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Trigonometry Chapter 5 Review
Chapter 5
Sine and Cosine Graphs
Periodic Functions:
A periodic function is a function whose graph repeats regularly over some interval of the
domain. The length of this interval is called the period of the function.
What are the basic characteristics of the sine and cosine graphs?
1.
2.
3.
4.
5.
6.
7.
The curve is periodic
The curve is continuous
The domain is real numbers
The range is – 1 ≤ y ≤ 1
The maximum value is 1
The minimum value is -1
The period of the graph is 360º or 2π.
Those are the similarities. What do you notice is different about the graphs? They are
shifted over by how much? 90 degrees. So, that changes their x and y-intercepts.
Sine:
x intercepts every 180º x = 180º n, nε I OR nπ, nε I
y-intercept: 0
Cosine:
x-intercepts start at 90º and occur every 180º, 90º + 180n, n ε I OR

 n , n I
2
y-intercept: 1
For the equation y = asinb(x + c) + d or y = acosb(x + c) + d:
Amplitude: half the distance between the maximum and minimum values of the
function. The basic amplitude of sine and cosine is 1.
amplitude = a 
max  min
2
You can find the amplitude by graphing or by reading the vertical stretch coefficient at
the beginning of the equation.
Period: the length of one cycle.
If y=a sin bx or y=a cos bx:
period =
360
(for degrees)
b
period =
2
(for radians)
b
If the question says transformations, we use the language from Chapter 1. If it says
parameters, we are looking for the new terms
Notice: vertical stretches affect amplitude, and horizontal stretches affect period.
Horizontal phase shift: Horizontal translation is now called phase shift.
Horizontal phase shift = c
*to the right if c > 0 (x – c)
*to the left if c < 0 (x + c)
Vertical Displacement: Vertical translation is called vertical shift , vertical
displacement or median.
Vertical displacement = d
*up if d > 0
*down if d < 0
To find vertical displacement, you can either draw a line on the graph that divides it
vertically in half….whatever y value that line occurs at, that’s your vertical displacement
OR ou can also use the formula:
d
Max  min
2
One last useful piece of information: the a and d values affect the range since they are
vertical parameters, so if you have an equation, the following are useful formulas:


Max: d + a
Min: d – a
Tangent function:




Period is π or 180º
there is no amplitude (or max/min)
Range is yεR
𝜋
Asymptotes at 𝑥 = 2 + 𝑛𝜋

Domain: 𝑥 ≠ 2 + 𝑛𝜋, 𝑥𝜖𝑅


x-intercepts: at x=nπ
y-intercept at 0
𝜋
According to the curriculum, you are not supposed to transform the tangent function!
The value of the tangent of an angle θ is the slope of the line passing through the origin
and the point on the unit circle (cosθ, sinθ). You can think of it as a slope of the
terminal arm of angle θ in standard position. Since y = sinθ and x = cosθ, we can use
the following formula:
𝑠𝑖𝑛𝜃
𝑡𝑎𝑛𝜃 =
𝑐𝑜𝑠𝜃
There are many real-life situations that are modeled by trigonometry. These are called
sinusoidal functions.
Sinusoidal function: A function whose graph resembles the sine or cosine curve.
NOTE….all application problems need to be done in radians.
In Chapter 4, we learned how to solve equations algebraically. Remember that you can
also solve equations graphically.
i.e. Solve 2 cos 2 x  1  0 . Write your answer in general form.
To solve using the graphing calculator, either put everything on one side into y1 and find
the x-intercepts or put one side of the equation into y1 and the other side into y2 and
find the intersection points.
We will use the first method to solve. In your calculator, enter 2 cos 2 x  1 into y1. If we
are doing general form, we start by finding the answers in the first rotation and then add
the period.
Using the 2nd trace #2 function, we find that the zeros are:
 3 5 7
, , ,
. How do we
4 4 4 4
get these answers from the calculator in exact value? After you solved for an answer
using the y1 method, go into the normal calculator screen and type in Ans/π. That will
give you a decimal that you can Math/Frac and convert into exact value. For example, if
I would have solved the first answer on the calculator, I would have got 0.785….. Then
I divide that by π. That gives me 0.25. That converts to ¼, so my answer is π/4. You
can do that for each angle if the question asks for radians in exact value.
Now, for the general solutions of the graph. The period of the graph is π. So the
general solution would be found by writing the first two angles plus the period.

3
 n , n I and
 n , n I . However, this is one time when that is actually
4
4

redundant. If you take a look at the equations, they are actually
apart. So we would
2
 n
, n I . General solutions can be tricky.
write instead: 
4 2
Problem Questions:
There are a few different types of problem questions.
1. You may be given an equation and asked questions about it. You need to be quite
familiar with your parameters and graphing calculator to solve these.
2. You may be given information (usually a graph) and then asked to make an
equation.
3. You may be asked to graph points and then come up with an equation.
For all application questions, you need to have your calculator in radians. DO NOT
FORGET!!!! Many of the questions can be solved with your graphing calculator so here
are some tips that may be helpful:
1. First, put your calculator in radians and then press Zoom Trig.
2. Enter the equation into y1.
3. You may need to adjust your window. To find out what the x-values should be, you
should figure out the period of the question and then put in at least one cycle. To
find out what the y-values should be, determine the max and min. by using d – a and
d + a.
4. Once your window is correct, your graph should show up. If you are given an xvalue and asked to find a y-value, you can use 2nd trace #1 (value). If you are given
a y-value and asked to find an x-value, put the y-value in y2 and then do 2nd trace #5
(intersect).
5. If you are asked to graph points, make sure that you enter them in Stat-Edit and also
turn Plot1 on and adjust your window based on those points.