Objectives:
• Convert between degree and radian measure
• Pythagorean identities
• Find the trigonometric values of any angle measure
• Unit circle
• Graph trigonometric functions
• Solve for lengths and angle measures of any type of triangle.
• Verify identities
• Application of trigonometry
Uses of trigonometry:
Used to describe the motion of any object that behaves in a circular,
oscillating or periodic manner.
Angle:
Consists of two rays or half lines that originate at a common point
called the vertex.
These two rays have names:
• Terminal side: The ray that rotates to generate the angle.
• Initial side: The ray that does not move.
Angles are commonly denoted using lower case Greek letters.
,  ,  , 
alpha, beta, theta, gamma (respectively)
To better describe the formation of angles we superimpose an xy plane on the
angle with the vertex at the origin.
y
Angles are generated by the terminal
side rotating counterclockwise or
clockwise.
B
A
0

x
If angles are generated by the
terminal side moving
counterclockwise the angle is
positive.
If angles are generated by the
terminal side moving clockwise
the angle is negative.
The direction of the arrow inside the angle will tell you if the terminal
side is generating a positive or negative angle.
y
B
If angles are generated by the
terminal side moving clockwise
the angle is negative.
A
0

x
If we have an angle of 390°, this is one revolution of the unit circle plus 30°.
So, we say that 390° is coterminal with 30°
Coterminal angles differ by multiples of 360°
Ex 1: Determine if the following angles are coterminal.
a.) 90° and 1170 °
YES
b.) 123 ° and 844 °
NO
Solution:
1170  90
3
360
If you get a whole number they are coterminal.
If the result is not a whole number then they
are not coterminal.
844  123
 2.002777778
360
To define the measure of an angle, we first add the unit circle centered at the
origin to the coordinate system.
y
This is called a unit circle
because the length of the radius
is 1.
1
-1
1
-1
x
The equation of the unit circle
is:
2
2
x  y 1
The domain is: [-1, 1]
To generate angles we must consider the terminal and initial sides.
The initial side is aligned with the x-axis and the terminal side starts at the x-axis
and rotates to generate the angle.
y
This point is P(t). P is the
function and t is the input value.

x
t
For every P(t) on the unit circle we can
define its measure by using degrees
or radians.
Radian Measure:
For any real number t, the angle generated by rotating
counterclockwise from the positive x-axis to the point
P(t) on the unit circle is said to have radian measure t.
t

3
2

2
y
(0,1)
…and so on…

(-1,0)
(0,-1)
1

(1,0)
-1
-2
-3
x

We can see that an angle that measures 90° is the same measure as
2
We can also see that 180° is 
We will use this fact to convert between degrees and radian measure.
To convert degree measure to radian measure you multiply the degree measure
by: 
180
Ex: Convert the following degree measure to radian measure.
a.) 150°
b.) 225°
   150 5
a.) 150


6
 180  180
c.) -72°
   225 45 5
b.) 225



36
3
 180  180
72
2
  
c.)  72





180
180
5


To convert radian measure to degree measure you multiply the radian measure
by: 180

Ex 2: Convert the following radian measures to degree measure.
7
a.)
6
12
b.)
5
a.)
7  180  1260
 210


6   
6
b.)
12  180  2160
 432


5   
5
11
c.) 
6
11  180 
1980
c.) 
 330


6   
6
3.2 The Sine and Cosine Functions
Recall: The terminal side rotates to generate the angles.
There are infinitely many points on the unit circle that the terminal
side “could” generate.
We will only “memorize” a few of them.
We will memorize all angles on the unit circle that are in increments of 30°
and 45°.
y
Unit Circle
90°
120°
60°
45°
135°
150°
30°
180°
0°
330°
210°
315°
225°
240°
270°
300°
x
An ordered pair has the form: (x, y)
The Sine and Cosine Functions:
Suppose that the coordinates of the point P(t) on the unit circle are
(x(t), y(t)). Then the sine of t, written sin t, and the cosine of t, written cos t,
are defined by
sint t = y(t)
and
cos t = x(t)
Our new ordered pairs are of the form: P(t) = (cos t, sin t)
Finding the cosine and sine values
y of the common angles on the unit circle.
120°
90°
60°
45°
135°
150°
30°
180°
0°
x
330°
210°
315°
225°
240°
270°
300°
We create right
triangles by drawing
lines perpendicular
to the x or y axis.
It does not matter
which axis.
To find the cosine and sine of 30° we must use the triangle we created.
1
60°
30°
1st: We know the third angle is 60° by the triangle sum theorem.
2nd: We know the length of the hypotenuse since this is a unit circle: r = 1
3rd: Using the properties of a 30-60-90 right triangle we can find the other two
sides.
Since the base of this triangle is on the x-axis this side would represent the
cos t. The height would represent the sin t.
Recall: * The side opposite the 30° angle is half of the hypotenuse.
* The side opposite the 60° angle is the product of the short leg
and the square root of 3.
If the hypotenuse has length 1, then the side opposite 30° is ½ .
The side opposite the 60° angle is
So, P(t) =
   3 1 
P    , 
 6   2 2
1
3
 3
2
2
This is how you will solve for all sine and cosine on
the unit circle dealing with 30 – 60 – 90 right triangles.
Finding the coordinates of a point using a 45 – 45 – 90 right triangle.
We know:
45°
• The third angle is 45°
• The hypotenuse has length 1.
1
• In a 45 – 45 – 90 right triangle the legs are
the same length. We can call both legs
the same variable since they are equal.
45°
a2  a2  1
Now solve for a.
2a 2  1
1
a 
2
1
a
2
1
a

2
1
a

2
2
1
2
2
2
2


2
4 2
We rationalized the denominator in this step
because we do not leave radicals in the
denominator.
Since the legs have the same length, the cosine and sine values are the same.
   2 2 
Pt   P   
, 
 4  2 2 
Pythagorean Identity:
For all real numbers t,
(sin t)2 + (cos t)2 = 1
Because of the Pythagorean Identity, sine and cosine have bounds:
For all real numbers t,
 1  sin t  1
The cosine function is even
For all real numbers t,
cos(-t) = cos t
and
1  cos t  1
The sine function is odd
For all real numbers t,
sin(-t) = - sin t
A Periodic Function is a function that repeats the same thing over and over again.
Trigonometric Functions are periodic because they repeat.
Sine and Cosine functions have a period of
2
From your unit circle you can see that the ordered pairs on your unit circle
do not begin to repeat until after one complete revolution of the unit circle
360°
sin(t  2 )  sin t
cos(t  2 )  cos t
and
Reference Number
For any real number t, the reference number r associated with t is the
shortest distance along the unit circle from t to the x-axis. The reference
number r is always in the interval   
Ex 2: What is the reference number of
0, 2 
11
6
Since the answer is always in
the above interval, it is 
6
Ex 3: Determine the values of t in [0,2 ] that satisfy
2(sin t ) 2  3 cos t  0
This is simply a quadratic function. Use what you know.
You can view this quadratic as…
2 y 2  3x  0
We cannot work with it this way; we can only have one unknown. We need
to turn both variables into y (sine) or x (cosine).
Pythagorean Identity: (sin t)2 + (cos t)2 = 1
Manipulate this identity so that it is something that you can use.
(sin t ) 2  (cos t ) 2  1
(sin t ) 2  1  (cos t ) 2
We can replace (sin t)2 with (1 – (cos t)2)
2(1  (cos t ) 2 )  3 cos t  0
Now distribute
2  2(cos t ) 2  3 cos t  0
Now factor and set each factor equal
to zero.
 2(cos t ) 2  3 cos t  2  0
To make things easy, and we don’t like our leading coefficient being negative,
multiply both sides by -1.
2(cos t ) 2  3 cos t  2  0
Now factor
2 x 2  3x  2  0
It may help you to view it like this to factor.
(2x + 1)(x – 2) = 0
Set each factor equal to zero and solve.
2x + 1 = 0
x
1
2
and
and
x–2=0
x2
1
cos t  
and
cos t  2
2
1
2
4
cos t   at : t 
and t 
2
3
3
Replace x with cos t since x
represents cos t.
Where on the unit circle is this true?
cos t = 2 will never happen; it is
outside the bounds of cosine.
Lesson 3.3: Graphs of the Sine and Cosine Functions
Sine Curve
We will use the values that we memorized from the unit circle to graph one
period of the sine function.
Sin t
1








4
4
-1

2
3
4


5
4
3
2

7
4
2
The more points you plot the more precise your graph will be. I will plot
only the 45° increments of the angles.
t
Cosine Curve
cos t
1 






4
4
-1

2


3
4


5
4
3
2
t
7
4
2
Ex 1: Use the graphs of y = sin x and y = cos x to sketch the graphs of
 
y  sin  x  
2



y  sin  x  
2

and
Trigonometric functions are no different than any other function when
“shifts” are involved… minus to the right …plus to the left.
Sin t

1
graph of y = sin x







4
4
-1

2
3
4


5
4
3
2

7
4
2
Again, we must be sure the leading coefficient is 1 before we try and see
the shifts.
t
Note: Your shift will help guide you as to what you should count by on the x-axis.
sin x
1




-1 

2

y  sin x  
2




3
2


2


x
5
2
Q: What is the last value going to be that we will write on the x-axis?
5
A:
2
NOW YOU TRY THE SECOND ONE!!! 

y  sin x  
2

sin x

1 







2
2
-1
You try…
x





y  cos x   and
2


3
2
2


y  cos x  
2

5
2
We can see from the previous 4 graphs that when we shift one graph to the
right or left that we obtain the graph of a different equation, such as:
The graph of:


sin x    cos x
2



cos x    sin x
2



sin x     cos x
2



cos x     sin x
2

These are all true only
because we are shifting by

2
When you shift by other
increments different
equalities occur.
When shifting by increments of pi…
y  sin( x   )  sin( x   )
y  cos( x   )  cos( x   )
Ex 1: Sketch the graph of


f ( x)  2 cos x  
2

All trig functions are of the form:
y = A sin(Bx + C)
The name of the trig function changes only.
|A| is the amplitude
Period is
Shift is
2
B
C
B
Don’t forget!!! The leading coefficient
must be 1 before you can see the shift!!!

f ( x)  2 cos x  
2

Amplitude is: 2; so the height will go to +2 and -2 on y-axis.
B = 1, so the period has not changed.
Shift: right

2
units
First, I will graph y = 2cosx, then shift this graph to the right

2
units.
Since the period did not change and the shift is  units, we will count by
 on the x-axis.
2
4
cos x
y = 2cos x
Notice that the zeros of this function do
2 

not change. Amplitude is just a vertical


elongation or compression of the graph.
x


-2
2





3
2
2
Now, we shift this graph!
cos x
2



-2

2
Ex 2: Sketch the graph of


f ( x)  2 cos x  
2

 




3
2



2
x
5
2


f ( x)  2 cos 3x  
2

B = 3, not 1 so we must find the length of the new period.
Recall: period is found by:
2
B
2
3
Always reduce if necessary.
Instead of this graph being graphed from 0 to 2π, the entire cosine curve
will be graphed between 0 and 2
3
Recall: To see a horizontal shift the leading coefficient must be 1 and it is not.
We must factor out the 3 from the quantity.


f ( x)  2 cos 3 x  
2



f ( x)  2 cos 3 x  
6

We will shift this graph to the right

units.
6
The main points of the cosine
curve (y = cos x) are:
cos x

2

-2


(0,1)
x



 2 5
6
3
2


3
6
 

7 4 3 5 11
2
6 3 2 3 6
 
  2 
,2 
3 
New points: (0,2),  6 ,0 ,  3 ,2 ,  2 ,0 , 
with
shift
 
 ,0 
2 
( ,1)
 3 
 ,0 
 2 
( 2 ,1)
Divide all of these
orginal x-coordinates
by 3 and multiply the
y-coordinates by 2,
these will be the
locations of your new
points.
    
  2   5 
 ,2 ,  ,0 ,  ,2 ,  ,0 ,  ,2 
6  3  2
  3   6 
Ex 3: Sketch the graph of
y  3 cos(2 x   )
cos x


2

x

 2  4
-2 


4
2
3
4

3.4 Other Trigonometric Functions
The Tangent, Cotangent, Secant, and Cosecant Functions:
The tangent, cotangent, secant and cosecant functions, written respectively as
tan x, cot x, sec x, and csc x are defined by the quotients
tan x 
sin x
cos x
1
sec x 
cos x
cot x 
cos x
sin x
1
csc x 
sin x
Note: Tangent and secant are only defined when cos x ≠ 0
cotangent is cosecant are only defined when sin x ≠ 0
Trig. function
Sin x
Cos x
Tan x
Csc x
Sec x
Cot x
Sign in quadrant…
I
II
III
IV
+
+
-
-
+
-
-
+
+
-
+
-
+
+
-
-
+
-
-
+
+
-
+
-
Ex:
Suppose sin x  23 and that 2  x   .
Determine the values of the other trigonometric functions.
Solution:
Since we know that csc x is the reciprocal of sin x, write the
reciprocal of sin x.
csc x 
1
3

sin x 2
Next we must find cos x because the remaining trig. functions
contain it.
What do you know that involves both sine and cosine that will
help you find cos x? Pythagorean Identity.
(sin x) 2  (cos x) 21
2
 2
2
   (cos x)  1
 3
 4
2
   (cos x)  1
9
4
(cos x) 2  1 
9
9 4
(cos x) 2  
9 9
5
(cos x) 2 
9
5
cos x 
9
5
cos x  
3
The interval for sin x is given.
Quadrant II
Cosine is negative in QII, so…
cos x  
5
3
2
sin x
2
3
6
2
5
2 5
tan x 
 3 5  




cos x  3 3
5
3 5
5 5
5
cot x = 
5
2
or you could have started
at the beginning to find this
solution.
sec x is the reciprocal of cos x. Write it and simplify:
sec x  
3
3
5
3 5



5
5
5 5
Note: You are now ready to write your answers. Make sure they have the correct
sign for quadrant II.
Given : sin x 
2
3
5
3
2 5
tan x  
5
3
csc x 
2
3 5
sec x  
5
5
cot x  
2
cos x  
Ex 2:
Given that cos t  
3
3
and that   t 
2
2
Find the values of the other trigonometric functions.
Solution:
We can see that we are in QIII. The given is
the cosine value for 7
6
7
1
sin x  sin

6
2
1
1
2
1
3
3
tan x  2 3   



3
3
3
3
 2 2
csc t  2
sec x  
2
2 3

3
3
cot x   3  3
The graph of the tangent function.
The tangent function is zero when the sine function is zero because sine is
in the numerator of the tangent function.
The tangent function is undefined when the cosine function is zero because
the cosine function is in the denominator of the tangent function.
The tangent function will be zero at:
The tangent function is undefined at:
tan t
1

0,  , 2
 3
2

2




3 6
-1

2


 
,



 2 5
6
3
2
3

6

t
y = tan x
Sine and cosine have periods of 2 , therefore, tangent will also repeat itself
on that same period.
Ex 3: Sketch the following graphs:
a. y  tan 2 x


b. y  tan  2 x  
3



c. y   tan  2 x    1
3

Solution:
a.) B = 2; set up your inequality:
0  2 x  2
0 x 
The length of your new period is  .
Since we divided the period by 2 we will also have to divide the
restrictions by 2:
Original restrictions:

2
and
3
2
New Restrictions:

4
and
3
4
a.) y = tan 2x
b.) y  tan 2 x   

3
c.) y   tan 2 x   

3
This graph is the graph from part b
reflected over the x-axis.
c. contd.)


y   tan 2 x    1
3

Now we shift it up one unit.
Note: The zeros from the graph are not obvious. To find them
we would set the function equal to zero.


0   tan 2 x    1
3

Which implies…


1  tan 2 x  
3



If we let x =  2 x  
3

We have… tan x = 1; where does this occur, when x = ?
We will find this later this chapter. 
Now that we have the graphs of sine, cosine and tangent we can
graph the remaining trig. Functions using the reciprocal technique.
Do you remember those properties of graphing reciprocals?
… as f(x) increases, its reciprocal… decreases!
The Cosecant function:
Graph: y = sin x.
y= csc x is undefined at:
Where sin x = 0.
There, we have
vertical asymptotes.
Now, use the fact that as y = sin x increases, y = csc x decreases and
vice versa.
Ex: Graph y = sec x
Ex: Graph y = cot x
Be careful with y = cot x.
The restrictions for y = tan x are different from y = cot x. They have
different vertical asymptotes.
1 

Ex: Sketch the graph of y  csc 3x  
2 
2
Note: There are several strategies for graphing:
1. You could start with the parent function and build
on it one change at a time.
2. You can find the most important pieces of information
of this new graph and make the changes to the pieces,
and then plot your new points and new vertical
asymptotes.
This time, lets make the changes to the parent function:
y = csc x
y = csc x has vertical asymptotes at: 0,  , 2
 2
4 5
, ,
,
y = csc 3x has vertical asymptotes at: 0, ,
3 3
3 3
Now, the amplitude changes to ½
Now, we shift the graph to the left

6
units.
The vertical asymptotes go with it!!! They get shifted too!
3.5 Trigonometric Identities
Using the first Pythagorean Identity we can manipulate it and obtain two more.
Pythagorean Identities:
(tan x) 2  1  (sec x) 2
(cot x) 2  1  (csc x) 2
Sum and Difference Formulas for Sine and Cosine
For every pair of real numbers x1 and x 2 , we have
sin( x1  x2 )  sin x1 cos x2  cos x1 sin x2
and
cos(x1  x2 )  cos x1 cos x2  sin x1 sin x2
Ex 1: Determine the following:
 
a.) sin  
 12 
 7 
b.) cos 
 12 
Using the values from the unit circle that we have committed to memory
we will find a combination of two that equal the function we are trying to
evaluate.
  ?

12
 
What equals 
We now label these two
   
  
 12  3 4
x1 and x2
Using the difference formula (we subtracted above and for sine you do
the operation of the combination that you used to equal your function), we
will evaluate
 
sin  
 12 




  
sin     sin cos  cos sin
3
4
3
4
 3 4




  
sin     sin cos  cos sin
3
4
3
4
 3 4
     3  2   1  2 
sin      
   

 3 4   2  2   2  2 
2
    3 2 
sin     
 
 3 4  4  4
Both terms have a common factor:
2
4
Factor it out!
2
  
 3  1
sin    
 3 4 4
You try part b. Hint: For cosine you use the opposite formula compared to the
sign you used to obtain your function.
 7 
If you subtracted to get cos
 then you will use the sum formula, and
 12 
visa versa.
 7 
  
cos

cos


  
b.)
 12 
 3 4




  
cos    cos cos  sin sin
3
4
3
4
3 4
3 2
   1 2
cos    


2 2
3 4 2 2
2
  
1  3
cos   
3 4 4
We can also find the tangent of functions on the unit circle…
tan( x1  x2 ) 
sin( x1  x2 )
cos( x1  x2 )

sin x1 cos x2  cos x1 sin x2
cos x1 cos x2  sin x1 sin x2
We can find tangent of t by using this formula or we can continue
to manipulate the formula to obtain a smaller one.
sin x1 cos x2  cos x1 sin x2
tan( x1  x2 ) 
cos x1 cos x2  sin x1 sin x2
We maintain equality as long as we do the same thing
to both the numerator and the denominator.
Lets divide the numerator and denominator by
cos x1 cos x2
tan( x1  x2 ) 
tan( x1  x2 ) 
sin x1 cos x2 cos x1 sin x2

cos x1 cos x2 cos x1 cos x2
cos x1 cos x2 sin x1 sin x2

cos x1 cos x2 cos x1 cos x2
tan x1  tan x2
1  tan x1 tan x2
Now simplify
Ex 2: Determine tan  
 12 

12


3


4
tan

 tan

  
3
4
tan   
 3 4  1  tan  tan 
3
4
To use this formula it is a
must that you can easily find
the tangent values of the
measures you have committed
to memory.
3 1
  
tan   
 3 4  1  3 1
3 1
  
tan   
3 1
3 4
YOU pick which is easier for you!!! 
Double angle formulas for sine and cosine
sin 2x = sin(x + x) = sin x cos x + sin x cos x = 2sin x cos x
cos 2 x  (cos x) 2  (sin x) 2  1  (sin x) 2  (sin x) 2  1  2(sin x) 2
Ex 3: Determine all the values of x in [0, 2π] that satisfy each
equation.
a.) cos x = sin 2x
Solution:
b.) sin x = cos 2x
c.) 1 = sin x + cos x
a.) We need to get everything into sin x and cos x
because that is what we know on the unit circle.
cos x = sin 2x = 2sin x cos x
cos x = 2sin x cos x
0 = 2sin x cos x – cos x = cos x(2sin x – 1)
0 = cos x(2sin x – 1)
cos x = 0
2sin x – 1 = 0
and
sin x = ½
cos x  0 when x 
sin x 

2
3
2
and
1

5
when x  and
2
6
6
b.) sin x = cos 2x
sin x  1  2(sin x) 2
2(sin x) 2  sin x  1  0
(2 sin x  1)(sin x  1)  0
sin x 
1
2
when x 

6
or
5
6
sin x  1 when x 
3
2
c.) 1 = sin x + cos x
Square both sides.
1  (sin x  cos x) 2
1  (sin x) 2  2 sin x cos x  (cos x) 2
Re-order expression
1  (sin x) 2  (cos x) 2  2 sin x cos x
1  1  2 sin xcos x
0  2 sin xcos x
0  sin 2x
In order for sine to ever be equal to 0, 2x must be
a multiple of pi.
x could be any of the following to make this statement true:
x  0, x 

2
, x  , x 
3
, x  2
2
CHECK!!!
x  0, x 

2
, x  , x 
3
, x  2
2
NOTE: We squared the equation earlier, so this could introduce
extraneous solutions. This must have happened because
the original expression, 1 = sin x + cos x is only
satisfied if the following is true:
x  0, x 

2
When x =
, x  2
3
2
the statement does not equal 1.
Half Angle Formulas
Half angle formulas come from double angle formulas involving “2x.”
Ex: Double Angle formula
half angle formula
 x
cos 
 2
cos 2 x
Instead of doubling the angle it is cut in half.
Half Angle Formula:
For any real number x we have
x  1  cos x
x  1  cos x


sin

and
cos



 
2
2
2
2


2
2
If we replace x with 2x we have
sin x    1  cos 2 x and (cos x)   1  cos 2 x
2
2
 7 
 7 
 7 
, cos , and tan .
 8 
 8 
 8 
Ex 4: Determine sin 
Solution:
 7 
1  cos 2 
 7 
 8 
sin    
2
 8 
 7 
sin  
 8 
 7 
1  cos 2 
 7 
 8 
sin    
2
 8 
 7 
1  cos 
 7 
 4 
sin    
2
 8 
 7 
sin    
 8 
1
2
2
2
Substitution for x
Reduce fraction
Now take cosine
Now use algebra and simplify expression
 7 
sin    
 8 
1
2
2
Now we must determine the sign.
2
2
2

 7 
sin     2 2
2
 8 
2 2
 7 
2
sin    
2
 8 
2 2 1
 7 
sin    

2
2
 8 
2 2
 7 
sin    
4
 8 
2 2
 7 
sin    
2
 8 
This function is in quadrant IIsine is positive here.
2 2
 7 
sin   
2
 8 
You try the next one!!! 
 7 
cos  
 8 
7
2
1
4 
2 
2
2
1  cos
Since
7
8
2
2

2 2 
2
2 2
2  2 2  2 2
2
4
2
is in the second quadrant, cosine is negative.
2 2
 7 
cos   
2
 8 
 7 
sin  
 7 
 8 
tan  
 8  cos 7 
 
 8 
2 2
2 2
2 2
2



2 2
2 2  2 2

2
Verifying Identities
Using all of the trigonometry we’ve learned so far we can use this information
to prove identities. (make one side of the equation look like the other)
Ex 5: verify the identity
(1  (cos x) 2 )(sec x) 2  (tan x) 2
It does not matter which side you start with to make it look like the other side.
Many times we only have to manipulate one side to look like the other.
Sometimes we work both sides together and as we go along they will be the
same.
Solution:
I will start with the left side. I notice that I can use the Pythagorean
Theorem.
(sin x) 2 (sec x) 2  (tan x) 2
1
2
(sin x)

(tan
x
)
(cos x) 2
2
(tan x) 2  (tan x) 2

Remember, your goal is to turn it into tangent.
(sin x ) 2
2

(tan
x
)
(cos x) 2
Now you try one!!!
Ex 6: Verify: cot x + tan x = sec x csc x
cos x sin x

 sec x csc x
sin x cos x
(cos x) 2 (sin x) 2

 sec x csc x
sin x
cos x
(cos x) 2  (sin x) 2
 sec x csc x
sin x cos x
1
 sec x csc x
sin x cos x
1
1

 sec x csc x
sin x cos x
sec x csc x  sec x csc x
3.6 Right Triangle Trigonometry
Trigonometric Functions of an Angle in a Right Triangle
For the angle θ in the right triangle shown, we have
b
sin   ,
c
c
csc  ,
b
a
cos  ,
c
c
sec  ,
a
b
tan   ,
a
a
cot  
b
Recall: SohCahToa
c
b
θ
a
opp
adj
opp
, cos 
, tan  
,
hyp
hyp
adj
hyp
hyp
adj
csc 
, sec 
, cot  
opp
adj
opp
sin  
3
Ex 1: Suppose that an acute angle θ is known to satisfy sin   . Determine the
5
other trigonometric functions of this angle.
Solution:
Lets use a right triangle and label what we know.
5
3
We must find a before
we can find the others,
except for csc θ
θ
a =4
The Pythagorean Theorem will help us find the value of a.
a2  b2  c2
a 2  32  5 2
a  9  25
2
a 2  16
a4
Now that we know all of the values for a, b, and
c, we can write all of the solutions:
3
sin   ,
5
5
csc  ,
3
4
cos  ,
5
5
sec  ,
4
3
tan   ,
4
4
cot  
3
Note: The last example said that the angle was acute which means it lies in the
first quadrant. If the angle was in the second, third, or fourth, we would
have solved the problem like we just did, and then change the signs as
necessary.
Ex 2: Find the value of the six trigonometric functions if the cos 
3
5
Solution:
Using the Pythagorean theorem we find that the opposite side equals 4.
4
sin   ,
5
5
csc  ,
4
3
cos  ,
5
5
sec  ,
3
4
tan   ,
3
3
cot  
4
Ex 3: A climber who wants to measure the height of a cliff is standing 35 feet from
the base of the cliff. An angle of approximately 60° is formed by the lines
joining the climber’s feet with the top and bottom of the cliff, as shown. Use
this information of approximate the height of the cliff.
cliff
x
tan 60 
opp x

adj 35
tan 60 
x
35
1.732050808 
60°
35 ft
Solution:
Solve for x.
x
35
x ≈ 60.6 ft.
When you have a right triangle situation and at least one angle and one
side is known, follow these steps to find the missing piece.
1. Ask yourself: What side do I have and what side do I want?
We have the adjacent side and we want the
opposite side.
2. Which trig function involves the two answers from question #1?
The tangent function will be used.
Ex 4: Two balls are against the rail at opposite ends of a 10 foot billiard table. The
player must hit the ball on the left with the cue ball on the right without
touching any of the other balls on the table. This is done by banking the cue
ball off the bottom cushion, as shown. Where should the cue ball hit the
bottom cushion, and what is the angle that its path makes with the bottom
cushion.
3
θ
x
2
2
3
 tan  
x
10  x
20 – 2x = 3x
2
3

x 10  x
x=4
2
Since tan   ;
x
θ
10 - x
2(10 – x) = 3x
tan  
2 1

4 2
Unfortunately, we do not know a value on our unit circle where this is true, but this
is the best we can do for now, even though we need the angle measure, not the
tangent of that angle.
We can answer where the cue ball should hit the bottom cushion: The cue ball
should hit the bottom cushion 2 feet from the bottom left or 8 feet from the bottom
right.
Ex 5: An engineer is designing a drainage canal that has a trapezoidal cross
section, as shown. The bottom and sides of the canal are each L feet long,
and the side makes an angle θ with the horizontal.
a.) Find an expression for the cross-sectional area of the canal in terms of
the angle θ with the horizontal.
b.) If the canal is S feet long, approximate the angle θ that will maximize
the capacity of the canal.
L
Lsin θ
L

Lcos θ
a.) Area of a trapezoid:
1
A  (b1  b2 ) h
2
L

Lcos θ
We know the length of one base (L) and can find the
height by making a right triangle.
We have angle θ and the hypotenuse; we want the
opposite side:
1
A  (b1  b2 ) h
2
A
1
2 L  2 L cos L sin 
2
A  ( L  L cos ) L sin 
A  L(1  cos ) L sin 
A  L2 (1  cos ) sin 
b.)
Capacity means volume, so a canal that is S feet long has the capacity…
V  A S
V  SL2 (1  cos ) sin  feet 3
To find the exact value of θ will have to wait until the next section! 
3.7 Inverse Trigonometric Functions
Recall properties of inverse functions:
Properties of Inverse Functions
Suppose that f is a one-to-one function. Then the inverse f-1 is
unique, and …
1. The domain of f-1 is the range of f.
2. The range of f-1 is the domain of f.
3. If x is in the domain of f-1 and y is in the domain of f, then
f-1(x) = y if and only if f(y) = x.
4. f(f-1(x)) = x when x is in the domain of f-1.
5. f-1(f(x)) = x when x is in the domain of f.
6. The graph of y = f-1(x) is the graph of y = f(x) about the line
y = x.
Sine
The sine function is not one to one because it is periodic. If fails the horizontal
line test for every horizontal line between -1 and 1 on the y-axis.
However, we can restrict the domain of the sine function so that it is one-to-one.
The sine function is one to one on the interval:
  
 2 , 2 
On this interval, the sine function does have an inverse and is denoted:
Arcsine function or simply: y = arcsin x
Note: csc x = (sin x)-1
The Arcsine Function
  
 ,
The arcsine function, denoted arcsin, has domain (-1,1) and range  2 2 
and is defined by
arcsin x = y if and only if
sin y = x
Arcsine Properties
sin(arcsin x) = x when x is in [-1, 1] and
arcsin(sin x) = x when x is in
  
 2 , 2 
The graph of y = arcsin x is the reflection of the restricted function of y = sin x
reflected about the line y = x.
Notice how steep the ends of
the curve are on the inverse.
This corresponds to the
flatness on the sine curve.
Ex 1: Find a.) arcsin
Solution:
1
2
a.) arcsin
 
b.) arcsin  sin 
3

1 

2 6
  
b.) arcsin  sin  
3 3

 3 
c.) arcsin  sin

4


 3 
c.) arcsin  sin

4 

Remember that we have a restricted interval
now for sine. So we ask ourselves: Where in
the restricted interval does sin = ½ ?
Recall the arcsine properties: arcsin(sin x) = x.
Here we have to be careful – our answer must
lie in the restricted interval.
3
is not in the restricted interval.
4
3
sin
Where in the restricted interval is the
the same value?
4
3

sin
 sin
So...
4
4
 3  
c.) arcsin  sin

4

 4
All of the other trigonometric functions are defined by making domain restrictions, too.
Cosine
The Arccosine Function
The arccosine function, denoted arccos, has domain [-1, 1] and range [0, π].
and is defined by
arccos x = y if and only if cos y = x
Arccosine Properties
cos(arccos x) = x when x is in [-1, 1]
in [0, π].
and
arccos(cos x) = x when x is
Ex 2: Find

 1 
a.) cos arccos  
 2 


 1 
b.) sin arccos  
 2 



c.) arccos cos 
3

   
d .) arccos cos  
  4 
Solution:
1

 1 
a.) cos arccos    
2
 2 

2
3

 1 
b.) sin arccos    sin

3
2
 2 

 

c.) arccos cos  
3 3




d .) arccos cos     except this is not in the restricted domain.
4
4


2
Where in the restricted domain does cos  
?
4 2

At
4
Tangent
The Arctangent Function
The arctangent function, denoted arctan, has domain (-∞, ∞) and
 
range   , , and is defined by
 2 2
arctan x = y if and only if tan y = x
Arctangent Properties
  
tan(arctan x) = x for x in (-∞, ∞) and arctan(tan x) = x for x in   ,  .
 2 2
The graph of y = arctan x is the graph of y = tan x reflected over the line y = x.
Graph together with students on board
12
3

Ex 3: Find cos arctan  arcsin 
5
5

We can see that this is of the form: cos( x1  x2 )
Solution:
We will use the difference formula for cosine.
12
3
12  
3
12  
3



cos arctan  arcsin   cos arctan  cos arcsin   sin  arctan  sin  arcsin 
5
5
5 
5
5 
5



We know the value of the very last identity.
12
3
12  
3
12  3



cos arctan  arcsin   cos arctan  cos arcsin   sin  arctan  
5
5
5 
5
5 5



The remaining identities must be found using a right triangle:
tan  
13  52  122
12
12
5
But this is not one of our identities…
To get θ by itself we take the inverse of both sides.
arctan(tan  )  arctan
θ
5
12
5
  arctan
12
5
Now, using this triangle find values of identities involving arctan 12
5
12
3
12  
3
12  3



cos arctan  arcsin   cos arctan  cos arcsin   sin  arctan  
5
5
5 
5
5 5



12
3
12  
3
12  3



cos arctan  arcsin   cos arctan  cos arcsin   sin  arctan  
5
5
5 
5
5 5



These values are:
5
13
12
13
12
3 5
3  12 3


cos arctan  arcsin    cos arcsin   
5
5  13
5  13 5


3

cos
arcsin


All we have to do now is draw another triangle to help us find
5


3
3
sin  
  arcsin
5
5
5
3
3 4

Now,
find
the
value
of
your
identity:
cos
arcsin


θ
5

 5
4
12
3  5 4 12 3

cos arctan  arcsin     
5
5  13 5 13 5

Now evaluate
12
3
16

cos arctan  arcsin   
5
5
65

Arcsecant
The arcsecant function has the same restricted interval as the cosine function.

Recall that y = sec x has a vertical asymptote at
2
The Arcsecant Function
The arcsecant function, denoted arcsec, has domain (-∞, -1] U [ 1, ∞) and
    
range 0,    ,   and is defined by
 2  2 
arcsec x = y if and only if sec y = x
Arcsecant Properties
Sec(arcsec x) = x when x is in (-∞, -1] U [1, ∞) and arcsec(sec x) = x when
    
   , 
 2  2 
x is in 0,
Ex: Graph y = arcsec x
Graph with students on board.
3.8 Applications of Trigonometric Functions
A Cessna Citation III business jet flying at 520 miles per hour is
directly over Logan, Utah, and heading due south toward
Phoenix. Fifteen minutes later an F-15 Fighting Eagle passes over
Logan traveling westward at 1535 miles per hour. We would like to
Determine a function that describes the distance between the planes
in terms of the time after the F-15 passes over Logan until it
reaches the California border 20 minutes later.
d = 1535t
Distance = rate x time
d (t )  (1535t ) 2  (520t  130) 2
d(t)
d = 520(t + 0.25) = 520t + 130
Recall: the distance formula here involves
time in hours: 20 minutes is 1/3 of an hour
t = 1/3
d (t )  (1535t ) 2  (520t  130) 2
1
1
1
d    (1535 ) 2  (520   130) 2
 3
 3
 3
1
d    595 miles
 3
It is more likely that the planes will not be traveling in paths that are
perpendicular. Lets change the problem a little.
A Cessna Citation III business jet flying at 520 miles per hour is
directly over Logan, Utah, and heading due south toward
Phoenix. Fifteen minutes later an F-15 Fighting Eagle passes over
Logan traveling 24° west of south at 1535 miles per hour toward
Nellis Airforce base 395 miles away.
Since this situation does not include
a right angle, we can not use the
Pythagorean Theorem…directly.
24°
1535t
520t + 130
d(t)
Law of Cosines
Suppose that a triangle has sides of length a, b, and c and
corresponding opposite angles α, β, and γ as shown. Then
a 2  b 2  c 2  2bc cos
A
c
α
B
β
a
γ
b
We also get…
b 2  a 2  c 2  2ac cos 
c 2  a 2  b 2  2ab cos 
C
Ex 1: A triangle has sides of length 6 and 8, and the angle between
these sides is 60°. What is the length of the third side?
6
x
60°
8
Solution:
Using the Law of Cosines…
x 2  6 2  82  2(6)(8) cos 60
1

x  100  96 
 2
2
x 2  52
x  2 13
Ex 2: If two sides of a non right triangle are of lengths 15 and 25
and the included angle measures 35°, find the missing side
and one of the other angles.
15
x
γ
35°
25
x  15 2  25 2  2(15)(25) cos 35
x  15.35
-----------------------------------------------------------------152  15.352  252  2(15.35)(25) cos 
225  235.6225  625  767.5 cos 
225  860.6225  767.5 cos 
 635.6225  767.5 cos 
0.8281726384  cos 
34.0085  
34  
We can now go back and solve the aircraft problem…
(d (t )) 2  (1535t ) 2  (520t  130) 2  2(1535t )(520t  130) cos 24
Lets let t = 0.333
(d (0.333)) 2  (511.155) 2  (173.16  130) 2  2(511.155)(173.16  130) cos 24
1
d    265 miles
 3
Ex 3: A picture in an art museum is 5 feet high and hung so that its
base is 8 feet above the ground. Find the viewing angle θ(x) of
a 6-foot tall viewer who is standing x feet from the wall.
APB and APC are both
right triangles. We can
Use the Pythagorean
Theorem.
AC  49  x 2
AB  4  x 2
Using the Law of Cosines…
25  (4  x 2 )  (49  x 2 )  2 4  x 2 49  x 2 cos ( x)
25  53  2x 2  2 4  x 2 49  x 2 cos ( x)
25  53  2 x 2
cos( ( x)) 
 2 4  x 2 49  x 2
2


14

x


 ( x)  arccos
2
2 
 (4  x )(49  x ) 
Using a graphing calculator we can see that the best viewing of the
painting is where the maximum occurs which is when x  3.7 feet.
This is the distance the viewer should stand from the wall.
When the angles and one side of a triangle are known we can use the
Law of Sines to find the other missing parts.
The Law of Sines
Suppose that a triangle has sides of length a, b and c with
corresponding opposite angles α, β and γ as shown. Then
B
β
sin  sin  sin 


c
a
a
A
α
γ
b
C
b
c
Ex 4: Find the missing angles.
α
6
2 13
β
60°
If we knew α we could
find β.
β = 180 – 60 – α
8
Solution:
We could use the Law of Cosines to find α…
82  6 2  (2 13) 2  2(6)(2 13) cos
but it is easier to use the Law of Sines.
sin  sin 60

8
2 13
2 3
  arcsin
13
8 sin 60
2 13
2 39
  arcsin
13
sin  
  arcsin
8 sin 60
2 13
  73.9;   46.1
Ex 5: The aircraft carrier Carl Vinson leaves the Pearl Harbor naval
shipyard and heads due west at 28 knots. A helicopter is
175 nautical miles from the carrier 35° south of west.
a.) On what course should the helicopter travel at its cruising
speed of 130 knots to intercept the aircraft carrier?
b.) How long will it take.
Solution: Draw a picture of the given information.
First, find θ which will give the
course the helicopter should fly.
sin  sin 35

28t
130t
sin  
28t sin 35
 0.1235
130t
  arcsin 0.1235  7
The helicopter should fly 42°
north of east.
Since we know θ = 7° we also know the third angle is 138°.
sin 138 sin 35

175
130t
130t sin 138  175sin 35
t
175 sin 35
130 sin 138
t  1.154 hours
It will take the helicopter about 1 hour and 9 minutes to intercept
the aircraft carrier.
Ex 6: Think Pair Share
A campground lies at the west end of an east-west road in a
relatively flat, but dense, forest. The starting point for a hike
lies 30 kilometers to the northeast of the campground. A hiker
begins at the starting point and travels in the general direction
of the campground, reaching the road after 25 kilometers.
Approximately how far is the campground from the road?
B
30
25
camp
A
45°
γ
C
First, we need to find γ:
γ'
C'
sin  sin 45

30
25
Lets assume triangle ABC gives
the correct solution. If the hiker
traveled along the line BC' would
mean the hiker was way off
course – BUT IT COULD
HAPPEN! This gives us an
isosceles triangle.
This could be the value
of γ or γ'
sin  sin 45

30
25
 or    58
Since γ is an obtuse angle it cannot be 58°, this must be the value of
γ'.
γ is part of a linear pair and the other angle is the same as γ' since they
are the base angles of an isosceles triangle.
γ = 122°
The angle at B is: 180 – 45 – 122 = 13°
sin 13 sin 45

AC
25
AC = distance of hiker from camp  7.95 km.
If the hiker was lost, but measured the distance from the road
correctly, B is 77°, therefore the distance would be…
sin 77 sin 45


AC
25
AC  
25 sin 77
 34.5 km
sin 45
If the hiker was badly off course he is about 34.5 km from camp.
Heron’s Formula
Heron’s formula is used to find the area of a triangle when
only the lengths of the sides of the triangle are known.
Heron’s Formula
A triangle with sides of length a, b and c has area given by

1
P( P  2a)( P  2b)( P  2c)
4
where P is the perimeter of the triangle, P = a + b + c
Ex 6: Find the area of the triangle.
7
9
12
P = 7 + 9 + 12 = 28
A  0.25 28(28  14)(28  18)(28  24) 
A  14 5