Download Relating Sines and Cosines : The Pythagorean Trig Identity

UNIT 5 – TRIGONOMETRIC RATIOS
Trigonometric Ratios in Right Triangles (Review)

The 3 primary trig ratios in a right triangle are :
sin  
opposite
hypotenuse
cos 
adjacent
hypotenuse
tan  
Remember : SOH CAH TOA
opposite
adjacent
B
Hypotenuse
Opposite

A
C
Adjacent
 To solve a triangle means to determine all unknown side lengths and angle measures.
 To determine a trig ratio on your calculator, you use the SIN, COS, or TAN function.
 To determine an angle measure on your calculator, you use the SIN-1, COS-1, or TAN-1 function.
Examples :
1. Calculate the following trig ratios, to 3 decimal places.
a) sin54
b) cos39
c) tan70
2. Calculate the following angle measures, to the nearest degree.
a) sinA = 0.241
b) cos = 0.823
c) tanP = 4/3
3. Solve each triangle. Round side lengths to one decimal place and angle measures to the nearest
degree.
a) PQR with P = 43, Q = 90, and p = 5.5cm
b) DEF with F = 90, d = 6.3cm, and f = 8.0cm
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5.1 RECIPROCAL TRIGONOMETRIC RATIOS
Recall :
* The three primary trig ratios are the sine, cosine, and tangent.
sin  
For any right triangle,
opposite
hypotenuse
cos  
adjacent
hypotenuse
tan  
opposite
adjacent
A
opposite
hypotenuse

C
B
Adjacent
The reciprocals of these ratios are respectively called the cosecant, secant, and cotangent and are
abbreviated as csc, sec, and cot.
The Reciprocal Trigonometric Ratios :
hypotenuse
opposite
1
csc  
sin 
csc  
For any right triangle,
hypotenuse
adjacent
1
sec  
cos 
sec  
adjacent
opposite
1
cot  
tan 
cot  
A
opposite
hypotenuse

C
B
Adjacent
2
* To use our calculators with the reciprocal trig ratios, we must use the reciprocal key,
1
or x 1 along
x
with the sine, cosine, and tangent keys.
Examples :
1. Determine the following trig ratios to 3 decimal places.
a) csc36
b) sec80
c) cot52
2. Determine the angle measure to the nearest degree.
a) csc = 2.564
b) sec = 3.723
c) cot = 1.149
3. Calculate the 6 trig ratios for  = 29 , to 3 decimal places.
4. Write the 6 trig ratios for the two acute angles.
A
17cm
8cm
C
B
15cm
5. Solve for x, to one decimal place. Use the most appropriate method.
P
42
x
R
Q
7cm
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5.2 EVALUATING TRIGONOMETRIC RATIOS FOR SPECIAL ANGLES
Certain angles have exact values that we can use in our trigonometric work.
This triangle is found by taking the diagonal
of a square with side length 1 unit.
This triangle is found by taking an equilateral
triangle with side length 2 units and bisecting
it.
From these triangles, we can calculate exact trig values for each of these angles.
We can also use these values to evaluate expressions.
Ex. Determine the exact value of (sin 45◦)(cos 45◦) + (sin 30◦)(sin 60◦).
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5.3 EXPLORING TRIGONOMETRIC RATIOS FOR ANGLES GREATER THAN 90◦

An obtuse angle is an angle between 90º and 180º, i.e. 90º    180º.

The equation of a circle with its center at the origin is : x 2  y 2  r 2

The Unit Circle :
 center is at O(0, 0)
 radius is 1 unit in length
 equation is : x 2  y 2  1
y
P(x, y)
r=1
O
(1, 0)
x
 Let P(x, y) be any point on the unit circle.
In Quadrant 1:
The coordinates of any point on the unit circle in quadrant 1 are : P(x, y) = P(cos, sin).
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In Quadrant 2 :
The coordinates of any point on the unit circle in quadrant 2 are : P(x, y) = P(-cos, sin).
Special Cases between 0º and 180º :
1)  = 0º
2)  = 90º
3)  = 180º
 sin0º =
 sin90º =
 sin180º =
cos0º =
cos90º =
cos180º =
Sine of Obtuse Angles :
 Any equation in the form sin = k, where 0º    180º, has 2 solutions:
1) 
2) 180º - 
ex. Find the values of  for 0º    180º, to the nearest degree: sin = 0.45
Relating Sines and Cosines : The Pythagorean Trig Identity
6
* PYTHAGOREAN TRIG IDENTITY :
sin2 + cos2 = 1
Examples : Find  to the nearest degree. Assume 0º    180º. Sketch a graph.
a) cos = 0.65
b) cos = -0.23
c) sin = 0.48
5.4 EVALUATING TRIG RATIOS FOR ANY ANGLE BETWEEN 0◦ AND 360◦
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Recall :

Point P(x, y) is a point that moves around the unit circle which has a centre at O(0, 0) and radius of 1
unit.

Point P starts at A(1, 0) on the x-axis.

For any position of P, an angle  represents the amount of rotation about the origin.
* We say that the angle  is in standard position where OA is the initial arm and OP is the
terminal arm.
* Direction of Rotation :
When   0, the rotation is counterclockwise.
*
When   0, the rotation is clockwise.
When P moves around the circle, the motion is repeated after P has rotated 360. So by adding or
subtracting multiples of 360 to , we can determine other angles for which the position of P is the
same. Since these angles have the same terminal arm as , they are called coterminal angles.
* Coterminal Angles :



Point P moves around a unit circle.
Two or more angles in standard position are coterminal angles when the position of P is the same for
each angle.
When an angle has measure  degrees, then the measure of any coterminal angle is ( + 360n), where
n is an integer.
Examples :
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1. For each given angle : a)  = 45 and b)  = -120
(i) Draw the angle  in standard position.
(ii) Determine three other angles coterminal with .
(iii) Write an expression to represent any angle coterminal with .
2.
a)
b)
c)
Suppose P has rotated -790 about O(0, 0) from A(1, 0).
How many full rotations have been made?
In which quadrant is P located?
Draw a diagram to show the position of P.
TRIGONOMETRIC FUNCTIONS OF ANGLES IN STANDARD POSITION
9

Let P(x, y) represent any point on the unit circle and  is the measure of the angle in standard position.
Recall :
 cos is the first coordinate of point P.
 sin is the second coordinate of point P.
 P(x, y) = P(cos, sin)
Refer to the right triangle in the diagram above.
opposite
adjacent
y

x
sin 

cos
tan  
 For any position of point P on the unit circle, we define :
 cos = x
 sin = y
sin 
 tan  
, where cos  0
cos 
Examples : For the given angles, use your calculators to determine the sine, cosine, and tangent of each
angle, to 3 decimal places. Draw each angle in standard position.
a)  = 50º

b)  = 115º
c)  = 220º
d)  = 330º
e)  = -95º
We can remember the sign of each trigonometric function in each quadrant by using the CAST rule.
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C  cos is positive in the 4th quadrant
A  all trig ratios are positive in the 1st quadrant
S  sin is positive in the 2nd quadrant
T  tan is positive in the 3rd quadrant

As P rotates around the unit circle, past 360, the same coordinates are encountered again.
ex. sin50  0.766, sin(50 + 360)  0.766, sin(50 + 720)  0.766, sin(50 + 1080)  0.766,…
These are examples of coterminal angles since the values are the same.
 In an equation with sin, cos, or tan, where you are solving for , there is an infinite
number of solutions.
When there are restrictions on , for example 0    360, you must use the CAST rule when
solving for .
Examples : Determine two angles between 0 and 360 that have each trigonometric value. Round angles
to the nearest degree.
a) sin = 0.672
b) cos = 0.291
c) tan = 1.573
d) sin = -0.875
* Solving sin = k, cos = k, and tan = k :
 When sin = k, the two values of  that satisfy this equation for 0    360 are : 1 and
2 = 180 - 1.
 When cos = k, the two values of  that satisfy this equation for 0    360 are : 1 and
2 = 360 - 1.
 When tan = k, the two values of  that satisfy this equation for 0    360 are : 1 and
2 = 180 + 1.
Examples : Solve each equation for 0    360. Round each angle to the nearest degree.
a) sin = 0.39
b) cos = -0.64
c) tan = -2.15
SINE, COSINE, & TANGENT OF SPECIAL ANGLES
11
Example: Determine the EXACT VALUES of the sin, cos, and tan of the following angles.
a)  = 135
b)  = -45
c)  = 225
Examples: Determine the EXACT VALUES of the sin, cos, and tan of the following angles.
a)  = 150
b)  = 240
c)  = -150
d)  = -300
* Other special angles are 0, 90, 180, 270, 360, and their multiples.
Recall :
Angle 0:
Angle 90:
Angle 180:
Angle 270:
sin0 = 0
sin90 = 1
sin180 = 0
sin270 = -1
cos0 = 1
tan0 = 0
cos90 = 0
cos180 = -1
tan90 = undefined
=
tan180 = 0
cos270 = 0
tan270 = 
y
P(0, 1)
P(1, 0)
P(-1, 0)
x
P(0, -1)
1. Solve for both values of  between 0 and 360. Round to the nearest degree.
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a) csc = 2.053
d) csc = -1.564
2.
b) sec = 1.491
e) sec = -3.976
c) cot = 0.882
f) cot = -3.157
Determine the EXACT VALUES for the following trig ratios.
a) sin330
d) csc120
g) csc405
b) cos240
e) sec-315
h) sec690
c) tan-210
f) cot225
i) cot-780
5.5 TRIG IDENTITIES
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TRIGONOMETRIC EQUATIONS

An equation that involves one or more trigonometric ratios of a variable is a trigonometric
equation.
ex. sin = 0.5
ex. 4cos + 1 = 0
ex. 2tan2 - 5tan - 3 = 0

To solve linear trigonometric equations:
1) Isolate for sin, cos, or tan.
2) Use your sin-1, cos-1, or tan-1 functions on your calculator to find .

Remember the special-angle triangles, the CAST rule, that x = sin and y = cos on the unit
circle, and the rules to finding a second angle between 0 and 360 for each trigonometric
function.
Examples:
1. Solve the following equations for 0    360. Round answers to 3 decimal places.
a) tan - 3 = 0
b) 5sin + 2 = 0
c) 4cos + 9 = 6
d) 8csc - 15 = 0
e) 7 – 2sec = 4
f) 1 – 6cot = 5
2. Solve the following equations for 0    360. Find the EXACT VALUE of 
1
1
3
a) cos = 
b) sin = 
c) tan =
2
2
3
d) sin = 0
e) tan = 
f) cos = -1
g) csc = 
h) sec = -1
i) cot = 0
14

k) sec =
m) csc =  2
n) sec = -2
l) cot = 1
3
o) cot =  3
To solve quadratic trigonometric equations:
1)
2)
3)
4)
5)
6)

2
j) csc = 2
Set one side equal to zero.
Let a = sin, or a = cos, or a = tan, then replace the trig function with a in the equation.
Factor the equation if possible. Then set each factor equal to zero and solve for a.
If it is not possible to factor, use the quadratic formula to solve for a.
Replace each a with the appropriate trig. function.
Solve for  using your calculator.
Sometimes, you may have to use the Pythagorean Trig Identity to help simplify the equation
first.
sin2 + cos2 = 1
or
sin2 = 1 – cos2
or
cos2 = 1 – sin2
Examples : Solve each equation for 0    360. Give EXACT ANSWERS where possible, or
round answers to 3 decimal places.
a) sin2 - cos2 = 1
b) sin2 - 1 = 0
c) 3tan2 - 2tan - 4 = 0
d) 2 sin2 - sin - 1 = 0
e) 6sin2 - cos - 4 = 0
TRIGONOMETRIC IDENTITIES
f) cot2 + 8cot - 3 = 0
15
*
Recall :
1) Pythagorean Trig Identity : sin 2   cos 2   1
2) Quotient Trig Identity : tan  
sin 
cos 
*
We can use these two trig identities to prove other trig identities.
*
We also use fraction skills, algebra skills (ex. Distributive Law), and factoring to help prove the
trig identities.
*
To prove an identity, we show that the left side of the equation equals the right side of the
equation. We must include the reasons in our work.
Examples : Prove the following identities.
sin 2 
a) 1  sin 2  
tan 2 
c)
1  sin 
cos 

cos 
1  sin 
e) tan  
1  cos 
b)

1  cos 
1
cos 
1
1
cos 
1
d) cos 4   sin 4   cos 2   sin 2 
1
1

tan  sin  cos
RECIPROCAL TRIGONOMETRIC IDENTITIES
16
Recall :
y
P(x, y)
r=1
O
(1, 0)
x
The Pythagorean Trig Identity : sin    cos    
The Reciprocal Trig Identities : csc  
The Quotient Trig Identities : tan  



and sec  
and cot  
sin 
cos 
tan 
sin 
cos 
and cot  
cos 
sin 
Examples :
1. Use the Pythagorean Theorem to prove the identity   tan    sec   .
2. Prove the following identities.
a)

 sin  sec 
cot 
b) tan  sin   cos   sec 
c) sin    cos   
  sec  
 sec  
MCR3U
17
RECIPROCAL TRIG IDENTITIES
HOMEWORK SHEET
1. Use the Pythagorean identity sin2 + cos2 = 1 to show that 1 + cot2 = csc2.
2. Prove that cos  cot  
3. a) To show that
1
 sin  by expressing the L.S. in terms of sin.
sin 
cos 2 
 1  sin  write cos2 in terms of sin.
1  sin 
b) What are the factors of 1 – sin2?
c) Prove the identity.
tan 2 
4. a) Prove the identity
 sin 2  .
2
1  tan 
b) Check the identity using  = 45.
tan 2 
 1  tan 2  .
2
sin 
b) Check the identity using  = 120.
5. a) Prove the identity
6. Prove the following identities.
cos  sin 
cot 
a) tan  cos  sin 
b) 1  cos 2  
c) sec cos  sec sin   1  tan 
d)
1  tan 2 
1
 1
2
tan 
tan   tan 
1
sin 
1


g)
sec  cot  cos 
cos  sin  sin   1


i)
csc  tan 
sec 
2
2
cos   sin 
cot   1

k)
2
cot 
cos   sin  cos 
2  sec 2 
2
2
m) sin   cos  
 sec 2 
f) cos 2   1  sin  1  sin  
e)
sin 2 
 1  cos 
1  cos 
1
1  sin 

cot 
cos 
2
sec 
1
1


j)
2
2
sin  sin  cos 2 
h) sec  
l) sin  tan   cos   sec   1  sec 2  cos 2 
n)
csc 
csc 
2 sin 


1  csc  1  csc 
cos 2 
5.6 THE SINE LAW
18
The Sine Law in Acute Triangles (Review)
A
* The Sine Law :
sinA = sinB = sinC b
a
b
c
c
c
C
a
B
 where A, B, and C are the angle measures and a, b, and c are the side lengths.

To use the Sine Law, you need to be given :
 the measure of 2 side lengths and 1 non-contained angle, or
 the measure of 2 angles and any 1 side lengths.
Examples : Solve each triangle. Round side lengths to one decimal place and angle measures to the
nearest degree.
a) ABC with A = 59, B = 76, and c = 5cm.
b) GHI with H = 62, g = 8.5cm, and h = 9.2cm.
* Recall :
 Given a right triangle, we use SOH CAH TOA.
 Given 2 side lengths and the contained angle, or all 3 side lengths, we use the Cosine Law.
 Given 2 side lengths and 1 of the non-contained angles, or 2 angles and any 1 side length, we use
the Sine Law.
 We use the Sine Law and/or Cosine Law to solve any triangle, acute or obtuse.
 An acute angle is an angle between 0 and 90.
 An obtuse angle is an angle between 90 and 180.
NOTA BENE: THE WORD “BEARING” HAS A VERY SPECIFIC CONNOTATION
IN TRIG. BEARING IS THE DIRECTION IN WHICH YOU HAVE TO MOVE IN
ORDER TO REACH AN OBJECT. A BEARING IS A CLOCKWISE ANGLE FROM
MAGNETIC NORTH.
* THE AMBIGUOUS CASE :
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 When we are given 2 side lengths and one of the non-contained angles, we use the Sine Law to
solve the triangle.
 However, there are sometimes 2 possible triangles that can be constructed given these
dimensions : an acute triangle and an obtuse triangle.
Ex. ABC : a = 4cm, c = 5cm, and A = 46
* To check for the Ambiguous Case :
adj(sin)  opp  adj
adjacent
opposite

adjacent
If this statement is true, then there are 2 triangles : one acute and one obtuse (in most cases).
Therefore, you must solve both triangles.
* To solve both the acute  and the obtuse  :
1) Use the Sine Law to find one of the missing angles ().
2) Solve the rest of the acute .
3) Use the angle you found in step 1) to determine the obtuse angle in the second triangle. (180 )
4) Solve the rest of the obtuse  using the obtuse angle (180 - ) found in step 3).
Examples :
1. Solve ABC given a = 4cm, c = 5cm, and A = 46.
2. Solve PQR given p = 9cm, q = 8.5cm, and Q = 54.
5.7 THE COSINE LAW
20
The Cosine Law in Acute Triangles (Review)
A
* The Cosine Law :
b
a2 = b2 + c2 – [2bc(cosA)], or
b2 = a2 + c2 – [2ac(cosB)], or
c2 = a2 + b2 – [2ab(cosC)]
C
c
a
B
 where A, B, and C are the angle measures and a, b, and c are the side lengths.

To use the Cosine Law, you need to be given :
 the measure of 2 side lengths and the contained angle, or
 the measure of all 3 side lengths.
Examples : Solve each triangle. Round side lengths to one decimal place and angle measures to the
nearest degree.
a) ABC with A = 58, b = 10cm, and c = 6cm.
b) PQR with p = 5.7cm, q = 5.0cm, and r = 4.1cm.
SOLVING PROBLEMS WITH TRIGONOMETRY
21
Recall:
Angle of Inclination/Angle of Declination:

These angles are ALWAYS measured from the HORIZONTAL.
declination
inclination
Examples :
1) A video camera is situated at the top of a 21m tower. Two individuals are standing on either side of
the tower. The camera sites person A at an angle of declination of 42˚ and sites person B at an angle
of declination of 38˚. How far apart are the teo individuals?
2) Two buildings are situated 32m apart. The shorter building is 60m tall. A camera sitting on the top
of the shorter building sites the top of the second building at an angle of 29˚. How tall is the taller
building?
3) A radio tower is supported by two wires on opposite sides. The wires form an angle of 60 at the
top of the post. On the ground, the ends of the wire are 15.0m apart, and one wire is at a 45 angle
to the ground.
a) How long will the wires be, to the nearest tenth of a meter?
b) How tall is the radio tower, to the nearest tenth of a meter?
4) A chandelier is suspended from the ceiling by two chains. One chain is 46cm long and forms an
angle of 60 with the ceiling. The other chain is 65cm long. What angle does the longer chain make
with the ceiling, to the nearest degree?
5) Two tracking stations, 20km apart, measure the angles of elevation of a rocket that was launched
with a weather satellite. From station A, the angle of elevation is 41, from station B, the angle of
elevation is 75. What is the altitude of the rocket, to the nearest tenth of a kilometer? (The stations
are on the same side of the rocket launch.)
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MCR3UR
EVEN MORE… SOLVING PROBLEMS WITH
TRIGONOMETRY
1) From an observation tower near Lake Winnipeg, the angle of elevation of a weather balloon
is 68. In the same plane, 35km away, the same balloon is sighted with an angle of elevation
of 47. The sighting points are on opposite sides of the balloons. Calculate the distance form
the weather balloon to the observation tower to the nearest tenth of a meter.
2) A house under construction is 10.0m wide. The two sides of the roof are to meet at an 80
angle. Each rafter is equal in length. To what length should the roof rafters be cut allowing
for 0.5 overhang?
3) Three towns A, B, and C are located so that B is 25.0km away from A and C is 34.0km away
from A. If the angle at B is 110, calculate the distance from B to C.
4) An architect is designing a solar heated house that will be 10.0m wide. The south side of the
roof containing the solar collectors is 8.0m long and rises at an angle of elevation of 60. At
what angle of elevation will the north side of the roof rise?
5) A football player is attempting a field goal. The angle formed by the player’s position of the
field and each line of sight to each upright is 33. If the distances to the uprights are 7.5m
and 10.0m, calculate the widths of the uprights.
6) Two points A and B are located on the same bank of a river 28.5km apart. A point C is
located on the other side of the river between point A and B. The angle at A is 62 and the
angle at B is 43. Determine the width of the river.
7) Two planes flying at the same altitude are 3000.0m when the spot a raft on the sea below
them (the raft is between the two planes). The angles of depression to the raft are 47 and
38 respectively.
a. Find the distance from the raft to the closest plane.
b. Determine the altitude that the planes are flying.
8) Steve and John are looking up at a cliff. The angle of elevation from Steve to the cliff is 32
and the angle of elevation from John to the cliff is 23. Steve and John are standing 50m
apart. Determine the height of the cliff.
9) A person standing on the roof of a 12m tall house spots a flagpole in the distance. The angle
of elevation from the person to the top of the flagpole is 25 and the angle of declination to
the bottom of the flagpole is 42. Determine the height of the flagpole.
ANSWERS
1) 28.2km
6) 17.8m
2) 8.3m
7) a) 1854 m
3) 16.0km
b) 1355.9m
4) 49
8) 66.2m
5) 5.5m
9) 18.2m
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5.8 SOLVING THREE-DIMENSIONAL PROBLEMS USING TRIGONOMETRY
Three-dimensional problems involving triangles can be solved using some combination of these
approaches:
 Trig ratios
 The Pythagorean Theorem
 Sine Law
 Cosine Law
Any and all of these tools may be used in solving three-dimensional problems. It is esssential to
sketch the given information, and determine any unknown angles using angle properties (parallel line
theorems, for example). You want to add as much information as possible before you begin solving
the question.
Ex. Emma is on a 50m high bridge and sees two boats anchored below. From her position, boat A
has a bearing of 230 degrees, and boat B has a bearing of 120 degrees. Emma estimates the angles
of depression to be 38 degrees for boat A and 35 degrees from boat B. How far apart are the boats
to the nearest metre?
Sol.
Ex. Manny is standing in front of a cliff and estimates the angle of elevation to the top of the cliff is
38 degrees. From another point, 68.5km away from Manny, Joe estimates that the angle between
the base of the cliff, himself, and Manny is 42 degrees, while Manny estimates the angle between the
base of the cliff, himself, and Joe to be 63 degrees.
What is the height of the cliff to the nearest tenth of a metre?
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