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# WEEK 9 PowerPoint PPT Presentation

WEEK 9. TRIGONOMETRIC FUNCTIONS RIGHT TRIANGLE TRIGONOMETRY. OBJECTIVES. At the end of this session , you will be able to: Use right triangles to evaluate trigonometric functions. Find function values for 30 °, 45°, and 60°. Recognize and use fundamental identities.

WEEK 9

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## WEEK9

TRIGONOMETRIC FUNCTIONS

RIGHT TRIANGLE TRIGONOMETRY

### OBJECTIVES

• At the end of this session , you will be able to:

• Use right triangles to evaluate trigonometric functions.

• Find function values for 30°, 45°, and 60°.

• Recognize and use fundamental identities.

• Use equal cofunctions of complements.

• Use right triangle trigonometry to solve applied problems.

### INDEX

• Six Trigonometric Functions

• Evaluating Trigonometric Functions

• Fundamental Identities

• Function Values for some Special Angles

• Trigonometric Functions and Complements

• Applications

• Summary

B

P

Side opposite q

Hypotenuse

q

O

Q

A

### 1. SIX TRIGONOMETRIC FUNCTIONS

• The word trigonometry means measurement of triangles. We begin our study of trigonometry by defining the six trigonometric functions.

• Recall that a right triangle is a triangle whose one angle measures 90°

• Let us take any acute angle AOB (Figure). We take a point P on ray OB and drop the perpendicular PQ on OA. We denote the angle POQ by Greek letter q(Theta). Then, we have a right triangle POQ in which angle QOP = q.

• We note that in the triangle POQ, OQ is the side adjacent to angle q and PQ is the

side opposite to angle q. The side opposite the right angle is known as the hypotenuse.

• For an acute angle q of the triangle POQ the ratio of the length of the side opposite angle q divided by the length of hypotenuse is called the sine of the angle.

• For example the sine of the angle q in the figure is the length of PQ divided

by the length of PO. This fact is denoted by

NOTE:sin  is an abbreviation for sine  and it is not the product of sin and .

The expression sin  means sin(), where sine is the name of the function

and , the measure of an acute angle. It is correctly read as ‘sine of angle  ‘.

B

P

Side opposite q

Hypotenuse

q

O

Q

A

### 1. SIX TRIGONOMETRIC FUNCTIONS(Cont…)

• For an acute angle  of the triangle POQ the ratio of the length of the side adjacent to angle  divided by the length of hypotenuse is called the cosine of the angle. For example the cosine of the angle  in the figure is the length of OQ divided by the length of OP. This fact is denoted by

• For an acute angle  of the right triangle POQ the ratio of

the length of the side opposite angle  divided by the

length of the adjacent side to angle  is called

the tangent of the angle. For example, for the angle  of

the triangle POQ in the figure, the tangent of the angle  is

PQ/OQ. This is denoted as

• We abbreviate the two functions respectively as cos , and tan . Each of these abbreviations is written in lower case letters.

B

P

Side opposite q

Hypotenuse

q

O

Q

A

### 1. SIX TRIGONOMETRIC FUNCTIONS(Cont…)

• Besides sine, cosine, and the tangent of an angle , there are three other trigonometric functions, namely cosecant (abbreviated as csc), secant (abbreviated as sec), and cotangent (abbreviated as cot) of the angle . We define them as follows:

• NOTE: Observe that csc, sec , and cot  are reciprocals of sin , cos , and tan  respectively.

Length of the hypotenuse

B

Length of the side opposite q

c

a

q

A

C

b

Length of the side adjacent to angle q.

### 1. SIX TRIGONOMETRIC FUNCTIONS(Cont…)

• Now let us summarize the six trigonometric functions in the following table:

We assume that be an acute angle in a right triangle as shown in the figure. The length of the side opposite  is a, the length of the side adjacent to  is b, and the length of the hypotenuse is c.

NOTE: We take the side opposite to angle A as a, side opposite angle B as b, and side opposite angle C as c.

a = 6

a = 3

b = 2

b = 4

tan  = a/b

= 3/2

tan  = a/b

= 6/4= 3/2

a =1.5

b = 1.5

tan  = a/b

= 1.5/1= 3/2

### 1. SIX TRIGONOMETRIC FUNCTIONS(Cont…)

• REMARKS:

• The figure shows three triangles of varying sizes. In each of these triangles the measure of angle  is the same.

• Notice that all these three triangles have the same shape and the lengths of their corresponding sides are in the same ratio. In each case the tangent function has the same value.

• Thus, the angle  remains the same measure. The lengths of the sides of the triangle change but the ratio of the sides will remain the same as before.

• Like tan , the other trigonometric functions for angle  will have

the same values regardless of the size of the triangle. However, the values of these functions will change if the angle  itself changes.

In general, the trigonometric function values of  depend

only on the measure of the angle , and not on the size of the triangle.

B

26

10

A

b

C

### 2. EVALUATING TRIGONOMETRIC FUNCTIONS

• Example: We have to find the value of each of the six trigonometric functions of  in the figure.

Solution Steps: In order to find the values of six trigonometric functions, first we need to find the

lengths of all the three sides of the triangle (a, b and c).

Given: Side BC, a = 10

Side AB, c = 26

We need to find side AC, that is, b. We find this side using the Pythagorean Theorem

Recall that Pythagorean Theorem: The sum of the squares of the lengths of the sides of a right triangle equals the square of the length of the hypotenuse. That is, if the sides have lengths a and b, and the hypotenuse has length c,

then

Substituting the values, we get,

(10)2 + (b)2 = (26)2

100 + (b)2 = 676

(b)2 = 676 – 100 = 576

b = 576 = 24.

Now that we know the lengths of all the three sides of the triangle, we apply the definitions of the six trigonometric functions of .

B

26

10

A

C

24

### 2. EVALUATING TRIGONOMETRIC FUNCTIONS(Cont…)

Now that we know the lengths of all the three sides of the triangle,

we apply the definitions of the six trigonometric functions of .

NOTE: The values of csc , sec , and cot  can be found by exchanging the numerator and the denominator

of sin , cos , and tan  respectively.

### 3. FUNDAMENTAL IDENTITES

• Many relationships exist among the six trigonometric functions. These relationships are described using trigonometric identities.

• For instance, we define

and

We observe that csc  can also be defined as a reciprocal of sin . Thus, this relationship can be defines as a reciprocal identity,

Hence, we have the reciprocal identities:

### 3. FUNDAMENTAL IDENTITES(Cont…)

• Next, we define

(Multiply and divide both sides by“length

of hypotenuse”)

(From the definitions of sin and cos)

Thus, we have the following Quotient identities:

This is known as a Quotient Identity

### 3. FUNDAMENTAL IDENTITES(Cont…)

• If the values of sin  and cos  are known then we can find the value of each of the remaining four trigonometric functions using reciprocal identities and the quotient identities.

• Example: Given sin  = (2/3) and cos  = (5/3), we have to find the values of the remaining four trigonometric functions.

Solution Steps: Step 1: Use Quotient Identity to find the value of tan .

(Rationalizing the denominator)

### 3. FUNDAMENTAL IDENTITES(Cont…)

Step 2: Now we have the values of sin , cos , and tan , next we find the values of the remaining

three functions using reciprocal identities.

(sin  = (2/3))

(cos  = (5/3))

(Rationalizing the denominator)

(tan  = 2 )

5

B

c

a

A

C

b

### 3. FUNDAMENTAL IDENTITES(Cont…)

• PYTHAGOREAN IDENTITIES:

Now we will find other relationships using the Pythagorean Theorem.

In the right triangle BAC, we have,

BC2 +AC 2 = AB2 (Using Pythagorean Theorem)

That is, we have a2 + b2 = c2(1)

(Dividing both sides by c2)

From the figure we observe, sin  = a and cos  = b ,

c c

Thus from the above equation we get,

(2)

NOTE: For convenience, we will use the notation sin2 for (sin )2 and cos2 for (cos )2.

We rewrite equation 2 using this notation,

(3)

B

c

a

A

C

b

### 3. FUNDAMENTAL IDENTITES(Cont…)

• Similarly, dividing both sides of equation 1 by b2, we get,

(Equation 1: a2 + b2 = c2)

(By definitions of tan  and sec )

(4)

Again dividing both sides of equation 1 by c2, we get,

(5)

Relations 3, 4 and 5 found above are called Pythagorean Identities.

### 3. FUNDAMENTAL IDENTITES(Cont…)

• Example: Given sin  = (1/2) use the Trigonometric Identity to find the value of cos , where  is an acute angle.

We will use the Pythagorean Identity sin2 + cos2 = 1, to find the value of cos .

Substituting the value of sin  in the above identity we get,

### 4. FUNCTION VALUES FOR SOME SPECIAL ANGLES

• Now we will learn how to calculate the values of trigonometric functions for 30° or radians,

45° or radians, and 60° or radians angles which occur frequently in trigonometry.

• TRIGONOMTERIC FUNCTIONS OF 45°

We construct a triangle ABC, right angled at B, in which angle A= angle C = 45°. Thus, the triangle is isosceles, that is, it has two sides equal. Hence, BC = AB.

Assume AB = BC = a

Then AC2 = AB2 + BC2 = a2 + a2 = 2a2 (By Pythagorean Theorem)

Thus, AC = a2.

Remembering that in triangle ABC, angle C = 45°, we get

A

a2

a

45°

C

B

a

2

30°

3

60°

1

A

C

### 4. FUNCTION VALUES FOR SOME SPECIAL ANGLES(Cont…)

• TRIGONOMETRIC FUNCTIONS OF 60°

We construct an equilateral triangle ABC (that is, a triangle in which all sides are equal). Assume that each side has length equal to 2. Observe that here we are considering angle A = 60°.

Now we consider half of the equilateral triangle. We have a right triangle, with

hypotenuse of length 2 and one side of length 1. The length of the third

side be a, we find the length of this side using Pythagorean Theorem.

a2 + 1 = 22

a2 + 1 = 4

a2 = 4 – 1 = 3

a = 3

Thus, we find the values of the trigonometric function as

B

### 4. FUNCTION VALUES FOR SOME SPECIAL ANGLES(Cont…)

• TRIGONOMETRIC FUNCTIONS OF 30°

We will consider the same figure for finding the values of the trigonometric functions of 30 °.

Note: Here we find the values of the trigonometric functions with respect to the

acute angle B = 30°

(Rationalizing the denominator)

NOTE: If a radical appears in the denominator, always rationalize the denominator.

A

2

60°

1

30°

B

3

C

### 4. FUNCTION VALUES FOR SOME SPECIAL ANGLES(Cont…)

• Let us now give the values of trigonometric functions of 30°, 45°, and 60° together in the form of the following table:

• Remark: Looking at the above table we make the following observations:

• The greater is the value of , the greater is the value of sin.

• The greater is the value of , the smaller is the value of cos.

• tan  is increasing with .

### 5. TRIGONOMETRIC FUNCTIONS AND COMPLEMENTS

• Consider a right triangle ABC, right angled at C. Assume that angle A is . Now as the sum of the angles of a triangle is 180°, thus in a right triangle the sum of the two acute angle is 90°. Hence, the two acute angles of a right triangle are complements of each other.

• So, if the degree measure of one acute angle is , then the measure of the other acute angle is (90° - ).

• Considering the acute angle , we already know that,

(1)

• Now considering the acute angle (90° - ), the length of the opposite side is b, and the length of the adjacent side is a.

• Because of this relationship, the sine and cosine are called cofunctions of each other. The name cosine is a shortened form of the phrase complement’s sine.

Thus any pair of trigonometric functions f and g for which

f()= g(90° - ) and g() = f(90° - ) are called cofunctions.

B

c

This angle is (90° - )

a

q

A

C

b

Using (1) above

### 5. TRIGONOMETRIC FUNCTIONS AND COMPLEMENTS(Cont…)

• Similarly we can show that the tangent and cotangent as well as secants and cosecants are also cofunctions of each other.

• Hence, we get the following definition,

Cofunction Identities:

The value of a trigonomteric function of  is equal to the cofunction of the complement of .

sin  = cos(90° - ) cos  = sin (90° - )

tan = cot (90° - ) cot  = tan (90° - )

sec =csc(90° - )csc  = sec (90° - )

NOTE: If  is in radians, replace 90 ° with

• Example: Find a cofunction with the same value as the given expression.

(a) sin 46°

• The value of the trigonometric function of is equal to the cofunction of the complement of , so we need to find the complement of each angle.

• Cofunction of sin  iscos (90° - )

### 5. TRIGONOMETRIC FUNCTIONS AND COMPLEMENTS(Cont…)

• Substituting the value of  in this expression, we get,

sin 46° = cos(90° - 46°) = cos 44°

(b) cot

• Cofunction ofcot  is

• Substituting the value of , we get

### 6. APPLICATIONS

• Many times we are required to find the height of a tower, building, tree, distance of a ship from a light house, width of a river, etc. Though we cannot measure them easily, we can determine these by using knowledge of trigonometric functions.

• Suppose we wish to determine the height of a tree without actually measuring it. We could stand on the ground at a point A at some distance, say 12m, from the foot B of the tree.

Suppose the measure of acute angle BAC is 30°. Then we find the height BC of the tree by using trigonometric functions.

• Thus, we have been able to find the height of the tree using

trigonometric functions.

• Whenever, an engineer faces problem in determining the

width of a river (or height of a tower etc.); which may not be easily possible to measure with a measuring tape, he imagines a big right triangle. One of the sides of this triangle is the line drawn across the river (or a line drawn vertically down to the ground from the top of the tower etc.) and such that any one of the other two sides of the triangle and one of the angles can be easily measured by using any surveying instrument.

• Knowing a side and angle, the engineer can use his knowledge of trigonometric functions to calculate the unknown side, that is, the width of the river (or height of the tower) as done above.

C

Tree

30°

B

A

12 m

### 6. APPLICATIONS(Cont…)

• Before we proceed to solve problems based on applications of trigonometric functions, let us first define a few terms:

• Line of sight: Suppose we are viewing an object standing on the

ground. Clearly, the line of sight (or the line of vision) to the object

is the line from our eyes to the object, we are viewing.

• Angle of Elevation: If the object is above the horizontal level

of the eyes(that is, it is above the eye-level), we have

to turn our heads upwards to view the object. In this process

our eyes move through an angle. Such an angle is

called the angle of elevation of the object from our eyes.

• Angle of Depression: Suppose a boy, standing on the roof

a building, observes an object lying on the ground at some

distance from the building. In this case, he has to move

his head downwards to view the object. In this process

his eyes again move through an angle. Such an angle

is called the angle of depression of the object from the

location of his eyes.

Object

Line of sight

Angle of Elevation

Horizontal Line

Angle of Depression

Horizontal Line

Line of sight

Object

14 m

10 m

### 6. APPLICATIONS(Cont…)

• Example: A flagpole that is 14 meters tall casts a shadow 10 meters long. Find the angle of elevation of the sun to the nearest degree.

Given that the height of the flagpole is 14 m and

the length of its shadow is 10 m. We are asked to

find angle . In order to find angle , we begin with the

tangent function.

From the definition of tangent function we know,

Substituting the values we get,

Using a calculator, we find that the value of the angle   54°

Thus, the angle of elevation is 54° approximately.

### 7. SUMMARY

Let us recall what we have learnt so far:

• There are six trigonometric functions which an be defined as

• Reciprocal Identities:

• Quotient identities:

### 7. SUMMARY(Cont…)

• Pythagorean Identities:

• Cofunction Identities:

sin  = cos(90° - ) cos  = sin (90° - )

sec =csc(90° - )csc  = sec (90° - )

tan = cot (90° - ) cot  = tan (90° - )