Trigonometry Handout

Trigonometry
Table of Contents:
Background Information………………………………………2
Pythagorean Theorem…………………………………2
Similar Triangles………………………………………2
Angles…………...………………………………....………….3
Standard Position……………………………………...3
Measurement of an Angle……………………………..3
Converting Measures………………………………….4
Angles and Arc Length………………………………..4
The Trigonometric Functions………………………………....5
Right Triangles and Acute Angles…………………….5
General Angles………………………………………...6
Conventional Notation…………………………………6
Common Angles……………………………………….7
Signs……………………………………………………7
The Unit Circle…………………………………………………8
Graphs of Trigonometric Functions……………………………9
Inverse Trigonometric Functions……………………………...11
Trigonometric Identities……………………………………….15
Consequences of Definition…………………………...15
Fundamental Identities…………………………………15
Addition Identities……………………………………..16
Subtraction Identities…………………………………..16
Double Angle Identities………………………….…….17
Half Angle Identities…………………………………..17
Product Identities………………………………………17
Trigonometric Problems……………………………………….17
The Law of Sines………………………………………………19
The Law of Cosines……………………………………………19
Extra Problems…………………………………………………21
Answers to Extra Problems…………………………………….23
Page 1 of 23

Trigonometry
Trigonometry is the study of how the sides and angles of a triangle relate to each other.
Background Information:
1) The Pythagorean Theorem:
Let Δ ABC be a right triangle (See Figure 1)
B
Let the length of AB = c (hypotenuse)
Let the length of BC = a (leg)
c
a
Let the length of AC = b (leg)
C
b
Figure 1
A
2
2
2
Then a + b = c . In other words, the sum of the squares of the legs of a right triangle
equals the square of the hypotenuse.
2) Similar Triangles
Definition: Two triangles are similar if corresponding angles are congruent (of equal
measure).
For a triangle, it is enough for two corresponding angles to be congruent, because
that would force the third set to be congruent.
It is also enough for the corresponding sides to be proportional:
a
B
b
d
E
e
Figure 2
A
c
C
D
f
F
If
Δ ABC is similar to Δ DEF , then ∠ A ≅ ∠D, ∠B
≅ ∠E,
∠C
≅ ∠F
and
a
d
=
b
e
=
c
f
From this, we also obtain that the ratio of any two sides of one triangle is equal to the
ratio of the corresponding sides of the similar triangle.
a d
For example: =
b e
Page 2 of 23

Trigonometry
Angles:
1) Standard Position:
Angle
Terminal Side
Figure 3
Vertex
Initial Side
The Standard Position of an angle occurs when the vertex of the angle is placed at the
origin of a coordinate system and its initial side is on the positive x-axis.
Angles in
Standard Position
θ
θ
Figure 4 Figure 5
- A positive angle is obtained by rotating the initial side counterclockwise until it
coincides with the terminal side (Figure 4).
- A negative angle is obtained by rotating the initial side clockwise until it coincides with
the terminal side (Figure 5).
2) Measurement of Angles:
Angles can be measured in degrees or radians (abbreviated rad).
A complete revolution is 360 or 2 π rad.
o
Thus: π rad = 180
o
o
⎛180
⎞ o o π
⇒ 1 rad = ⎜ ⎟ ≈ 57.3 or 1 = rad ≅ 0. 017
⎝ π ⎠
180
rad
Page 3 of 23

Trigonometry
3) Converting from degrees to radians and from radians to degrees
Example 1:
o
a) Find the radian measure of 90
o ⎛ π ⎞
o π
Solution: 90( 1 ) = 90⎜
⎟rad
⇒ 90 = rad
⎝180
⎠
2
7π
b) Express rad
6
in degrees
7π
1
6
Solution: ( )
o
rad
o
7π
⎛180
⎞ 7π
= ⎜ ⎟ ⇒ rad = 210
6 ⎝ π ⎠ 6
The following table gives some conversions between radians and degrees of some
common angles:
Degrees 0 30 45 60 90 120 135 150 180 270 360
Radians 0 π π π π 2π 3π 5π π 3π 2π
6 4 3 2 3 4 6 2
Figure 6 shows some angles in standard position:
θ
θ
θ
θ
π
θ =
4
θ = −
7π
4
Figure 6
9π
θ =
4
5π
θ = −
4
4) Angles and Arc Length:
Let O be a circle with radius r (Figure 7)
Let θ be a central angle.
r
Let a be the length of the arc subtended by θ
a
O θ
Figure 7
Page 4 of 23

Trigonometry
The length of the arc is proportional to the measure of the angle and since the entire circle
a θ
has circumference 2 πr
and central angle measure 2π
⇒ =
2π r 2π
a a
⇒ a = rθ , r = , θ =
θ r
Note: These are only valid when θ is measured in radians.
Example 2:
a) If the radius of a circle is 7 cm, what angle is subtended by an arc of 21 cm?
a 21
Solution: θ = = = 3 rad
r 7
b) A circle has a radius of 4 cm. What is the length of the arc subtended by an
o
angle of 90 ?
o π
Solution: Step 1: Angle must be in radians 90 = rad
2
π
Step 2: a = rθ = 4 ⋅ = 2π
cm
2
The Trigonometric Functions:
1) Right Triangles and Acute Angles
⎛ π ⎞
Let θ be an acute angle ⎜0
≤ θ < ⎟
⎝ 2 ⎠
hypotenuse The trigonometric functions are defined as ratios of
opposite
lengths of sides of a right triangle
θ
adjacent
Note: Any right triangle with angle θ is similar to any other right triangle with angle θ .
Since the ratios of sides in similar triangles are equal, then the trig functions will remain
the same regardless of the lengths of the sides.
sinθ
=
cosθ
=
opp
hyp
adj
hyp
cscθ
=
secθ
=
hyp
opp
hyp
adj
A tool to remember SOHCAHTOA
tanθ
=
opp
adj
cotθ
=
adj
opp
π
NOTE: These ratios only apply if 0 ≤ θ < .
2
Page 5 of 23

Trigonometry
2) General Angles
Let θ be any angle in standard position
Let P ( x,
y)
be any point on the terminal side of θ
Let r be the distance from the origin to P (See Figure 9)
P ( x,
y)
Then,
sinθ
=
y
r
cscθ
=
r
y
r
θ
cosθ
=
x
r
secθ
=
r
x
O
Figure 9
tanθ
=
y
x
cotθ
=
x
y
Note: If y=0, then csc θ and cot θ are not defined. If x=0, then tan θ and sec θ are not
defined. These definitions are consistent with the previous definition if θ is an acute
angle.
3) Conventional Notation
If θ is a number, then by convention
of the angle whose radian measure is θ .
≈ where as sin( 5 ) ≈ 0. 0876
So, sin5
−9.
589
o
sin θ (or any trig function) means the sine
When using your calculator to compute trigonometric functions, you need to make sure
that your calculator is set to the correct mode. If you are computing radian measures, you
need to have your calculator in radians, and in degrees if you are computing degree
measures.
4) Some Common Angles:
Here is a table of common angles and the trig functions computed at them.
θ 0
sin θ 0
cos θ 1
tan θ 0
π
6
π
4
π
3
π
2
1
3
2 22
2
1
1
23
22
2
0
3 1 3 Und.
3
Page 6 of 23

Trigonometry
5) Signs
The signs of the trig functions for angles in each quadrant can be remembered by
the following saying:
All Students Take Calculus
Quad I: All ratios are positive
S
A
Quad II: Sine and cosecant are positive
Rest are negative
Quad III: Tangent and cotangent are positive
Rest are negative
T
C
Quad IV: Cosine and secant are positive
Rest are negative
Figure 10
3π
Example 3: Let P = (−1,1 ) be a point on the terminal side of the angle θ = . Find the
4
exact trigonometric ratios for θ .
Solution: Since P = (−1,1 ) is a point on the terminal side of the angle
2 2
( −1)
+ (1) = 2 ⇒ r = 2
1
-1
P
θ
sinθ
=
1
2
=
2
2
cscθ
=
2
1
=
2
cosθ
=
−1
2
=
−
2
2
secθ
=
2
−1
= −
2
tanθ
=
1
−1
= −1
−1
cotθ
= = −1
1
Page 7 of 23

Trigonometry
π
Note: You could have also formed the right triangle with θ = and made the appropriate
4
sign changes for the quadrant you were in.
Example 4: Let sin − 5
3π
θ = where π < θ < . Find the values of the other trigonometric
6
2
functions.
Solution:
5
6
x =
11
θ
We know that θ is in the 3 rd Quadrant.
Thus, only tangent and cotangent have
positive signs. Every thing else will be
negative. For now we will just drop the
signs to form a right triangle with hyp = 6
and opp = 5 since sine is -5/6
By the Pythagorean Theorem, we know that
2 2 2 2
2
x + 5 = 6 ⇒ x + 25 = 36 ⇒ x = 11 ⇒ x =
11
So, we get the following ratios for the other 5 trig functions
11 5
6
6
cos θ = − tanθ
= cscθ
= − secθ
= − cotθ
=
6
11 5
11
11
5
The Unit Circle:
The unit circle is a circle centered at the origin with radius 1, hence the name unit
circle. It allows us to easily find the values of sine and cosine, and thus the rest of the
trigonometric functions quickly and easily. There is a picture of the unit circle on the
following page (Figure 11). The x-coordinates are the values of cosine and the y-
coordinates are the values of sine.
Page 8 of 23

Trigonometry
Figure 11: The Unit Circle
Graphs of the Trigonometric Functions:
1) f ( x)
= sin x
Domain of f: All real numbers
x-intercepts: Let y = 0 , sin x = 0 ⇒ x = nπ
where n is an integer.
y-intercept: Let x = 0 , sin 0 = 0
π
−
2
π
2
Figure 12:
f ( x)
= sin x
Page 9 of 23

Trigonometry
Range of f: −1 ≤ sin x ≤ 1
The function is odd (symmetric with respect to the origin)
The period is 2 π
2) f ( x)
= cos x
Domain of f: All real numbers
nπ
x-intercepts: Let y = 0 , cos x = 0 ⇒ x = where n is an odd integer
2
y-intercept: Let x = 0 , cos 0 = 1
π
−
2
π
2
Figure 13: f ( x)
= cos x
Range of f: −1 ≤ cos x ≤ 1
The function is even (symmetric with respect to y-axis)
The period is 2 π
The graphs of the remaining trigonometric functions are as follows. It is left to the reader
to determine the characteristics of the graphs. (x-scale is 2 π units)
Figure 14:
f ( x)
= tan x
Figure 15:
f ( x)
= csc x
Figure 16: f ( x)
= sec x
Page 10 of 23

Trigonometry
Figure 17:
f ( x)
= cot x
Inverse Trigonometric Functions
The trigonometric functions allow us to find the ratio of sides of a triangle if we
know a given angle. What happens if we know the ratio and want to find the angle?
1
For instance, we know that sin θ = and we want to find θ
2
This is similar to when you want to know the value of x when x
2 = 5 . In that
case, you use an inverse function – namely, the square root function – which undoes the
operation of the squaring function. In the same way we want to build an inverse
trigonometric function which undoes the operation of the trigonometric function.
Now, in order for a function to have an inverse function, the original function
must pass the Horizontal Line Test. This means that any horizontal line drawn intersects
the graph in at most one place. This is a problem for the trigonometric functions since
they are periodic. In order to get inverse functions, we must restrict our domains so that
we are able to define a unique inverse.
We must meet the following criteria for the trig functions to have an inverse
function:
1. Each value of the range is taken on only once so that we can pass the
horizontal line test.
2. The range of the function with its restricted domain is the same as the range of
the original function.
π
3. The domain includes the most commonly used numbers (or angles) 0 < x <
2
1) Inverse Sine
The range of sine is −1 ≤ y ≤ 1, so this will become the domain of our inverse
sine function. Now we need to find our restricted domain. If we look at the graph of
sine, we see that the range is satisfied and we pass the horizontal line test if we take only
π π
the values − ≤ x ≤ . This will become the range of our inverse sine function.
2 2
We now define our inverse sine function as follows:
−1
If −1 ≤ x ≤ 1, then f ( x)
= sin x = arcsin x if and only if sin f ( x)
= x
π
π
and − ≤ f ( x)
≤ .
2 2
Page 11 of 23

Trigonometry
π
2
Figure 18:
f ( x)
= sin
−1
x
2) Inverse Cosine
−1
If −1 ≤ x ≤ 1, then f ( x)
= cos x = arccos x
0 ≤ f ( x)
≤ π .
if and only if
cos f ( x)
= x and
π
2
Figure 19:
f ( x)
= cos
−1
x
3) Inverse Tangent
If x is any real number, then
π
π
and − ≤ f ( x)
≤ .
2 2
f ( x)
= tan
−1
x = arctan x
if and only if
tan f ( x)
= x
π
2
Figure 20:
f ( x)
= tan
−1
x
Page 12 of 23

Trigonometry
4) Inverse Cosecant
If x ≥ 1 , then =
− 1
f ( x)
csc x = arc csc x if and only if csc f ( x)
= x and
−
f , f ( x)
≠ 0
π
≤ ( x)
≤
π
2 2
π
2
Figure 21:
f ( x)
= csc
−1
x
5) Inverse Secant
If x ≥ 1 , then =
− 1
f ( x)
sec x = arcsec x if and only if sec f ( x)
= x and
0 ≤ f ( x)
≤ π , f x)
≠
2
(
π
π
2
Figure 22:
f ( x)
= sec
−1
x
6) Inverse Cotangent
If x is any real number, then =
− 1
f ( x)
cot x = arccot x if and only if cot f ( x)
= x
and 0 < f ( x)
< π
π
2
Figure 23:
f ( x)
= cot
−1
x
Page 13 of 23

Trigonometry
Now, back to our problem, If
1
sin θ = , then what is θ ?
2
One way to come up with the solution is to look at the unit circle and find out
which points have a y-coordinate of 1 π
2
. You will see that this occurs when θ = and
6
5π
when θ = . Since sine is periodic, we know that we have infinite solutions to our
6
problem, namely θ = π + 2 nπ
and θ 5 π
= + 2n
π .
6 6
1
Now, say instead that we wanted to find (
1
sin − )
. This will only give us one answer, but
2
1
which one will it be? Well, by our restrictions, we need to find θ such that sin θ = and
2
π π
−1
− ≤ θ ≤ . Looking at all of our possible solutions, only one works, (
1
)
π
θ = sin =
2 2
2 6
In general, there are infinitely many solutions to problems where you know the value of
the trigonometric function but not the angle. Using inverse trigonometric functions gives
us what is called the principal value of the relation. The principal value of the relation is
the angle that is a solution to the problem and that has the smallest absolute value. If
positive and negative values both satisfy, then we use the positive angle.
Example: Find tan −1
1
π π
Solution: We want to find x such that tan x = 1 and − ≤ x ≤ . If we use the
2 2
5
unit circle, tan x = 1 when cos x = sin x , which occurs when x =
π , π . Only one of these
4 4
−
values is in our interval, so 1 π
tan 1 = .
4
.
Trigonometric Identities:
1) Consequences of the Definitions
1
1
1) csc θ = 2) sec θ = 3)
sinθ
cosθ
cot θ =
1
tanθ
4)
sinθ
tan θ = 5)
cosθ
cosθ
cot θ =
sinθ
2) Some fundamental identities
2 2
6) cos θ + sin θ = 1
Proof: Refer back to Figure 9. The distance formula gives that
2 2 2
2 2 ⎛ x ⎞ ⎛ y ⎞ x y x + y r
x + y = r . Thus, cos θ + sin θ = ⎜ ⎟ + ⎜ ⎟ = + = = = 1
2 2 2 2
⎝ r ⎠ ⎝ r ⎠ r r r r
2
2
2
2
2
2
2
Page 14 of 23

Trigonometry
2
2
7) tan θ + 1 = sec θ
2 2
Proof: Take sin θ + cos θ = 1 and divide by cos
2 θ
2
2
sin θ cos θ 1
2
2
+ = ⇒ tan θ + 1 = sec θ
2 2
2
cos θ cos θ cos θ
2
2
8) 1 + cot θ = csc θ
2 2
Proof: Take sin θ + cos θ = 1 and divide by sin
2 θ
9a) sin( − θ ) = −sinθ
(sine is an odd function)
9b) cos( − θ ) = cosθ
(cosine is an even function)
Proof: You can refer to figures 12 and 13, or
By definition:
(x, y)
y
sinθ
=
r
θ
− y
sin( −θ
) = = −sinθ
− θ
r
x
cosθ
=
r
(x, -y)
cos( −θ
) =
10a) sin( θ + 2π
) = sinθ
10b) cos( θ + 2π
) = cosθ
True because θ + 2 π = θ
This implies that sine and cosine are
x
r
= cosθ
2π
− periodic
3) Addition Formulas
11a) sin( x + y)
= sin xcos
y + sin y cos x
11b) cos( x + y)
= cos xcos
y − sin xsin
y
Proof: Refer to Figure 25
B
y
l(
OB)
= 1
l(
AE)
= d
l(
BA)
= a
l(
DA)
= c
l(
OC)
= f
l(
DC)
= d
l(
BD)
= g
l(
OA)
= b
l(
BC)
= e
l(
CE)
= c
O
x
y
D
F
A
C
E
∠ OAD ∠AEC
∠OCB
∠BDC
are right angles
m∠AFO
= 90 − y = m∠DFC
m∠DBC
= 90 − m∠DFC
= 90 − (90 − y)
=
y
Figure 25(Not to Scale)
Page 15 of 23

Trigonometry
From
From
From
From
Δ AOB , we get that sin( x + y)
= a and cos( x + y)
= b
Δ OCB , we get that sin x = e and cos x = f
c
Δ OCE , we get that sin y = ⇒ c = cos xsin
y and
f
b + d
cos y = ⇒ d = cos x cos y − cos( x + y)
f
d
g
Δ BDC , we get that sin y = ⇒ d = sin xsin
y and cos y = ⇒ g = cos y sin x
e
e
Putting everything together, we get that
d = sin xsin
y = cos x cos y − cos( x + y)
⇒ cos( x + y)
= cos x cos y − sin xsin
y
And
Since a = g + c , then sin( x + y)
= sin x cos y + sin y cos x
4) Subtraction Formulas
12a) sin( x − y)
= sin x cos y − sin y cos x
12b) cos( x − y)
= cos x cos y + sin xsin
y
Proof: Replace y by –y in Addition Formulas.
5) Addition and Subtraction with Tangent
tan x + tan y
13a) tan( x + y)
=
1−
tan x tan y
tan x − tan y
13b) tan( x − y)
=
1+
tan x tan y
sin( x + y)
sin x cos y + cos xsin
y
Proof of 13a: =
Left-hand side is
cos( x + y)
cos x cos y − sin xsin
y
tan( x + y) . Divide the top and bottom of the right-hand side by cos x cos y to get
sin x cos y cos xsin
y
+
cos x cos y cos x cos y tan x + tan y
tan( x + y)
=
=
cos x cos y sin xsin
y 1−
tan x tan y
−
cos x cos y cos x cos y
6) Double Angle Formulas
14a) sin 2x
= 2sin x cos x
2 2
14b) cos2x
= cos x − sin x
Proof of 14a) sin 2x
= sin( x + x)
= sin x cos x + sin x cos x = 2sin x cos x
Page 16 of 23

Trigonometry
7) Alternate forms of Double Angle Formulas for Cosine
2
15a) cos2x
= 2cos x −1
2
15b) cos2x
= 1−
2sin x
2 2
Proof of 15a: Use sin θ + cos θ = 1
2 2 2
2
2
cos2x
= cos x − sin x = cos x − (1 − cos x)
= 2cos x −1
8) Half Angle Formulas
1 cos2
16a) cos 2 + x
x =
2
1 cos2
16b) sin 2 − x
x =
2
Proof of 16a:
cos2x
= 2cos
cos2x
+ 1 = 2cos
cos
2
2
x −1
2
1+
cos2x
x =
2
9) Product Formulas
17a) sin x cos y =
1 [ sin( x + y)
+ sin( x − y )]
2
17b) cos x cos y =
1 [ cos( x + y)
+ cos( x − y )]
2
17c) sin x sin y =
1 [ cos( x − y)
− cos( x + y )]
2
Proof of 17a:
1
1
sin( x + y)
+ sin( x − y)
= sin xcos
y + sin y cos x + sin x cos y − sin y cos x
2
2
1
= 2sin xcos
y = sin xcos
x
[ ] [ ]
[ ] y
2
Trigonometric Problems
Example 5: Find all values of x in the interval [ 0 ,2π ] such that cos x = cos2x
.
2
2
Solution: We know that, cos2x = 2cos x −1, so we get cos x = 2cos x −1
2
⇒ 2cos x − cos x −1
= 0
Let u = cos x
2
⇒ 2u
− u −1
= 0
⇒ (2u
+ 1)( u −1)
= 0
⇒ u = −
1
2
⇒ cos x = −
⇒ x =
u = 1
cos x = 1
1
2
2π 4π
x =
3 3
x = 0 x = 2
π
Page 17 of 23

Trigonometry
Example 6: Find the period of the function y = 17 sin(9t
+ 21)
Solution: We know that the period of sine is 2 π . sin(bt ) is going through the
radians b times faster, so the values of sin(bt ) repeat b times sooner, or in b1
th the time.
This gives us that the period is 2 π
. For )
b
sin( bt + c , the c does not change how quickly
we go through the radians, it just affects where we start, so it has no impact on the period
(It produces a horizontal shift, so it just moves the entire graph right or left, but does not
change the shape, which would change the period). Thus the period of
y = 17 sin(9t
+ 21) is 2π 9
.
Example 7: The wheels of a bicycle have radius 55 cm and are rotating at 60 rpm. Find
the speed of the bicycle in km/hour. Round your answer to the nearest tenth.
Solution: In one minute, the angle changes by 60 ⋅ 2π = 120π
radians
In one hour, the angle changes by 60 ⋅ 120π = 7200π
radians
Hence, the linear speed in cm/hr is 55⋅ 7200π
55⋅7200π
So in km/hr we get ≈ 12. 4 km/hr
100,000
Example 8: The angle of elevation of tree 230 feet away is 12 degrees. Find the number
of feet in the height of the tree. Round to the nearest integer.
Solution: If we look at this problem as a right triangle, then we are looking for the
length of the side opposite the 12 degree angle. We know that the adjacent side is 230
o x
o
feet. Thus, tan( 12 ) = ⇒ x = 230tan(12 ) ⇒ x ≈ 49 feet tall.
230
1 1
Example 9: The trigonometric function
−
is equal to which of the
1−
cos(7x ) 1+
cos(7x)
following?
a) 2csc(7x )cot(7x)
b) 2sec(7x )cot(7x)
c) 2cot(14x )csc(7x)
d) 2sec(7x
csc(7x)
Solution: We will start by getting a common denominator
[ 1+
cos(7x)
] − [ 1−
cos(7x)
] 1+
cos(7x)
−1+
cos(7x)
2cos(7x)
cos(7x)
1
=
= = 2⋅
⋅
2 2
[ 1−
cos(7x)
][ 1+
cos(7x)
] 1−
cos (7x)
sin (7x)
sin(7x)
sin(7x)
= 2 cot(7x )csc(7x)
= a
5
Example 10: If sin p = , cos = 12
3
4
p , sin q = and cos = sin p + q .
13
13
5
5
56
Solution: sin( p + q)
= sin p cosq
+ sin qcos
p =
5 ⋅
4 +
3 ⋅
12 =
20 +
36 =
13 5 5 13 65 65 65
q , find ( )
Example 11: What are the solution(s) of sin 2 − 4sin x + 3 = 0
− π ,π ?
2
Solution: Let u = sin x ⇒ u − 4u
+ 3 = 0 ⇒ ( u − 3)( u −1)
= 0 ⇒ u = 3, 1
sin x = 3
π
⇒ ⇒ x = Note: sin x will never be 3 because the range of sine is −1 ≤ y ≤ 1
2
sin x = 1
x in the interval ( )
Page 18 of 23

Trigonometry
The Law of Sines
In any triangle ABC with sides a, b, and c (as in Figure 26),
a b c
= =
sin A sin B sinC
Proof: B
h
From Δ ADB , sin A = ⇒ h = csin
A
c
h
a
From Δ CDB , sin C = ⇒ h = asinC
c h
a
Thus, csin A = asinC
a c
⇒ =
A D b
C sin A sinC
Figure 26
Other ratios can be found by constructing the perpendiculars from other vertices.
The Law of Sines is used when we know more angles than sides.
o
Example 12: Solve Δ ABC if A = 32 , C = 81.8 , and a = 42. 9 cm
Solution: We want to find the measure of angle B and the length of sides b and c.
Step 1: We know two of the three angles. The sum of angles in a triangle
o
o o o o
is 180 . So m ∠B = 180 − 32 − 81.8 = 66.2
Step 2: We will use the Law of Sines to compute the sides.
42.9 b 42.9 c
= and =
o
o
o
o
sin(32 ) sin(66.2 ) sin(32 ) sin(81.8 )
With some computation, we find that b = 74. 1cm and c = 80. 1cm
The Law of Cosines
In any triangle ABC with sides a, b, and c,
2 2 2
a = b + c − 2bccos
A
2 2 2
b = a + c − 2accos
B
2 2 2
c = a + b − 2abcosC
Proof: Let ( x , y)
be the coordinates of A
A
y
x
sin B = and cos B =
c
c
y
b
c
⇒ y = csin
B and x = ccos
B
⇒ A = ( ccos
B,
csin
B)
Also, C = (a,0 D ) and x l ( AC B ) = b . a By the distance C Formula,
2
2
b = ( ccos
B − a)
+ ( csin
B)
Figure 27
o
Page 19 of 23

Trigonometry
By squaring both sides of the equation, we get
2
2
2
b = ( ccos
B − a)
+ ( csin
B)
2 2
2 2 2
= c cos B − 2accos
B + a + c sin B
2 2 2 2
= a + c (cos B + sin B)
− 2accos
B
2 2
= a + c − 2accos
B
The other equations may be obtained by placing the other vertices at the origin.
The Law of Cosines is used when we know more sides than angles.
Example 13: A surveyor wishes to find the distance between two inaccessible points A
and B on opposite sides of a lake. While standing at point C, she finds that l ( AC)
= 259
o
meters, l ( BC)
= 423 meters, and m ∠ACB = 125 . Find the distance from A to B. See
Figure 28.
Solution:
The Law of Cosines can be used here since
A
we know the length of two sides and the
measure of the included angle.
B
2 2 2
o
[ l ( AB)
] = 259 + 423 − 2(259)(423)cos(125 )
= 371688.63
⇒ l( AB)
= 609.66meters
C
Figure 28
Page 20 of 23

Trigonometry
Extra Problems
1. A circle has a radius of r cm. If θ = 3 rad, and the arc a = 33 cm, then what is the
radius of the circle, r?
2. The angle of elevation of a tree 219 feet away is 17 degrees. Find the number of feet
in the height of the tree. Give your answer to the nearest integer.
3. In a right triangle, a = 39 cm, b = 36 cm and c = 15 cm. Find cot θ .
c
a
b
4. Using the same right triangle from problem 3, but now with a = 37 cm, b = 35cm, and
c = 12 cm, find sec θ .
1 1
5. The trigonometric function −
1−
sin(3x ) 1+
sin(3x)
is equal to which of the
following:
a) 2sec(3x )cot(3x)
b) 2cot(6x )csc(3x)
c) 2sec(3x
)csc(3x)
d) sec( 3x ) tan(3x)
e) 2sec(3x
) tan(3x)
1+
sin(6x)
1−
sin(6x)
6. The trigonometric function − is equal to which of the
1−
sin(6x)
1+
sin(6x)
following:
a) 2sec(6x ) tan(6x)
b) 4cot(12x )csc(6x)
c) 4sec(6x
)cot(6x)
d) sec( 6x ) tan(6x)
e) 4 tan(6x
)sec(6x)
8
15
5
7. If sin p = , cos p = , sin q = , and cos q = 12 . Find sin( q − p)
. Express your
17
17
13
13
answer as a common fraction.
8. Two straight railways cross each other at angle θ = 33 . Two trains A and B leave the
intersection O at the same time. The velocity of train A is 29 mph and the velocity of
train B is 57 mph. Find the distance between the two trains after 5 hours. Give your
answer to the nearest integer. The picture is not drawn to scale.
A
θ
o
O
θ
B
Page 21 of 23

Trigonometry
15
9. If cos p = and sin p < 0. Find cot p
17
15
10. If cos p = and sin p < 0. Find csc p
17
11. The period of the function y = 15sec(7t)
is?
12. The period of the function y = 13 cos(11t
+ 15)
is?
13. The radius of the wheels of your car measure 27 cm. Suppose that each wheel is
rotating at 964 rpm. Find the speed of your car in km/h. Round your answer to the
nearest tenth.
14. The rectangle below has a diagonal AC = 31cm. The angle θ = 27 . How many
square centimeters are in the area of the rectangle. Picture not to scale.
D
C
o
A
θ
B
5
15. If sin p = , find sin( π − p)
.
7
16. What are the solutions of 2sin + sin a −1
= 0
17. What are the solutions of 2sin b − 9sinb + 4 = 0 ?
2
2
a in the interval ( 2π ,0)
− ?
18. Find
−
⎛ 3 ⎞
sin 1 ⎜ ⎟
−
.
⎝ 2 ⎠
19. Find sec − 1 (2)
.
o
20. Erin wishes to measure the distance across Eagle River. She find that C = 102 ,
o
A = 25 and b = 345 feet. Find the required distance. Round your answer to the nearest
foot.
A
b
B
a
C
Page 22 of 23

Trigonometry
Answers to Extra Problems
1. 11 cm
2. 67 feet
12
3. cm
5
37
4. 35
cm
5. e
6. e
39
7. 221
8. 181 miles
15
9. −
8
17
10. −
8
11.
2π
7
2π
12.
11
13. 98.1 km/hr
14. 388.73 square centimeters
5
15. 7
16.
11 π 7π
a = − , − , −
π
6 6 2
π 5π
17. b = + 2nπ
, + 2nπ
6 6
π
18. −
3
π
19.
3
20. 183 feet
Page 23 of 23