.5
28
11.5
A  tan 1
28
tan A 