Download The Unit Circle

Trigonometry and the Unit Circle
Lesson One: Angles and Angle Measures
Specific Outcome 1: Demonstrate an understanding of angles in standard position, expressed in degrees and
radians.
Look at the explanation of Trigonometry on p. 162. Show the students how trig has real
life examples.
To start our lesson, consider a circle that has a centre C, and an angle PCA that has an
arc length and radius of the same length. If you were to measure the angle, it would be
60 ˚.
P
C
A
About how many radius lengths are there in one complete circumference of your circle?
Should be around 6 lengths (because we know there are approximately 360˚ in a
circle.
Now, the circumference, C, of a circle is given by the formula C = 2πr where r
represents the radius. Explain how this formula relates to your answer.
For example, take the circle with radius of 5. It has 6 equal radius lengths for its
circumference, therefore the circumference is equal to 6 × 5 = 30𝑐𝑚. If you solve
for the circumference using the formula, you get 𝐶 = 2𝜋(5) = 31.42. Pretty
close!
What does the circumference formula find for us?
Circumference finds the distance around a circular object.
This actually introduces an important concept….the concept of radians. The measure
of ∠PCA is considered to be 1 radian.
So what is a radian?
Radian: the measure of the central angle (∠PCA) of a circle subtended by an arc that is
the same length as the radius of the circle. In other words, you get a radian when the
arc length and radius are the same length!
P
C
θ
A
∠θ = 1
What degree benchmark can we use for 1 radian?
1
Since 1 radian is about 6 𝑡ℎ 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒, we say it’s about equal to 60˚ (the actual
measurement is 57.3˚ which we’ll see later)
So, again, one radian is the measure of the angle at the centre of a circle subtended by
a
an arc equal in length to the radius of the circle. This brings us the formula:   .
r
We can measure angles, therefore, in degrees and radians.
One full rotation is 360˚ or 2π radians because 𝜃 =
𝜋
or π radians. One quarter rotation is 90˚ or
radians. So, one radian is equal to:
180
𝜋
2
2𝜋𝑟
𝑟
= 2𝜋. One half rotation is 180˚
radians. One eight rotation is 45˚ or
𝜋
4
which is approximately 57.3 degrees!
Angle measures without units are considered to be in radians.
Conversion Chart
Degrees to Radians multiply by

180
Radians to Degrees multiply by
180

You will need to be able to convert back and forth from degrees to radians and vice
versa. We will also be looking at putting angles on coordinate planes.
Do you remember what an angle in standard position is from Math 20-1? It has its
centre at the origin and its initial arm along the positive x-axis. The terminal arm then
rotates about the origin. Angles in a counterclockwise direction are positive and angles
in a clockwise direction are negative.
Ex #1: Draw each angle in standard position. Change each degree measure to radian
measure and each radian measure to degree measure.
a. 30˚

  
 30 
 rad  rad
6
 180 
b. -210˚
7
  
 210 



6
 180 
2
c.
2 180 
2
rad 
 120
3
3
2
3
d. -1.2
1.2rad 
1.2 180 

 68.75
Coterminal Angles
Sketch an angle of 60˚ and 420˚ on the same grid. What do you notice?
The terminal arms coincide so these are coterminal angles. 420 – 60 = 360, which
makes sense because a full circle is 360. Therefore, if you want a coterminal angle,
you can either add or subtract 360 to find another coterminal angle. If the angle is in
radians, one full rotation is 2  so we will add or subtract 2  to get coterminal angles.
The principal angle is the angle in the group that is between 0 and 360 or 0 and 2  .
Ex #2: For each angle in standard position, determine one positive and one negative
angle measure that is coterminal with it.
a. 270˚
270 + 360 = 630
270-360 = -90
b. −
5𝜋
4
5𝜋
5𝜋 8𝜋 3𝜋
+ 2𝜋 = −
+
=
4
4
4
4
5𝜋
5𝜋 8𝜋
13𝜋
−
− 2𝜋 = −
−
=−
4
4
4
4
−
c. 740˚
Need to find principle angle (between 0 and 360)
740-360=380-360=20˚
20 + 360 = 380˚
20 – 360 = -340˚
You may have noticed that there doesn’t seem to be an end….you could keep rotating
around to infinity.
3
General Form:
By adding or subtracting multiples of one full rotation, you can write an infinite number
of angles that are coterminal with any given angle.
To find the general form of an angle of 40˚, you will basically add multiples of 360.
How can we write this as a general term?
40˚±(360˚)n, n ε N
If we wanted to find a general term for the coterminal angles of
it?
2𝜋
± 2𝜋𝑛
3
2𝜋
3
, how would we write
Any given angle has an infinite number of angles that are coterminal, since each time
you make one full rotation from the terminal arm, you arrive back at the same terminal
arm. Angles coterminal with any angle θ can be described using the expression:
𝜃 ± (360°)𝑛 𝑜𝑟 𝜃 ± 2𝜋𝑛,
Where n is a natural number. This way of expressing an answer is called the general
form.
Ex #3: Write an expression for all possible angles coterminal with each given angle.
Identify the angles that are coterminal that satisfy −720° ≤ 𝜃 ≤ 720°, 𝑜𝑟 − 4𝜋 ≤ 𝜃 ≤ 4𝜋
a. -500˚
-500+360=-140
-140 + 360 = 220
220 + 360 = 580
−500° ± 360𝑛
b. 650˚
650 – 360 = 290
290 – 360 = -70
-70 – 360 = -430
650° ± 360𝑛
4
c.
9𝜋
4
9𝜋
9𝜋 8𝜋 𝜋
− 2𝜋 =
−
=
4
4
4
4
𝜋 8𝜋
7𝜋
−
=−
4
4
4
7𝜋 8𝜋
15𝜋
−
−
=−
4
4
4
Remember when we were talking about radians, we discovered that one radian is when
the arc length and radius are the same length. This gives us a formula and we earlier
a
saw it in this form:   . We can rearrange it to solve for a.
r
Arc Length:
a  R
where a= arc length
R = radius
 = angle in radians
Ex #4: A Pendulum 30 cm long swings through an arc of 45 cm. Through what angle
does the pendulum swing? Answer in degrees and in radians.
a  R
a = 45 cm
  1.5 
45 = 30 
 = 1.5 radians
r = 30 cm
180

 85.9
Ex #5: Calculate the arc length of a sector of a circle with diameter 9.2m if the sector
angle is 150 .
a  R
  150  150 

180

5
rad
6
 5 
 4.6  
 6 
 12.0m
1
1
d   9.2   4.6m
2
2
Assignment: pg. 175-179 #1ad, 2ad, 3ef, 5ace, 6ad, 7ad, 8ac, 9ab, 10, 11aeg, 12bc,
13d, 15, 18, 21, 27
r
5
Lesson Two: The Unit Circle and Lesson Three: Trigonometric ratios Part 1
Specific Outcome 2: Develop and apply the equation of the unit circle.
Specific Outcome 3: Solve problems, using the six trigonometric ratios for angles expressed in radians and
degrees.
A unit circle is a circle with radius 1 unit:
y (0,1)
(-1,0)
(1,0)
0
x
(0,-1)
Then you want them to be able to draw a triangle on the unit circle and find the equation
of the unit circle using the Pythagorean Theorem. Consider a point P on the unit circle.
Let P have coordinates (x, y).
y
OP = 1
PA = |𝑦|
OA = |𝑥|
(𝑂𝑃)² = (𝑂𝐴)2 + (𝑃𝐴)²
1² = 𝑥² + 𝑦²
1 = 𝑥² + 𝑦²
P(x, y)
1
0
A
x
The equation of the unit circle is 𝑥² + 𝑦² = 1
Use this information to show them that this formula works for any circle on the origin.
Ex. Determine the equation of a circle with centre at the origin and radius 6.
𝑥² + 𝑦² = 36
Have them use the formula s 𝑥 2 + 𝑦 2 = 1 to see if they can answer the following
questions:
6
Ex. Determine the missing coordinates(s) for all points on the unit circle that satisfy the
conditions given. Draw a diagram in each case.
2
a. the x-coordinate is 3
2
(3 , 𝑦)
Use 𝑥² + 𝑦² = 1
2 2
(3) + 𝑦² = 1
4
𝑦2 = 1 − 9
√5
3
𝑦=±
Why are there two answers?
The x is positive, which means the y-value could be in quadrant 1 and IV
(positive or negative)
2 √5
)
3
Two points satisfy the conditions: (3 ,
2
And (3 , −
√5
)
3
𝑖𝑛 𝑄𝑢𝑎𝑑𝑟𝑎𝑛𝑡 1
𝑖𝑛 𝑄𝑢𝑎𝑑𝑟𝑎𝑛𝑡 𝐼𝑉
We will now make the unit circle with all the angles on it. (using paper plate and
markers)
Before we can complete the unit circle, we need to remind ourselves of information from
Math 20-1: Special Triangles from Math 20-1. We are going to use these to help in our
unit circle. Instead of just drawing the triangles, remind the students how they are built
from the right isosceles triangle and equilateral triangle.
Also, review SOHCAHTOA with the students.
Then, go through and list out the special angles of 30, 45, 60 as well as radians.
sin 60˚ =
√3
sin 30˚ =
1
2
2
cos 60˚ =
cos 30˚ =
1
2
√3
2
tan 60˚ =
tan 30˚ =
√3
1
1
√3
7
Also, review with the students the fact that the radical could be either in the numerator
or denominator. Review the fact that “exact value” means fraction answer only….no
decimals allowed. Show the students the MODE button on the calculator and remind
them of degree vs. radian. Also, compare the decimal answer to the fraction answer
and focus on why you can’t math frac the decimal answer.
See if you can get the students to also think of SOHCAHTOA in terms of x, y, r rather
than opp, adj, hyp. This will help us with the next investigation. Once they have it
figured out, write the formulas down on the board.
Express the sin, cos and tan ratios in terms of x, y and r:
Once the triangles have been recalled, redraw a unit circle that looks like the circle from
the beginning of the lesson. See if you can get the students to identify the sin, cos and
tan for 0, 90, etc. Draw the points (1, 0), (0, 1), etc….
Once we have gone through 0, 90, 180…..have the students see that all the answers
are 0, 1, -1 and undefined….therefore the calculator can be used instead of
memorization. “If the question is on the arm of the axes, use your arm to punch the
calculator”
Recall:
Suppose θ is any angle in standard position, and P(x, y) is any point on its terminal arm,
at a distance r from the origin.
If given the values (x, y) of the two smaller legs of a right
angled triangle, what formula would you use to find the r
value?
Pythagorean Theorem, 𝑟 = √𝑥 2 + 𝑦 2
Have the students come up with….what happens on a unit circle?
With a unit circle with radius of 1:
𝑥
cos 𝜃 = = 𝑥 , 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒
1
𝑦
sin 𝜃 = = 𝑦 , 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒
1
8
Therefore, you can describe the coordinates of any point 𝑃(𝜃) as (cos 𝜃, sin 𝜃). This is
true for any point 𝑃(𝜃) at the intersection of the terminal arm of an angle θ and the unit
circle. Make a note of this on your paper plate unit circle.
So, using our special triangles, put in the points for 0 to 90 so far.
Now, have them use their calculators to investigate:
sin40º = 0.64
sin140º = 0.64
sin220º = -0.64
sin320º = -0.64
Draw these angles on the board and see if the students can come up with how these
are related. This will bring up the concept of reference angles….
Reference Angles: the acute angle whose vertex is the origin and whose arms are the
terminal arm of the angle and the x-axis.
Have them figure out the quadrant rules and write them down for both radians and
degrees.
QI
reference angle
QII
π – r.a. or 180° - r.a.
QIII
π + r.a. or 180° + r.a.
QIV
2π – r.a. or 360° - r.a.
Once the quadrant rules are determined, go through the CAST rules. Have the
students figure out where the CAST rule comes from.
CAST Rule:
Sine ratio
positive
All ratios
positive
Tangent
ratio
positive
Cosine ratio
positive
Once we have determined the CAST rule, figure out how these work on the unit circle.
Then, have the students fill out the rest of the unit circle.
9
The unit circle will be key in diploma questions. Make sure that you know how it works.
Any tricks for memorizing it? Also, it is very important that you know it in both radians
and degrees.
The Unit Circle:
Assignment: pg. 186-190 #1 – 2 (1st and last letter), 3ae, 4abij, 5acgi, 6, 9 –13, 16, 18,
C2, C3
10
Reciprocal Trigonometric Ratios:
Each primary trig function has its corresponding reciprocal function.
cosecant ratio  csc 
1
sin 
secant ratio  sec 
cotangent ratio  cot  
1
cos
1
tan 
Ex. The point (15, 8) lies on the terminal arm of  as shown. Calculate the
value of r and the exact values of the primary and reciprocal ratios.
r x y
2
2
2
r  152  82
r  17
8
17
15
cos  
17
8
tan  
15
sin  
17
8
17
sec 
15
15
cot  
8
csc 
1 2√2
Ex. The point B(− 3 , 3 ) lies at the intersection of the unit circle and the terminal arm
of an angle θ in standard position.
a. In what quadrant does point B lie?
Quadrant 2 (how do you know this?)
b. Determine the values of the six trigonometric ratios for θ. Express your
answers in lowest terms.
1
cos 𝜃 = − , sec 𝜃 = −3
3
2√2
3
3√2
𝑠𝑖𝑛𝜃 =
, csc 𝜃 =
𝑜𝑟
3
4
2√2
2√2
𝑦
2√2 −3
6√2
tan 𝜃 = = 3 =
×
=−
= −2√2
1
𝑥
3
1
3
−3
1
√2
cot 𝜃 =
=−
4
−2√2
**you can have the radical on either the numerator or denominator.
11
Exact values for the trigonometric ratios can be determined using special triangles or
the unit circle. With the knowledge of the special triangles, the CAST rule can be used
to tell us where each trigonometric ratio is positive.
The trigonometric ratios of any angle can be written as the same function of a positive
acute angle called the reference angle with the sign of the ratio being determined by the
CAST rule. Do the following examples using both special triangles and unit circle.
Ex. Determine the exact value for each.
a. sin 225
**if you know the unit circle, you can just read the answer off the circle.
**if you’re using the special triangles, use the following steps:
1. What quadrant is the angle in? 3
2. What is the reference angle?  225 180  45
1
2
4. Use the CAST rule to determine the pos/neg sign: it will be negative
5. Answer:
rotation angle of 225 , has a reference angle of  225 180  45
3. Find the sin45 from the special triangles. sin 45 
from special triangles sin 45 
sin 225  
b. tan
5
3
1
. In QIII the sine ratio is negative so
2
1
2

 
  from special triangles  3 .
3
 3
5
 3
In Quadrant 4 cos is positive so cos
3
5

reference angle of  2 
3

c. 𝑐𝑜𝑠
𝜋
2
Use your calculator!
d. sec 30°
It’s primary ratio is cos, find cos30˚. Cos30 =
sec30 =
2
3
12
3
, therefore
2
e. sin(-300˚)
√3
Terminates in Quadrant 1, reference angle of 60˚. Sin60˚= 2 . The cast
rule states that sin is positive in quadrant one so, sin(−300°) =
Assignment: pg. 201-205 #1 (odd letters), 3 – 6 (odd letters), 8
13
√3
2
Lesson 4: Trig Ratios Part 2
Specific Outcome 3: Solve problems, using the six trigonometric ratios for angles expressed in radians and
degrees.
You can determine approximate values for sine, cosine, and tangent using a calculator.
Not all answers will have fraction form. Most calculators can determine trigonometric
values for angles measured in degrees or radians. You will need to set the mode to the
correct angle measure. On the graphing calculator, it is under MODE.
Check:
Cos 60˚=0.5 (degree mode)
Cos 60 = -0.952412980 (radian mode) (To use radians to comply with cast rule
the 60 degrees needs to be converted to radians, use cosine π/3 = 0.5)
Your calculator can also compute trigonometric ratios for negative angles. However,
you should use your knowledge of reference angles and the CAST rule to check that
your calculator display is reasonable.
Cos (-200˚) = -0.939692620…
Why is this value negative? In what quadrant does the angle terminate?
In Quadrant II, cosine is negative.
You can find the value of trigonometric ratio for cosecant, secant, or cotangent using the
correct reciprocal relationship:
1
sec 3.3 =
= −1.0127
cos 3.3
Ex. What is the approximate value for each trigonometric ratio? Round your answers
to four decimal places. Be sure to state the quadrant the angle terminates in and why
the value is positive or negative.
a. sin 1.92
In radians mode: 0.9396
b. tan (-500˚)
500-360 = 140 (in quadrant 3, recall the negative sign means direction)
In degree mode: 0.8391, in quadrant 3 (tan positive)
14
c. sec 85.4˚
1
= cos 85.4 = 12.5, in quadrant 1, all are positive.
d. 𝑐𝑜𝑠
7𝜋
5
In radians mode: -0.31, Quadrant 3, cos is negative.
How can you find the measure of an angle when the value of the trigonometric ratio is
given?
Use inverse trig function 𝑠𝑖𝑛−1
If sin θ = 0.5, what is the measure of its angle?
𝑠𝑖𝑛−1 (0.5) = 30˚
0° < 𝜃 < 360°
1
Note: 𝑠𝑖𝑛−1 ≠ 𝑠𝑖𝑛𝜃
The inverse calculator keys return one answer only, when there are often two angles
with the same trigonometric function value in any full rotation. In general, it is best to
use the reference angle applied to the appropriate quadrants containing the terminal
arm of the angle. So the second angle is in QII 180 – 30 = 150°
Remember: Normal trig button when you have the angle, second trig button when you
want the angle.
Ex. Determine the measures of all angles that satisfy each of the following.
a. cos θ = 0.843 in the domain 0° ≤ 𝜃 < 360°. Give approximate answers to the
nearest tenth.
𝑐𝑜𝑠 −1 (0.843) = 32.5°, cos is positive in quadrant 1 and IV. Given the domain, there will
two answers, 32.5˚ in QI, and 327.5˚ in QIV.
b. sin θ = 0 in the domain 0° ≤ 𝜃 ≤ 180°. Give exact answers.
In Quadrant 1: 𝑠𝑖𝑛−1 (0) = 0˚. For quadrant II, 180 – 0 = 180˚
c. cot θ = -2.777 in the domain −𝜋 ≤ 𝜃 ≤ 𝜋. Give approximate answers.
1
1
𝑡𝑎𝑛𝜃 = −2.777, to find the angle 𝑡𝑎𝑛−1 (−2.777) = −0.35. Tan is negative in
QII and IV.
15
Use the reference angle 0.35 (the negative means direction in QIV) and
apply to the various quadrants.
For 180 to 0: answer in QII is π – 0.35 = 2.79
For 0 to -180: answer in QIV = -0.35 (Recall direction of rotation)
d. 𝑠𝑒𝑐𝜃 =
2
√3
𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 − 2𝜋 ≤ 𝜃 ≤ 2𝜋. Give exact answers.
Secant is positive in quadrants I and IV. The domain includes four
quadrants in both the positive and negative directions. There are two
positive and two negative answers. 𝐶𝑜𝑠𝜃 =
In the domain 0 - 2π, the answers are
𝜋
6
√3
2
𝑎𝑛𝑑
𝜋
𝜋
= 6.
11𝜋
6
.
In the domain 0 – (-2π), the answers are − 6 𝑎𝑛𝑑 −
11𝜋
6
.
Ex. The point (-4, 3) lies on the terminal arm of an angle A in standard position. What
is the exact value of each trig ratio for A?
Draw the angle in standard position and draw the triangle to the x-axis. Then use
pythagorus to find the hypotenuse and write the ratios.
Assignment: pg. 202-205 #2acjl, 7, 9 – 12 (odd letters), 13 – 17, 18ab, 19bc, 21, C1,
C2
16
Lesson Five: Trig Equations
Specific Outcome 5: Solve, algebraically and graphically, first and second degree trigonometric equations with the
domain expressed in degrees and radians.
**good idea to show them how to draw the unit circle quickly since they will need
to draw it for their tests if they are using it.
Investigate:
1
1. What are the exact (fraction answers only) measures of θ if 𝑐𝑜𝑠𝜃 = − 2 , 0 ≤ 𝜃 < 2𝜋?
How is the equation related to 2 cos 𝜃 + 1 = 0?
𝜋
Answers will be in Quadrant II and III. 𝐶𝑜𝑠 −1 (0.5) = 60° 𝑜𝑟 3 .
𝜋
In quadrant II: 𝜋 − 3 =
𝜋
2𝜋
3
4𝜋
In Quadrant IV: 𝜋 + 3 =
3
The equations are related because the terms in equation 1 have been brought
over to one side.
2. What is the answer for step 1 if the domain is given as 0° ≤ 𝜃 < 360°?
120˚, 240˚
3. What are the approximate measures for θ if 3 cos 𝜃 + 1 = 0 and the domain is
0 ≤ 𝜃 < 2𝜋?
1
1
cos 𝜃 = − 3, Cosine negative in II and III. 𝐶𝑜𝑠 −1 (3) in radians mode is: 1.23
In Quadrant II: π-1.23 = 1.91
In Quadrant III: π + 1.23= 4.37
(if it isn’t special triangle or unit circle, you should use calculator)
17
4. Complete the chart below. In the left column, show the steps you would use to solve
the quadratic equation 𝑥 2 − 𝑥 = 0. In the right column, show similar steps that will lead
to the solution of the trigonometric equation cos 𝜃² − cos 𝜃, 0 ≤ 𝜃 < 2𝜋.
Quadratic Equation
𝑥 2 − 𝑥 = 𝑥(𝑥 − 1), 𝑥 = 0, 1
Trigonometric Equation
cos 𝜃² − cos 𝜃 = 𝑐𝑜𝑠𝜃(𝑐𝑜𝑠𝜃 − 1), 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 cos 𝜃
= 0 𝑎𝑛𝑑 cos 𝜃 = 1
5. Identify similarities and differences between solving a quadratic equation and solving
a trigonometric equation that is quadratic.
6. When solving a trigonometric equation, how do you know whether to give your
answers in degrees or radians?
The notation [0, 𝜋] represents the interval from 0 to π inclusive and is another way of
writing 0 ≤ 𝜃 ≤ 𝜋. (make sure they understand interval notation)
Ex. Solve each trigonometric equation in the specified domain.
a. 5 sin 𝜃 + 2 = 1 + 3𝑠𝑖𝑛𝜃 , 0 ≤ 𝜃 < 2𝜋 (for first degree equations, get the trig
function on one side and the numbers on the other)
1
Simplify expression : 2 sin 𝜃 = −1, 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒: sin 𝜃 = − 2
𝜋
𝑠𝑖𝑛−1 (0.5) = 30° 𝑜𝑟
6
Sin is negative in quadrants III and IV
𝜋
Quad III: 𝜋 + 6 =
𝜋
7𝜋
6
Quad IV: 2𝜋 − 6 =
11𝜋
6
b. 4 sec x + 8 = 0, 0 ≤ 𝜃 < 360˚
1
Simplify: sec 𝑥 = −2, to solve cos 𝑥 = − 2
𝑐𝑜𝑠 −1 (0.5) = 60°
Cos is negative in QII: 120˚
And QIII: 240˚
18
Ex. Solve for θ:
a. tan²θ – 5tanθ + 4 = 0, 0 ≤ 𝜃 < 2𝜋 (for second degree equations, move
everything to one side and either factor or use the quadratic formula)
Need to factor: (tanθ – 1)(tanθ – 4)
Need to solve for each factor:
tan 𝜃 = 1, 𝑡𝑎𝑛−1 (1) = 45° 𝑜𝑟
𝜋
4
𝜋 5𝜋
Positive in QI and QIII: ,
4
4
Second factor: tan 𝜃 = 4, 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 𝑚𝑜𝑑𝑒 𝑡𝑎𝑛−1 (4) = 1.33
Postive in QI, and QIII: 1.33 and 4.467 (π + 1.33)
b. 𝑐𝑜𝑠 2 𝜃 − cos 𝜃 − 2 = 0, 0 ≤ 𝜃 < 360˚
Factor: (cosθ – 2)(cosθ + 1)
First factor does not exist, there is no value for cosθ = 2 (does not cross
graph)
Second factor : cos θ = -1
Negative in QII and QIII: 𝑐𝑜𝑠 −1 (1) = 0
In QII: 180 – 0 = 180˚
In QIII: 180 + 0 = 180˚
Only answer 180˚
For each one of the equations above, is there another way to solve each equation other
than algebraically?
Graphically and look for intersections
Recall what a general solution is?
Solutions that satisfy a given equation regardless of the amount of rotations
apply in either the positive or negative direction.
19
Ex. a. Solve 𝑐𝑜𝑠 2 𝑥 − 1 = 0 𝑓𝑜𝑟 0 ≤ 𝜃 < 360˚
𝑐𝑜𝑠𝑥 = ±1, 𝑐𝑜𝑠 −1 (1) = 0
Positive in QI and QIV:
QI: 0 (not 360 because of domain)
QIV: 360 (not part of domain therefore not a solution)
For 𝑐𝑜𝑠 −1 (1) = 0
Negative in QII and QIII
In QII: 180 – 0 = 180˚
In QIII: 180 + 0 = 180˚
Only answer 180
Solutions are 0 and 180˚
b. Determine the general solution for the function where the domain are real
numbers measured in degrees.
If we use coterminal angles for each (add 360 degrees to each):
0: 360, 720, 1080
180: 540, 900, 1260
We notice a pattern, each solution is 180 degrees from each other.
Therefore the general solution is: 𝑥 = 0° + 180°𝑛. We find solutions every 180˚
Assignment: pg. 211-214 #1- 7 (odd letters), 8 – 12, 15, 16, 18, 19, 21, 22, C4
(leave #15 and 19 for the end of chapter 5….come back to it after having taught that
lesson)
20
Chapter Five: Trigonometric Functions and Graphs
Lesson 6: Graphing Sine and Cosine Functions
Specific Outcome 4: Graph and analyze the trigonometric functions sine, cosine and tangent to solve problems.
Many sinusoidal and periodic patterns occur within nature. We will be looking at the
sine, cosine and tangent graphs. We will also be doing transformations….except they
will get their own special names
Investigation: What do you think the sine and cosine graphs look like?


Have the students use their unit circle to come up with a graph for sine and
cosine. Have them draw one rotation for each. Put degrees on the x-axis for
sine and exact value radians on the x-axis for cosine.
What do they notice about the two graphs? Go through and write down some of
their conclusions. Once they notice that these graphs repeat, introduce the idea
of periodic functions.
Show them how to graph on the calculator. Make sure they are in Zoom Trig and also
discuss the issues with radians vs. degrees.
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 to 1
The maximum value is 1
The minimum value is -1
The period of the graph is 360º or 2π.
21
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 180 degrees (go through the fancy definition)
y-intercept: 0
Cosine: x-intercepts start at 90 degrees and go every 180 degrees
y-intercept: 1
What do they notice about the vertical distance of the graph? Make sure they identify
the symmetry about the x-axis. Once this is discussed, go through and give them the
information.
The amplitude of a periodic function is defined as half the distance between the
maximum and minimum values of the function.
amplitude 
max  min
2
What is the basic amplitude of sine and cosine? The basic amplitude is 1.
Let’s investigate what happens when we put a coefficient at the beginning. You
probably have a good idea based on your information from transformations.
Have the students graph y = 2sinx and y = 0.5sinx. Have them come up with a
conclusion as to what happens. Explain that it follows our transformation rules, but now
we call it amplitude!
Is it the same for cosine? Yes, it is. So in conclusion, you can find the amplitude by
graphing or by reading the vertical stretch coefficient at the beginning of the equation.
Let’s investigate what happens when we put a coefficient in front of the angle. Again,
you probably have a good idea based on transformations.
Have the students graph y=cos x and y=cos 2x, for 0    360
What do they notice?
y=cos x : 1 complete cycle in 360
y= cos 2x: Period = 2 complete cycles in 360 , 1 in 180
Now graph y=cos 3x and y=cos0.5x over the same interval. What did you notice?
y=cos 3x: 3 complete cycles in 360 , 1 cycle in 120
y=cos 0.5x: 1 complete cycle in 720 (360/0.5)
22
These are all examples of horizontal stretches. What else changed? The period.
There is a formula for period as well.
In general:
To find the period of a function in the form y=a sin bx or y=a cos bx:
period =
360
(for degrees)
b
period =
2
(for radians)
b
Ex. State the amplitude and period of each:
a. y  2 cos 3 : amplitude = 2, period =
b. y 
360
2
 120or
radians
3
3
1
1
360
2
cos  : amplitude = 0.5, period =
 1080or
 6
1
1
2
3
3
3
What transformations are occurring in the following examples? Watch how the
question is worded. If it actually says transformations, we use our old language. If it
says parameters, we are looking for the new language.
a.
b.
c.
d.
y= 2 sin x : vertical stretch by factor 2.
y=sin 2x: horizontal stretch by factor 1/2.
y=-3sin x: vertical stretch by factor 3, reflection in x-axis.
y=sin (-3x): horizontal stretch by factor 1/3, reflection in y-axis
Notice: vertical stretches affect amplitude, and horizontal stretches affect period.
Ex. List the steps involved in graphing the function y = 3 sin 4x.
Now, of course you can use your calculator to help you. However, it is useful to know
the information without calculator because sometimes the units on the axis are difficult
to know from the graph.
If you were to graph this by hand, you would notice that the amplitude is 3. That means
all the y-values are to be multiplied by 3 (a vertical stretch by 3). You would also notice
that the period of the graph is 360/4 = 90 degrees. The horizontal values are multiplied
by ¼. So take the normal sine graph and adjust the x and y values accordingly.
What would the max and min be?
Max: 3, min: -3
What would the x and y intercepts be?
23


sin x has x-intercepts at 0, 180, 360, divide all by 4, so x-intercepts will be at 0,
45, 90 in one cycle
y-intercept is 0
What would the Domain and range be?


𝑑𝑜𝑚𝑎𝑖𝑛: 𝜃𝜖𝑅
Range: −3 ≤ 𝑦 ≤ 3
How many of the parameters above will change if we compare to the graph of y = cos x
Max/min value: same
Amplitude/period: same
Domain/range: same
y-intercept will now be 1
x-intercepts will change to 90, 270
Assignment: pg. 233-237 #4ad, 5bc, 7d, 8d, 9cd, 10, 11cd, 12a, 14, 15, 18a, 20, 21,
23, C2
24
Lesson 7: Transformations of Sinusoidal Functions
Specific Outcome 4: Graph and analyze the trigonometric functions sine, cosine and tangent to solve problems
Remember transformations?
Ex. Describe how the graph of the functions below compare with y = sinx.
a. y  sin( x  30) : horizontal translation 30 right.
b. y = sinx + 2: vertical translation 2 units up.
c. y  sin  x  60 1 : horizontal translation 60 left and a vertical
translation 1 down.
d. y  45  sin( x  45) : horizontal translation 45 right and a vertical
translation 45 units up.
The rules for sine and cosine follow the same rules for transformations. However, they
have different names, of course.
The horizontal translation is now called phase shift. The vertical translation is called
vertical shift or vertical displacement or median.
There is a formula for vertical displacement. Draw a graph for the students and see if
they can figure out what the vertical displacement formula is. Once they do, also talk
about drawing the horizontal line across that divides the graph in half vertically.
d
Max  min
2
There will be many types of questions. You may be asked info about an equation, you
may be asked to graph or you may be asked to make an equation. Some of this can be
done with your calculator, but some of it you need to know. Unfortunately, none of this
is on your formula sheet….you need to know this stuff.
Another important piece of info is which of the 4 parameters we’ve studied so far affect
the x-values and which ones affect the y-values. Have the students discuss this. See if
they can come up with a formula for max and min.
25
Summary of the Effect of the Parameters a, b, c and d:
For:
y  a sin b  x  c    d
y  a cos b  x  c    d
amplitude = a 
Period =
max  min
2
360
(for degrees)
b
Period =
2
(for radians)
b
Horizontal phase shift = c
 to right if c > 0
 to left if c < 0
Vertical displacement = d

up if d > 0, down if d < 0
max  min
 d
2
 Max: d + a
 Min: d – a
Ex. Write the equations of the following.
a) a cosine function having a horizontal phase shift of 75 right:
y  cos  x  75
b) a sine function having a horizontal phase shift of
vertical displacement 4 units up.
3

y  sin  x 
5


4

26
3
5
radians left, and a
Ex. Name the amplitude, period, phase shift and displacement of each. How does
each transformation affect the table of values?
2
1
 
a. y   cos  x    3
3
4
12 
2
2
(multiply y-values), period =
 8 ( multiply x-values by 4),
1
3
4

horizontal phase shift:
right (add to each x-value), vertical displacement: 3
amplitude:
12
units up


b. y  2sin  3x     4  y  2sin 3  x    4
3

2

amplitude: 2, period =
, phase shift: left , displacement: 4 down
3
3
How would you graph b)? Can you figure out how to do it by hand? Is it necessary?
Talk through the steps with the students and focus on the advantages of understanding
graphing by hand.
Ex. Consider the graph shown, find the sine equation of the graph
Amplitude = 3
Period = 
a = 3, b=
𝜋
Window [0,2𝜋, 4 ] [−4,4,1]
2
y=3 sin 2x
27

2
Ex. Consider the graph shown:
Write the equation represented by the graph if a) a > 0 and b) a < 0
For a > 0: 1st find the median (at y = -1)
a = 3, b = 2 (1 cycle in π), d = -1 (vertical displacement)
𝜋
c = − (phase shift left)
4
𝜋
𝑦 = 3 cos 2 (𝑥 + ) − 1
4
Window 𝑥: [−
3𝜋 5𝜋 𝜋
4
,
4
, ] , 𝑦: [−5, 5, 1]
4
For a < 0: 1st find the median (at y = -1)
a = 3, b = 2 (1 cycle in π), d = -1 (vertical displacement)
𝜋
c = (phase shift left)
4
𝜋
𝑦 = −3 cos 2 (𝑥 − ) − 1
4
Need to visualize a reflection in the x-axis (upside down)
There are many real-life situations that are modeled by trigonometry. These are called
Sinusoidal Functions: A function whose graph resembles the sine or cosine curve is
called sinusoidal. By the way….all application problems need to be done in radians.
Ex. The minimum depth of water in a harbour can be approximated by the function d(t)
= 12 +5 cos 0.5t where 0  t  24
a. Determine the max and min values: Graph the function or use formula.
**from graph max = 17, min = 7
b. Period of the function:
2
 4 hours
0.5
c. Suitable Window to view graph (this can be done from the max/min/period)
x : 0, 24, 2 y : 0, 20,5
d. What is the depth of water, at 2:00 am?
**use value function on calc: x=2  y=14.7
28
e. A ship requires 8.5 m of water in the harbour to dock safely. It is midnight,
when will the boat have to move to prevent grounding?
y2  8.5  intersect = 4.6923876
multiply (0.6923876 by 60) to get time in minutes
4:41 a.m.
f. When can the ship return to the harbour?
same graph , use intersect on way up: 7.873983
multiply (0.873983 by 60) to get time in minutes
7:53 a.m.
Assignment: pg. 250-255 #1ef, 2ef, 3a ii, 5, 6bc, 7c, 8 –10, 11c, 13, 14bc, 15c, 16c,
18, 21, 23, 24, 27cd, C2
29
Lesson 8: The Tangent Function
Specific Outcome 4: Graph and analyze the trigonometric functions sine, cosine and tangent to solve problems.
Have the students use their unit circle to come up with the graph of tangent. See if they
can come up with period, max/min values, the range, domain, vertical asymptotes x and
y intercepts.
Once they think they have it, draw it on the board.




Period is π
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
𝜋
Window: 𝑥 [−2𝜋, 2𝜋, 4 ] 𝑦: [−5,5,1]
Can the students explain why tangent has asymptotes? (use the unit circle)
How did we find tangent using the unit circle in an earlier lesson? We knew that tanA
was y/x so we took the y from sinA = y/r and the x from cosA = x/r to get our answer.
What else do we know about tangent? Draw a triangle in the first quadrant that
identifies x, y and r. See if they can come up with the slope = rise/run formula. Once
they have, go through the following info:
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
When sinθ = 0, what is tan θ?
The value is 0 (x-intercept), sin 180 = 0, x-intercept at 180 degrees or π
When cosθ = 0, what is tan θ?
30
A vertical asymptote occurs when cosθ=0, this occurs when cos has x-intercepts
at 90 and 270
See if they can come up with the following information: For tangent graphs, the
distance between any two consecutive vertical asymptotes represents one complete
period.
   
Ex. Graph y= tan 3x for   , ,   3,3,1 and state the amplitude and period.
 2 2 4



No amplitude
sin x
since tan x 
, cos x  0, undefined for x  2  n
cos x
1

horizontal stretch by , so period is now
3
3
Summary of the Effect of the Parameters a, b, c and d:
For y  a tan b  x  c    d
According to the curriculum, you are not supposed to transform tangent function!
amplitude – not applicable
a value represents a stretch:
180
Period =
(degree measure)
b
Period =

b
(radian measure)
horizontal phase shift = c
 right if c > 0, left if c < 0
vertical displacement = d
 up if d > 0
 down if d < 0
31
However, we can do problem questions.
Ex. A small plane is flying at a constant altitude of 6000m directly toward an observer.
Assume that the ground is flat in the region close to the observer.
a. Determine the relation between the horizontal distance, in metres, from the
observer to the plane and the angle, in degrees, formed from the vertical to the
plane. (the triangle is actually standing up vertically)
d
plane
𝑡𝑎𝑛𝜃 =
6000 m
𝑑
6000
observer
𝑑 = 6000 tan 𝜃
b. Sketch the graph of the function.
The graph represents the horizontal distance between the plane and the
observer. As the plane flies toward the observer, that distance decreases. As
the plane moves from directly overhead to the observer’s left, the distance value
becomes negative. The domain of the function is: −90° < 𝜃 < −90°
c. Where are the asymptotes located in the graph? What do they represent?
Located at 90˚ and -90˚. They represent when the plane is on the ground to the
right or left of the observer, which is impossible, because the plane is flying in a
straight line at a constant altitude of 6000m.
d. Explain what happens when the angle is equal to 0˚?
When the angle is equal to 0˚, the plane is directly over the head of the observer.
The horizontal distance is 0.
Assignment: pg. pg. 262-265 #1ac, 2ef, 6, 8 – 9, 12, C1
32
Lesson 9: Equation and Graphs of Trig Functions
Specific Outcome 4: Graph and analyze the trigonometric functions sine, cosine and tangent to solve problems
One of the most useful characteristics of trigonometric functions is their periodicity.
With a partner, name as many situations in the world around us that can be represented
in a sinusoidal graph.
Look for cycles that are predictable, sunsets, sunrise, comet appearances, ferris wheel
height, seasonal temperature changes, the movement of the waves in the ocean, quality
of musical sound.
Mathematics and scientists use the periodic nature of trigonometric functions to develop
mathematical models to predict many natural phenomena.
You can identify a trend or pattern, determine an approximate mathematical model to
describe the process, and use it to make predictions (interpolate or extrapolate)
You can use graphs of trigonometric functions to solve trigonometric equations that
model periodic phenomena, such as the swing of a pendulum, the motion of a piston in
an engine, motion of a ferris wheel, variations in blood pressure, the hours of daylight
throughout a year, and vibrations that create sounds.
First, let’s practice solving equations.
Ex. Determine the solutions for the trigonometric equation 4𝑠𝑖𝑛2 𝑥 − 3 = 0 for the
interval 0° ≤ 𝑥 ≤ 360° both graphically and algebraically.
Graphically: graph and look for x-intercepts:60˚, 120˚, 240˚, 300˚
Algebraically:
3
4𝑠𝑖𝑛2 𝑥 = 3, 𝑠𝑖𝑛2 𝑥 = 4, sin 𝑥 = ±
√3
2
√3
𝑠𝑖𝑛−1 ( ) = 60° = 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑛𝑔𝑙𝑒
2
We are looking for 4 answers (one in each quadrant) because of the ± after we square
rooted.
60˚, 120˚, 240˚, 300˚
33
𝜋
Ex. Determine the general solutions for the trigonometric equation 10 = 6 sin 4 𝑥 + 8.
𝜋
Solve graphically and put in radians mode: y = 6 sin 4 𝑥 − 2.
0.43, 3.57, the period of the function is 8 radians (
2𝜋
𝜋
4
= 8)
Add 8 to each value to see multiple values.
The period of the function is 8 radians, then the solutions repeat in multiples of 8
radians from each original solution.
𝑥 ≈ 0.43 + 8𝑛 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 𝑎𝑛𝑑 𝑥 ≈ 3.57 + 8𝑛 𝑟𝑎𝑑𝑖𝑎𝑛𝑠, where n is an integer.
Ex. In some Caribbean countries, the current makes 50 complete cycles each second
and the voltage is modeled by 𝑉 = 170 sin 100𝜋𝑡.
a. Graph the voltage function over two cycles. Explain what the scales on the axes
represent.
x-axis represents the time passed, the y-axis represents the number of volts
34
b. What is the period of the current in these countries?
2𝜋
100𝜋
1
= 50, there are 50 complete cycles each second.
c. How many times does the voltage reach 110V in the first second?
50 cycles in one second, need to see how many times it reaches 110 in one
cycle.
Graph y = 110 in the second y plot:
2 times each cycle x 50 = 100 times
Ex. Windsor, Ontario, is located at altitude 42˚N. The table shows the number of hours
of daylight on the 21st day of each month as the day of the year on which it occurs for
this city.
Hours of Daylight by Day of the Year for Windsor, Ontario
21
52
80
111
141
172
202
233
264
9.6 10.8 12.2 13.6 14.7 15.2 14.8 13.6 12.2
2
7
0
4
9
8
1
4
2
294
10.8
2
325
9.5
9
355
9.0
8
a. Draw a scatterplot for the number of hours of daylight, h, in Windsor, Ontario on the
day of the year, t. (remind them how to do this on calculator)
radians mode
b. Which sinusoidal function will best fit the data without requiring a phase shift: sin t, sin t, cos t, or –cos t? (because it starts at the minimum)
-cos t
35
c. Write the sinusoidal function that models the number of hours of daylight.
𝑎=
𝑑=
𝑚𝑎𝑥 − 𝑚𝑖𝑛 15.28 − 9.08
=
= 3.1
2
2
𝑚𝑎𝑥 + 𝑚𝑖𝑛 15.28 + 9.08
=
= 12.18
2
2
𝑃𝑒𝑟𝑖𝑜𝑑 =
2𝜋
365
Phase shift, for y = -cos t, the minimum value occurs at x = 0 (y-intercept). For the
daylight curve, the actual minimum occurs on day 355, which represents a 10-day shift
to the left, c = -10.
ℎ(𝑡) = −3.1 𝑐𝑜𝑠 (
2𝜋
(𝑡 + 10)) + 12.18
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d. Graph the new function.
e. Estimate the number of hours of daylight on each date:
i. March 10
day 69, substitute 69in for t
11.53
ii. July 24
day 205
10.59
iii. December 3
day 337
9.23
Assignment: pg. 275-279 #1, 3, 4cd, 5cd, 6cd, 9, 10, 12, 13, 15, 16, 19
(and go back and do p. 213 #15, 19)
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