TRIGONOMETRY
AP Calculus AB
Summer Review
TRIGONOMETRY AND CALCULUS



AP Calculus AB, SJHS 2014-2015
Calculus requires a thorough knowledge of the basic
functions, which includes the transcendental
functions
Thus, you are expected to understand polynomial,
exponential, logarithmic, and trigonometric functions
Trigonometry occurs very often in calculus, and
without knowing the algebra of trig functions several
calculus concepts will be very difficult
2
ALGEBRA OF TRIGONOMETRY

The most basic trigonometry is understanding the
algebra trig functions
AP Calculus AB, SJHS 2014-2015
This means the methods of rearranging trig functions (as
a whole) in addition/subtraction, multiplication/division,
and exponents to solve equations
 This also includes the use of “inverse trig functions” to
pull information from a trigonometric equation


Several trig equations require a calculator to solve
angles in either degrees or radians
3
DEGREES
Degree measurement is a standard measurement for
angles in some applied sciences, such as mechanics or
optics
 It is based on the idea that a circle turns across 360°
for one revolution
 Degrees are a completely made up unit; they have no
physical basis, but are still a useful reference unit

AP Calculus AB, SJHS 2014-2015
4
RADIANS

The radian is based on the proportion of a circle
length to its radius
AP Calculus AB, SJHS 2014-2015

Every circle has this relationship, and it helps to define a
circle

This is not the definition of a circle!
By definition, there 2π radians in one revolution of
any circle
 You can also find the length of a sector, area of a
sector, and area of a circle (as well as several other
quantities) using angles in radians
 Most math (and theoretical science) is done in radians

5
CONVERSION OF DEGREES TO RADIANS

Since we have the following formulae for 1 revolution:


We can deduce that the following is true:


AP Calculus AB, SJHS 2014-2015

1rev  360
Radians: 1rev  2
Degrees:
2  360
Notice that radians have no units when written down, but if you
wish you may write “rad” as a unit (Ex: 2π rad)
Thus, if we ever need to convert, we can use either of the
following factors
360
or
2

2
360
Note: you may have learned a less significant, but still correct,
relationship of 180°=π; this proportion works, but it loses the
definition of a revolution

It is better to think of angles and radians in terms of revolutions, not
just a meaningless simplified proportion
6
EXACT VALUES
There are few things that need to be memorized in
math, and exact values of some angles are a necessity
 You should become familiar with the following degree
measurements in terms of radians:
30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°,
240°, 270°, 300°, 315°, 330°, 360°
 Although these degree measurements appear to have
no significant connections, they are very closely
connected in radians

AP Calculus AB, SJHS 2014-2015
7
CONNECTIONS OF THE 30N° ANGLES

In radian measurement, the 30n°angles take on

Notice that 30° is one-sixth of a semi-circle (pi radians)

This makes 60° one-third of a semi-circle
AP Calculus AB, SJHS 2014-2015
n
reduced values of
6
n
 By this logic, all 60n° angles take on reduced values of
3
8
CONNECTIONS OF THE 45N° ANGLES

In radian measurement, the 45n°angles take on

AP Calculus AB, SJHS 2014-2015
n
reduced values of
4
45° is one-fourth of a semicircle
n
 By this logic, all 90n° angles take on reduced values of
2
 90° is half of a semi-circle

By extension, all angles 180n° take on reduced values
of n
9
SUMMARY OF EXACT VALUES
0
30
45
60
90
120
135
150
180
Radian
0
π/6
π/4
π/3
π/2
2π/3
3π/4
5π/6
π
Degree
210
225
240
270
300
315
330
360
Radian
7π/6
5π/4
4π/3
3π/2
5π/3
7π/4
11π/6
2π
AP Calculus AB, SJHS 2014-2015
Degree
10
GRAPHICAL REPRESENTATION OF EXACT
VALUES 
2
3
4
2

0 
5
4
3
2
Fourths
7
4

6
AP Calculus AB, SJHS 2014-2015
 5
4 6
3

3
0
7
6
5
3
4
3
Thirds
11
6
11
WHAT SHOULD WE USE: DEGREES OR
RADIANS?
RADIANS!
The only exception is if the problem explicitly uses
degrees in its statement, or asks for an answer in
degrees
AP Calculus AB, SJHS 2014-2015

12
TRIANGLES IN THE COORDINATE PLANE



When you draw this triangle, draw an angle arrow on the
x-axis, draw a ray starting at the origin in the direction of
that angle with length r, and drop a perpendicular to the xaxis
Note: r is always positive, but x and y can be positive or
negative, depending on the quadrant
AP Calculus AB, SJHS 2014-2015
We are given a right triangle of hypotenuse r
superimposed onto an x,y-plane; there is an angle θ
between the hypotenuse and the x side
13
EX: DRAW A TRIANGLE IN THE X,Y-PLANE
WITH HYPOTENUSE 2 AT AN ANGLE AT 50°.
50
AP Calculus AB, SJHS 2014-2015
2
14
EX: DRAW A TRIANGLE IN THE X,Y-PLANE
WITH HYPOTENUSE 3 AT AN ANGLE AT 290°.
AP Calculus AB, SJHS 2014-2015
290
15
TRIGONOMETRIC FUNCTIONS

For a triangle in the x,y-coordinate plane, the three
trigonometric functions are defined as:
AP Calculus AB, SJHS 2014-2015
y
sin  
r
x
cos 
r
y
tan  
x
16
THE RECIPROCAL FUNCTIONS

Each trig function has an associated reciprocal
function:
AP Calculus AB, SJHS 2014-2015
r
csc 
y
r
sec 
x
x
cot  
y
17
RIGHT TRIANGLE GEOMETRY

We may use any (x,y) coordinate to solve for the
hypotenuse by the Pythagorean Theorem
x2  y 2  r 2
24
26
-10
24 12

26 13
10
5
cos 

26
13
24
12
tan  

10
5
sin  
13
12
13
sec  
5
5
cot   
12
csc 
AP Calculus AB, SJHS 2014-2015
Ex: Given point (-10,24), find all six trig
x functions.
 10
2
 r  x 2  y 2   10   242  26
y  24

18
QUADRANTS AND SIGNS




Sine is related to the y-coordinate; thus, any triangle
pointing up (above the x-axis) has a positive sine and
anything down is a negative


Cuz positive is to the right, and negative is to the left for x
Cuz up is positive and down is negative for y
Tangent is a fraction of the y over x; fractions are positive if
top and bottom are both positive or negative
So triangles in the (+,+) and (-,-) regions have a positive tangent;
this is Quadrants I and III (respectively)
 Since the others are mixed, that is QII(-,+) and QIV(+,-), then
tan is negative in those quadrants
AP Calculus AB, SJHS 2014-2015
To analyze what sign to expect, just consider the following:
Cosine is related to the x-coordinate; thus, any triangle to
the right (of the y-axis) has a positive cosine and anything to
the left is negative

19
SUMMARY OF QUADRANT SIGNS
I
II
III
IV
Sin
+
+
-
-
Cos
+
-
-
+
Tan
+
-
+
-
AP Calculus AB, SJHS 2014-2015
Quadrant
20
DOMAIN AND RANGE

The trig functions all have an infinite domain (except
some undefined points); their ranges significantly differ,
but you can likely see the relationships in this table:
Domain
Range
Sin
  x  
1  y  1
Cos
  x  
1  y  1
Tan
 3 5
x
,
,
, 
  y  
Csc
2 2 2
x     
  y  
Sec
 3 5
  y  
Cot
x
,
,
, 
2 2 2
x     
  y  
AP Calculus AB, SJHS 2014-2015
Function
21
SPECIAL RIGHT TRIANGLES
60
30

2
3
45
1
45
1
1
30-60-90
45-45-90
AP Calculus AB, SJHS 2014-2015
2
22
EX: USE THE SPECIAL RIGHT TRIANGLES TO
EVALUATE EACH TRIG FUNCTION
60
sin 60 
tan30 
sec30 
2
3
30
1
2
45
1
AP Calculus AB, SJHS 2014-2015
cos45 
3
2
1
2
1
3
2
3
45
csc45  2
1
23
WAIT, SO WHAT’S THE POINT OF EXACT VALUES IF
WE JUST NEED TO KNOW TRIG FUNCTIONS?



The only difference is the sign!
Notice that the angles of the exact values apply to
special right triangles
AP Calculus AB, SJHS 2014-2015
Exact values are great to work with since each class
of exact values gives the same answer for a trig
function
24
EX: EVALUATE SIN(30°), SIN(150°),
SIN(210°), SIN(330°).

These seem to be a random collection of angles in
degrees
AP Calculus AB, SJHS 2014-2015
If we switch to radians we see they fall into the pattern of
 We already know sin π/6 = ½; so each of these will have
the value ½, and the sign depends on the quadrant

25
EX: EVALUATE TAN(45°), TAN(135°),
TAN(225°), AND TAN(315°)

Switching to radians, we see there all are fractions of
fourths, and thus must have the same value
We know tan π/4=1 , so all the values are 1 and the signs
are dependent on the quadrant
AP Calculus AB, SJHS 2014-2015

26
NEGATIVE ANGLES
A negative angle means the direction of sweep from
0° is clockwise
 These follow the same rules as exact values, in that
the values of your function is the same but the sign
has to do with the quadrant rules

AP Calculus AB, SJHS 2014-2015
27
3
sin  120   
2
1
cos120  
2
1
cos  120   
2
120
 3
2
AP Calculus AB, SJHS 2014-2015
EX: EVALUATE SIN(120°), SIN(-120°),
COS(120°), AND COS(-120°).
2
3
3
sin120 
1 120
2
28
“TRIANGLES” OF ONLY ONE LINE
A triangle by definition has three lines and three
angles, with angles that add up to 180°
 The situation arises of what happens if two sides have
the same length


If you are given a point that does not make a proper
triangle, then look at the coordinates to help you
evaluate the trig functions
AP Calculus AB, SJHS 2014-2015
Then it’s not technically a triangle since you would have
two angles of 90°
 This is a purely theoretical case, but let’s just look at it in
terms of x and y

29
EX: 90° ANGLE
Suppose we want to find the sin 90°, cos90°, and
tan90°
 If we draw a 90° angle and an arbitrary hypotenuse r
in that direction, we get a line pointing up (not a
triangle)
 We still have enough information to evaluate, since
we know a few things:

1. In this case, there is no x-coordinate

2. In this case, if did have a “triangle”, x=0 and the
“hypotenuse” r is the same as our y side  r 
AP Calculus AB, SJHS 2014-2015
 x0

y
30
EX: 90° ANGLE

Now we can plug in this information:
90
AP Calculus AB, SJHS 2014-2015
y y
sin 90    1
r y
x 0
cos90    0
r r
y y
tan 90    undefined
x 0
31
EX:

SIN, COS, AND TAN OF 180°

2. The “hypotenuse” r has the same length as x, but x is
clearly a negative number since the ray points backwards
Now we can formulate:
y 0
sin180    0
r r
x
x
cos180     1
r
x
y 0
tan180    0
x x

180
AP Calculus AB, SJHS 2014-2015
This is the same problem, but instead our angle gives
the following information:
 1. There is no y-coordinate, so  y  0
32
TRIG IDENTITIES

Since trig functions are a result of triangles, it is worth
relating them to the Pythagorean Theorem; we write in
terms of x, y, and r
2
2
2
x  y r
x2 y 2 r 2
 2  2
2
r
r
r
x2 y 2 r 2
 2  2
2
x
x
x
x2 y 2 r 2
 2 2
2
y
y
y
 cos   sin   1
2
2
 1  tan   sec 
2
2
AP Calculus AB, SJHS 2014-2015
 If we rewrite this by dividing by x, y, or r, we get our three
trig identities:
 cot 2   1  csc2 
 Do not memorize these! Just remember how to get them if
you need them!
33
PROOFS INVOLVING TRIG FUNCTIONS
Since trig functions are related to one another by x, y,
and r, it is an instructive exercise to rearrange a given
expression of trig functions into another equivalent
expression
 To do this, use your basic definitions and your exact
values

AP Calculus AB, SJHS 2014-2015
34
EX: PROVE THE FOLLOWING STATEMENT BY
REARRANGING THE LEFT SIDE ONLY.
1  sin x
 sin x
csc x  1
sin x 1  sin x 
sin x 1  sin x 


1
1  sin x
 sin x  sin x
sin x
AP Calculus AB, SJHS 2014-2015
sin x 1  sin x 
1  sin x 1  sin x sin x



csc x  1 csc x  1 sin x csc x  sin x  sin x
 sin x
35
TRIGONOMETRIC EQUATIONS AND INVERSE
TRIGONOMETRY
A trigonometric equation is an equation containing
one or more trig functions
 Simple trig equations can be solved analytically


Equations containing several different trig functions are
often only possible to solve with a computer
Once the equation is in the form that the trig function
has a value, you may use the “inverse trig function”
to get the answer
The inverse trig function simply means the operation (not
function) used to convert your output value to your input
angle
 Algebraically, it is more correct to use an inverse trig
function to solve an equation; logically, you may just infer
your answer from the information
 These are sometimes referred to as “arc” (like arcsine)

AP Calculus AB, SJHS 2014-2015

36
DEFINITION OF AN INVERSE TRIG FUNCTION



Because inverse trig functions “undo” the function,
we get the property: sin 1  sin    


Thus, the function and inverse switch their domain (x)
and range (y) coordinates
1
sin
sin
 A  A
This works both ways!
Ex: Cosine vs Arccosine
cos  0   1
Cos 1 1  0
AP Calculus AB, SJHS 2014-2015
By definition, an inverse trig function is defined at the
points its parent function has a value at; its value is
the angle input of the parent function
37
DOMAIN OF INVERSE TRIG FUNCTIONS


Ex: sin 0  1, sin 2  1, sin 4 
arcsin 1  ?
Which value do we pick? Is the answer 0, 2π, 4π,
or something else
1
 Actual answer is Sin
1 0
AP Calculus AB, SJHS 2014-2015
Since these functions must have well defined values, it
its essential to restrict the domains
 Why? Because otherwise we wouldn’t know which
value to choose!


38
DOMAIN AND RANGE OF INVERSE TRIG
FUNCTIONS
Domain
Sin 1  x 
1  x  1
Cos 1  x 
1  x  1
Tan
1
 x
  x  
Range


 y 
2
2
0 y 
2
 y 
2
AP Calculus AB, SJHS 2014-2015
y  f  x
39
SIMPLE EXAMPLE OF AN INVERSE TRIG
FUNCTION

For what value of θ does sin θ = 1?
We know that sin π/2 = 1, so logically θ= π/2
 Algebraically, the process goes:

 sin    sin 1
  sin 1 1
sin
1
  2
1
AP Calculus AB, SJHS 2014-2015
sin   1
40
IS IT NECESSARY TO USE INVERSE TRIG
FUNCTIONS
No, not if you can logically deduce the answer
yourself
 Because of this tedious process, and the fact that
inverse functions are difficult to conceptualize, most
people prefer to use the logical approach
 You do need to use inverse trig functions if you are
solving for angles not included in the exact values set

In that case, you would need to use your calculator’s
inverse trig button
AP Calculus AB, SJHS 2014-2015

41
NOT SO SIMPLE INVERSE TRIG PROBLEM
Simplify :

 2 
cos  Sin 1    
 5 

So the answer is just the cosine of this
triangle!
We need to solve for x, but other than that it’s
just writing
the cosine.
2
x 2   2   52
21
5
-2
AP Calculus AB, SJHS 2014-2015
This is tricky! Arcsine is defined only in Quadrants I and IV.
Since the arcsine is negative, we assume it’s a fourth quadrant
angle.
The question asks: what is the cosine of the
angle defined by the inverse sine of (-2/5).
 x  25  4  21

21
 2 
cos  Sin1     
5
 5 

42
EX: 11 SIN X = 2

Let’s get this into a useable form:
2
sin x 
11
2
sin  sin x   sin  
 11 
x  0.183
1
1
AP Calculus AB, SJHS 2014-2015
11sin x  2
So our answer is 0.183 radians, or 10.5°
43
EX: 2cos  x     1


12 


 

1  1 
Cos  cos  x     Cos  
12  

2

1
Cos 1 undoes cos; both go away
x
x

12

3


this is an exact value !

3

12
 x
AP Calculus AB, SJHS 2014-2015
  1

cos  x   
12  2


4
44
EX: 4sin  3x  1  5

First step is to isolate the sine:

This doesn’t make sense. No matter what, sine is
between -1 and 1. Since 5/4 is out of our range for
sine, this doesn’t have a solution.
AP Calculus AB, SJHS 2014-2015
5
sin  3x  1 
4
45
1 2
EX: tan x  sec x  1
2

tan 2 x sec 2 x 1
1
2
sec
x  1  sec x  1

2
AP Calculus AB, SJHS 2014-2015
Anytime we mix trig functions, the equation is more
difficult. Always check which functions they are,
because some are related by the trig identities. In this
case, we can see that tan and sec are related
(1+tan2x=sec2x). You should replace the squared
term and see where that takes you.
46
EXAMPLE CONTINUED
sec x  1
AP Calculus AB, SJHS 2014-2015
This is a difference of two squares now, so separate
into factors. You’ll see it simplifies quickly:
1
 sec x  1 sec x  1  sec x  1
2
1
 sec x  1 sec x  1 sec x  1
2

sec x  1
sec x  1
1
 sec x  1  1
2
sec x  1  2
1

 1  cos x  1
cos x
Cos 1  cos x   Cos 1 1
x0
47
EX: 2sin   sin   1  0
2

As stated this is difficult to see

a  sin 
Let’s make a simple variable substitution
2a 2  a  1  0
Our new equation is easy; just factor and solve
 2a  1 a  1  0


AP Calculus AB, SJHS 2014-2015

1
a   ,1
2
This solves “a” but we need to solve for our
original variable θ

Let’s replace with our substituted variable
1
sin   
2
sin   1
 
 
7 11
,
6 6

2
 
 7 11
,
2 6
,
6
48
EX: 3cos x  6sin x cos x  0

Consider each trig function to be its own
variable!
a  sin x
b  cos x
 3b 1  2a   0
b  0


1
a  2
x 
 3b  0

1  2a  0
cos x  0


1
sin x  2
  5
, ,
6 2 6


 x  2

 x   , 5

6 6
AP Calculus AB, SJHS 2014-2015
3b  6ab  0
49
SOLVE THE TRIANGLE

This means to solve all parts, typically given only
two parts

In this class, we will focus on right triangles

Given two sides
Solve third side using Pythagorean Theorem
 Solve angles using inverse functions


Given a side and angle
Given right triangles, the last angle is obvious
 Use trig functions to solve another side
 Solve for last side with:

AP Calculus AB, SJHS 2014-2015

No Law of Sines or Cosines!
Pythagorean Theorem (long way)
 Any other trig function (much faster)

50
SOLVE THE TRIANGLE
37
14.3
11.4
53
sin 53 
11.4
r
r
11.4
 14.3
sin 53
We can use Pythagorean Theorem for
the last side, or another trig function
- let’s use cosine
cos53 
x
 x  14.3cos53  8.6
14.3
AP Calculus AB, SJHS 2014-2015
The third angle completes the
triangle:
180 ° – 90 ° – 53 ° = 37°
To solve the sides, choose a trig function
that includes your given side
- here, you are given a y-coordinate
51
SOLVE THE TRIANGLE
54.2°
54.6
31.9
44.3
We can use the Pythagorean Theorem for the last side:
x 2  31.92  54.62
 x  54.62  31.92  44.3
For the angles, choose any trig function and take its
inverse
31.9
tan  
44.3
 31.9 
   Tan 
  35.8
 44.3 
1
AP Calculus AB, SJHS 2014-2015
35.8°
The last angle is simple: 180° - 90° - 35.8° = 54.2 °
52
END
AP Calculus AB, SJHS 2014-2015
53
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