This document provides an overview of trigonometry topics for preparing for the SAT II exam. It begins by explaining that trigonometry originates in right triangles but can be applied more broadly using trigonometric functions. The document then outlines key trigonometry topics like radian measure, the unit circle, trigonometric functions and their graphs, identities, and solving trigonometric equations. Examples are provided for each topic to demonstrate concepts like determining amplitude, frequency, and phase shift from equation definitions of trig functions.

1. Trigonometry
Preparing for the SAT II

2. ©CarolynC.Wheater,2000
2
Trigonometry
Trigonometry begins in the rightTrigonometry begins in the right
triangle, but it doesn’t have to betriangle, but it doesn’t have to be
restricted to triangles. Therestricted to triangles. The
trigonometric functions carry thetrigonometric functions carry the
ideas of triangle trigonometry into aideas of triangle trigonometry into a
broader world of real-valuedbroader world of real-valued
functions and wave forms.functions and wave forms.

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Trigonometry Topics
Radian Measure
The Unit Circle
Trigonometric Functions
Larger Angles
Graphs of the Trig Functions
Trigonometric Identities
Solving Trig Equations

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Radian Measure
To talk about trigonometric functions, it is
helpful to move to a different system of
angle measure, called radian measure.
A radian is the measure of a central angle
whose intercepted arc is equal in length to
the radius of the circle.

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Radian Measure
degrees
360
radians

=
2π
There are 2π radians in a full rotation --
once around the circle
There are 360° in a full rotation
To convert from degrees to radians or
radians to degrees, use the proportion

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Sample Problems
Find the degree
measure equivalent
of radians.
degrees
360
radians
210
360
r



=
=
=
= =
2
2
360 420
420
360
7
6
π
π
π
π π
r
r
degrees
360
radians
360
3 4



=
=
=
=
2
2
2 270
135
π
π
π
π π
d
d
d
3
4
π
Find the radian
measure equivalent
of 210°

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The Unit Circle
Imagine a circle on the
coordinate plane, with its
center at the origin, and
a radius of 1.
Choose a point on the
circle somewhere in
quadrant I.

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The Unit Circle
Connect the origin to the
point, and from that point
drop a perpendicular to
the x-axis.
This creates a right
triangle with hypotenuse
of 1.

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The Unit Circle
sin( )θ = =
y
y
1
cos θbg= =
x
x
1
x
y
1
θ is the
angle of
rotation
The length of its legs are
the x- and y-coordinates of
the chosen point.
Applying the definitions of
the trigonometric ratios to
this triangle gives

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The Unit Circle
sin( )θ = =
y
y
1
cos θbg= =
x
x
1
The coordinates of the chosen point are the
cosine and sine of the angle θ.
 This provides a way to define functions sin(θ)
and cos(θ) for all real numbers θ.
 The other trigonometric functions can be
defined from these.

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Trigonometric Functions
sin( )θ = y
cos θbg= x
tan θbg=
y
x
csc θbg=
1
y
sec θbg=
1
x
cot θbg=
x
y
x
y
1
θ is the
angle of
rotation

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Around the Circle
As that point
moves around the
unit circle into
quadrants II, III,
and IV, the new
definitions of the
trigonometric
functions still
hold.

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Reference Angles
The angles whose terminal sides fall in
quadrants II, III, and IV will have values of
sine, cosine and other trig functions which
are identical (except for sign) to the values
of angles in quadrant I.
The acute angle which produces the same
values is called the reference angle.

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Reference Angles
The reference angle is the angle between
the terminal side and the nearest arm of the
x-axis.
The reference angle is the angle, with vertex
at the origin, in the right triangle created by
dropping a perpendicular from the point on
the unit circle to the x-axis.

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Quadrant II
Original angle
Reference angle
For an angle, θ, in
quadrant II, the
reference angle is
π−θ
In quadrant II,
 sin(θ) is positive
 cos(θ) is negative
 tan(θ) is negative

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Quadrant III
Original angle
Reference angle
For an angle, θ, in
quadrant III, the
reference angle is
θ-π
In quadrant III,
 sin(θ) is negative
 cos(θ) is negative
 tan(θ) is positive

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Quadrant IV
Original angle
Reference angle
For an angle, θ, in
quadrant IV, the
reference angle is
2π−θ
In quadrant IV,
 sin(θ) is negative
 cos(θ) is positive
 tan(θ) is negative

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All Star Trig Class
 Use the phrase “All Star Trig Class” to
remember the signs of the trig functions in
different quadrants.
AllStar
Trig Class
All functions
are positive
Sine is positive
Tan is positive Cos is positive

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Sine
 The most fundamental sine wave, y=sin(x),
has the graph shown.
 It fluctuates from 0 to a high of 1, down to –1,
and back to 0, in a space of 2π.
Graphs of the Trig Functions

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The graph of is
determined by four numbers, a, b, h, and k.
 The amplitude, a, tells the height of each peak
and the depth of each trough.
 The frequency, b, tells the number of full wave
patterns that are completed in a space of 2π.
 The period of the function is
 The two remaining numbers, h and k, tell the
translation of the wave from the origin.
Graphs of the Trig Functions
y a b x h k= − +sin b gc h
2π
b

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Sample Problem
Which of the following
equations best describes
the graph shown?
 (A) y = 3sin(2x) - 1
 (B) y = 2sin(4x)
 (C) y = 2sin(2x) - 1
 (D) y = 4sin(2x) - 1
 (E) y = 3sin(4x)
−2π −1π 1π 2π
5
4
3
2
1
−1
−2
−3
−4
−5

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Sample Problem
Find the baseline between
the high and low points.
 Graph is translated -1
vertically.
Find height of each peak.
 Amplitude is 3
Count number of waves in
2π
 Frequency is 2
−2π −1π 1π 2π
5
4
3
2
1
−1
−2
−3
−4
−5
y = 3sin(2x) - 1

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Cosine
 The graph of y=cos(x) resembles the graph of
y=sin(x) but is shifted, or translated, units to
the left.
 It fluctuates from 1
to 0, down to –1,
back to 0 and up to
1, in a space of 2π.
Graphs of the Trig Functions
π
2

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Graphs of the Trig Functions
y a b x h k= − +cos b gc h
Amplitude a Height of each peak
Frequency b Number of full wave patterns
Period 2π/b Space required to complete wave
Translation h, k Horizontal and vertical shift
The values of a, b, h, and k change the
shape and location of the wave as for the
sine.

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 Which of the following
equations best describes
the graph?
 (A) y = 3cos(5x) + 4
 (B) y = 3cos(4x) + 5
 (C) y = 4cos(3x) + 5
 (D) y = 5cos(3x) +4
 (E) y = 5sin(4x) +3
Sample Problem
−2π −1π 1π 2π
8
6
4
2

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 Find the baseline
 Vertical translation + 4
 Find the height of
peak
 Amplitude = 5
 Number of waves in
2π
 Frequency =3
Sample Problem
−2π −1π 1π 2π
8
6
4
2
y = 5cos(3x) + 4

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Tangent
 The tangent function has a
discontinuous graph,
repeating in a period of π.
Cotangent
 Like the tangent, cotangent is
discontinuous.
• Discontinuities of the
cotangent are units left of
those for tangent.
Graphs of the Trig Functions
π
2

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Graphs of the Trig Functions
y=sec(x)
Secant and Cosecant
 The secant and cosecant functions are the
reciprocals of the cosine and sine functions
respectively.
 Imagine each graph is balancing on the peaks and
troughs of its reciprocal function.

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Trigonometric Identities
An identity is an equation which is true for
all values of the variable.
There are many trig identities that are useful
in changing the appearance of an
expression.
The most important ones should be
committed to memory.

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Trigonometric Identities
Reciprocal Identities
sin
csc
x
x
=
1
cos
sec
x
x
=
1
tan
cot
x
x
=
1
tan
sin
cos
x
x
x
=
cot
cos
sin
x
x
x
=
Quotient Identities

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Cofunction Identities
 The function of an angle = the cofunction of its
complement.
Trigonometric Identities
sin cos( )x x= −90
sec csc( )x x= −90
tan cot( )x x= −90

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Trigonometric Identities
sin cos2 2
1x x+ =
1 2 2
+ =cot cscx x
tan sec2 2
1x x+ =
Pythagorean Identities
 The fundamental
Pythagorean identity
 Divide the first by sin2
x
 Divide the first by cos2
x

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Solving Trig Equations
Solve trigonometric equations by following
these steps:
 If there is more than one trig function, use
identities to simplify
 Let a variable represent the remaining function
 Solve the equation for this new variable
 Reinsert the trig function
 Determine the argument which will produce the
desired value

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Solving Trig Equations
To solving trig equations:
 Use identities to simplify
 Let variable = trig function
 Solve for new variable
 Reinsert the trig function
 Determine the argument

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Solve
 Use the Pythagorean
identity
• (cos2
x = 1 - sin2
x)
 Distribute
 Combine like terms
 Order terms
Sample Problem
3 3 2 0
3 3 2 1 0
3 3 2 2 0
1 3 2 0
2 3 1 0
2
2
2
2
2
− − =
− − − =
− − + =
− + =
− + =
sin cos
sin sin
sin sin
sin sin
sin sin
x x
x x
x x
x x
x x
c h
3 3 2 02
− − =sin cosx x

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 Let t = sin x
 Factor and solve.
Sample Problem
Solve 3 3 2 02
− − =sin cosx x
2 3 1 02
sin sinx x− + =
2 3 1 0
2 1 1 0
2 1 0 1 0
2 1 1
1
2
2
t t
t t
t t
t t
t
− + =
− − =
− = − =
= =
=
( )( )

37. ©CarolynC.Wheater,2000
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Sample Problem
Solve 3 3 2 02
− − =sin cosx x
x =
π π
6
5
6
or
x =
π
2
x =
π π π
6
5
6 2
, ,
 Replace t = sin x.
 t = sin(x) = ½ when
 t = sin(x) = 1 when
 So the solutions are