Pre – Calculus Math 40S: Explained!
www.math40s.com
9
Trigonometry Lesson 2
Part One — The Unit Circle
The Unit Circle
What you see here is the unit circle. This is a useful tool in:
a) Comparing angles in degrees & radians.
b) Finding exact values of the six trigonometric ratios.
It is very important you memorize the unit circle
as it will not be provided on the diploma.
The x-coordinate is the cosine of the angle
The y-coordinate is the sine of the angle.
Pre – Calculus Math 40S: Explained!
www.math40s.com
10
Trigonometry Lesson 2
Part One — The Unit Circle
Example 1: Find the exact value of cos 135º
From the unit circle, we can see that the x-coordinate of 135º is
-
2
2
Example 2: Find the exact value of sin 750º
First, find the principal angle for 750º.
750º - 360º = 390º
390º - 360º = 30º
From the unit circle, sin 30º =
First convert the given angle
to a principal angle, then use the unit
circle to find the exact value.
1
2
Example 3: Find the exact value of cos(—1020º)
First, find the principal angle for cos(-1020º)
It is considered proper form to keep
negative angles inside brackets.
-1020º + 360º + 360º + 360º = 60º
From the unit circle, cos 60º =
1
2
Example 4: Find the exact value of sin2 5π
6
First convert the radian fraction to degrees.
5π 180o
×
=150o
6
π
From the unit circle, sin 150º =
Now square that answer =
1
2
1
4
Example 5: Find the exact value of cos
16π 180o
×
= 960o
π
3
2
2
sin θ is the same thing as ( sinθ )
Evaluate what is inside the brackets
first, then square the result to get
the answer.
16π
3
960º - 360º - 360º = 240º
From the unit circle, cos 240º =
-
1
2
⎛ 9π ⎞
⎟
⎝ 2 ⎠
Example 6: Find the exact value of -sin ⎜ -
-
9π 180o
×
= -810o
2
π
-810º + 360º + 360º + 360º = 270º
Finally, -sin 270º = - (-1) = 1
Pre – Calculus Math 40S: Explained!
www.math40s.com
11
Trigonometry Lesson 2
Part One — The Unit Circle
The other four trigonometric ratios can be found from the unit circle using the following formulas:
⎛ π⎞
⎟
⎝ 3⎠
Example 7: Find the exact value of sin ⎜ -
-
π 180o
×
= -60o
π
3
-
-60º + 360º = 300º
sin 300o = -
23π
6
23π 180o
×
= -690o
6
π
-690º + 360º + 360º = 30º
3
2
tan30o =
⎛ 20π ⎞
⎟
⎝ 3 ⎠
Example 8: Find the exact value of cot ⎜
20π 180o
×
=1200o
3
π
1200º - 360º - 360º - 360º = 120º
cot120o =
⎛
⎝
Example 9: Find the exact value of tan ⎜ -
cos120o
sin120o
1
= 2
3
2
1 2
=- ×
2
3
1
=Now Rationalize The Denominator
3
sin30o
cos30o
1
= 2
3
2
1 2
= ×
2
3
1
=
Now Rationalize The Denominator
3
=
1
3
×
3
3
=
3
3
-
=-
1
3
×
3
3
=-
3
3
⎛ 3π ⎞
⎟
⎝ 2⎠
Example 10: Find the exact value of sec ⎜
3π
= 270o
2
sec 270o
1
=
cos 270o
1
= = undefined
0
Pre – Calculus Math 40S: Explained!
www.math40s.com
12
⎞
⎟
⎠
Trigonometry Lesson 2
Part One — The Unit Circle
Evaluate each of the following:
1) sin
5π
4
2) sin180
16) cot 2
D
3) cos(−240D )
⎛ 35π ⎞
17) sec ⎜ −
⎟
⎝ 6 ⎠
4) sec (−420 )
⎛ −7π ⎞
5) csc ⎜
⎟
⎝ 6 ⎠
19) sec 0
⎛ −10π ⎞
6) cot ⎜
⎟
⎝ 3 ⎠
⎛ 13π ⎞
7) − tan ⎜
⎟
⎝ 6 ⎠
8) tan
π
⎛ −3π ⎞
9) csc ⎜
⎟
⎝ 4 ⎠
7π
6
⎛ −11π ⎞
11) sec ⎜
⎟
⎝ 2 ⎠
12) − sec180
13) cot 240
D
21) cot ( −10π )
22) tan
13π
4
π
6
24) − tan 0
25) cos
11π
6
26) sec
5π
3
27) − csc180D
28) cot
D
⎛ −5π ⎞
14) csc ⎜
⎟
⎝ 4 ⎠
15) sec
20) − csc1380D
⎛ 4π ⎞
23) − sec ⎜ −
⎟
⎝ 3 ⎠
2
10) tan
Express all fractions
with rationalized
denominators.
29π
4
18) tan
D
2
20π
3
29) sin
π
2
π
3
30) − tan
π
3
Pre – Calculus Math 40S: Explained!
www.math40s.com
13
Trigonometry Lesson Two
Part I - The Unit Circle
1) −
2
⎛
3⎞ 3 1
16) ⎜⎜ −
⎟⎟ = =
⎝ 3 ⎠ 9 3
2
2
2) 0
3) −
17)
1
2
2 3
3
18) 1
4) 4
19) 1
5) 2
20)
6) −
7) −
3
3
21) undefined
3
3
22) 1
8) undefined
23) 2
24) 0
9) − 2
10)
2 3
3
25)
3
3
3
2
26) 2
11) undefined
27) undefined
12) 1
13)
3
3
28) 0
14)
2
29)
15)
2 3
3
3
2
30) − 3
Pre – Calculus Math 40S: Explained!
www.math40s.com
14
Trigonometry Lesson Two
Part II — Solving Basic Equations
In the last section, we were looking for the exact value when given an angle.
In this section, we’ll look for the angle(s) given the exact value.
Example 1: What is the solution to: sinθ = -
3
2
( 0 ≤ θ ≤ 2π )
3
on the
2
unit circle and see what angles correspond to it.
4π
5π
The solution is θ =
and
3
3
To solve this, look for y-coordinates equal to −
Example 2: Where is secθ undefined? ( 0 ≤ θ ≤ 2π )
1
, so secθ is undefined
cosθ
whenever the denominator becomes zero.
π 3π
cosθ =0 when θ = ,
2 2
We know that secθ =
In the previous two questions, we were given ( 0 ≤ θ ≤ 2 π ) , which means we only need to go
around the circle once. Most of the time, you will want the general solution, which includes
all possible co-terminal angles.
Example 3: Find the general solution to cosθ =
1
2
From the unit circle, we have an x-coordinate
of
1
π
5π
when θ = and
2
3
3
π
π
± n(2π )
3
3
5π
5π
The general solution for
is
± n(2π )
3
3
The general solution for
is
In Lesson 1, we used 360˚ for
general solutions. If you are dealing
with radians, use 2π instead.
Pre – Calculus Math 40S: Explained!
www.math40s.com
15
Trigonometry Lesson Two
Part II — Solving Basic Equations
Watch out for questions involving tan θ. It has special rules.
tanθ = 1 at
π
4
and
5π
4
Places
where
tanθ = 1
2
, since sinθ and cosθ are both
2
5π
2
−
sin
5π
4
2
example : tan
=
=
= 1
5π
4
2
cos
−
4
2
tanθ = -1 at
3π
4
and
7π
4
have a magnitude of
, since both sinθ and cosθ
2
2
Places where
tanθ = -1
but differ by a negative.
7π
2
sin
−
7π
4 =
2 = −1
example : tan
=
7π
4
2
cos
4
2
I.n the special case of tanθ = 1 or tanθ = -1, we can combine all answers into one general solution.
The reason for this is that each solution is exactly π units away from the other solution.
Example 4: Find the general solution to tanθ = -1
We know tanθ = -1 at
3π
4
and
7π
4
. The angles are drawn on the right.
3π
In this case, simply write the general solution as:
± nπ
π
4
When n = 0 , we will have
3π
4
When n = 1 , we will have
7π
4
Common Denominator:
3π
4
+π =
3π
4
+
π
1
=
3π
4
+
4π
4
=
7π
4
π
As you can see by trying out a few values of n, we will always obtain a
difference of π . Thus, we can account for all co-terminal angles using
3π
± nπ without having to write two separate general solutions.
4
Pre – Calculus Math 40S: Explained!
www.math40s.com
16
Trigonometry Lesson Two
Part II — Solving Basic Equations
For each of the following, find the solution for ( 0 ≤ θ ≤ 2 π )
3
2
1) sin θ =
2) sin θ = 0
Memorize these general solutions:
3) cos θ = −1
tanθ = 1 or cotθ = 1 →
2
4) sin θ = −
2
π
4
tanθ = -1 or cotθ = -1 →
± nπ
3π
± nπ
4
5) tan θ = 0
6) csc θ = undefined
7) tan θ = −1
For each of the following, find the general solution. (Watch for solutions that are separated by π)
8) cos θ =
1
2
9) cos θ = 0
10) sin θ = −
1
2
11) cos θ =
3
2
12) cos θ =
2
2
13) tan θ = undefined
14) cot θ = 0
15) tan θ = 1
16) cot θ = −1
Pre – Calculus Math 40S: Explained!
www.math40s.com
17
Trigonometry Lesson Two
Part II — Solving Basic Equations
1) θ =
π 2π
3
,
θ=
3
12)
2) θ = 0, π , 2π *Notice that ( 0 ≤ θ ≤ 2π ) has a
π
± n(2π )
4
7π
± n(2π )
θ=
4
less than/equals sign for 2π, so include the 2π.
13) tan θ = undefined when
3) θ = π
4) θ =
undefined, and this happens when the
denominator is zero
5π 7π
,
4 4
8)
2
2
15) θ =
π
16) θ =
π
2
cos θ
is zero,
sin θ
π
θ = ± nπ is the solution to cosθ = 0
3π 7π
,
4 4
± n(2π )
whenever
and this happens when the numerator
is zero.
3
5π
± n(2π )
θ=
3
9) θ =
± nπ is the solution to cos θ = 0
14) cot θ = 0
1
is
6) csc θ = undefined whenever
sin θ
undefined. sin θ = 0 when θ = 0, π , 2π
θ=
π
θ=
5) θ = 0, π , 2π
7) θ =
sin θ
is
cos θ
π
4
± nπ
3π
± nπ
4
± nπ *Note the combined general solution,
since both angles where cos θ = 0 are separated by π
7π
± n(2π )
6
10)
11 π
± n(2π )
θ =
6
θ=
θ=
π
± n(2π )
6
11)
11π
± n(2π )
θ=
6
Pre – Calculus Math 40S: Explained!
www.math40s.com
18
Trigonometry Lesson Two
Part III — Graphically Solving Equations
In the last section, all the solutions could be found on the unit circle.
Now we will look at equations involving solutions not on the unit circle.
Example 1: Find the angles where sinθ = -0.4235,
( 0 ≤ θ ≤ 360 )
o
To solve this, we graph in the TI-83 (use degree mode):
y1 = sin θ
y2 = −0.4235
1
X min = 0
X max = 360
When the domain is in degrees,
the answer should be in degrees also.
0.5
X scl = 90
Y min = -1
Y max = 1
Y scl = 1
90 °
180 °
270 °
When the domain is in radians,
the answer should be in radians also.
360 °
-0.5
Use 2nd Æ Trace Æ Intersect to find the
points of intersection of the two lines.
-1
The x-coordinates will give you the
angles that solve the equation.
θ = 205º and 335º
Example 2: Find the angles where cosθ =
5
,
7
( 0 ≤ θ ≤ 2π )
When the domain is given in radians, the answer should be in radians too.
To solve this, we graph in the TI-83 (use radian mode):
y1 = cos θ
y2 =
5
7
1
Window for
radian mode
Use 2nd Æ Trace Æ Intersect to
find the points of intersection
of the two lines.
0.5
X min = 0
X max = 2π
X scl =
1
π
2
Y min = -1
Y max = 1
Y scl = 1
2
3
4
5
6
-0.5
The x-coordinates will give you
the angles that solve the
equation.
θ = 1.25 and 5.04 radians
-1
Pre – Calculus Math 40S: Explained!
www.math40s.com
19
Trigonometry Lesson Two
Part III — Graphically Solving Equations
Solve the following equations for the domain 00 ≤ x ≤ 3600
*In some of the graphs, you will see vertical lines. These represent asymptotes, and do not
contribute to the solution. Don’t try to obtain intersection points at asymptotes.
1) cos x = -0.1288
2) sin x = 0.5
3) tan x = -1
4) sec x = -1.3342
5) cot x = 2
6) csc x = 3.4219
7) tan x = -0.5397
8) cos x = 0.3994
9) sec x = 1
10) sin x = -0.4398
Solve the following equations for the domain 0 ≤ x ≤ 2π
11) cos x =
3
The following answers
are in radians.
2
12) tan x = 0
13) sin x = 2
14) cot x = 0.1123
Answers:
15) csc x = 0.5
The following answers
are in degrees.
16) sec x = -1
17) sin x = -0.5
18) tan x = 1.7321
19) cos x= -0.7071
20) cot x = -1
1) 97.4 & 262.6
2) 30 & 150
3) 135 & 315
4) 138.5 & 221.5
5) 26.6 & 206.6
6) 17 & 163
7) 151.6 & 331.6
8) 66.5 & 293.5
9) 0 & 360
10) 206.1 & 333.9
π 11π
,
6 6
12) 0, π , 2π
13) Undefined
14) 1.46 & 4.60
15) Undefined
16) π
7π 11π
17)
,
6 6
π 4π
18) ,
3 3
3π 5π
19)
,
4 4
3π 7π
20)
,
4 4
11)
Pre – Calculus Math 40S: Explained!
www.math40s.com
20