Lecture 14 section 5.3 trig fcts of any angle

MATH 107

Section 5.3

Unit Circle Approach

Properties of the Trigonometric Functions

y

x

x

yx

r

r

x

y

r

r

y

cottan

seccos

cscsin

Remember the six trigonometric functions

defined using a point (x, y) on the terminal

side of an angle, .

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TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE

A circle with radius 1 centered at the origin of a rectangular coordinate system is a unit circle.

In a unit circle, s = rθ = 1·θ = θ, so the radian measure and the arc length of an arc intercepted by a central angle in a unit circle are numerically identical.

The correspondence between real numbers and endpoints of arcs on the unit circle is used to define the trigonometric functions of real numbers, or the circular functions.


Let t be any real number and let P = (x,y) be the point on the unit circle associated with t. Then

UNIT CIRCLE DEFINITIONSOF THE TRIGONOMETRIC FUNCTIONS

OF REAL NUMBERS

yt sin )0( 1

csc yy

t

xt cos )0( 1

sec xx

t

)0( tan xx

yt )0( cot y

y

xt

We will first look at the special angles called the quadrantal angles.

90

180

270

0

The quadrantal anglesare those angles that lie on the axis of the Cartesian coordinate

system: 0° , 90°, 180°, 270° and 360° .

We can count the quadrantal angles in terms of . radians2

radians2

radians2

2

radians2

3

radians

360,0

90

180

270

radians57.1

radians14.3

radians71.4

radians 2

radians, 0

radians,0

radians28.6

radians2

0 radians

radians2

3

radians

radians2

0

90

180

270

360

(1, 0)

or

undefinedis0cot01

00tan

10sec10cos

undefinedis0csc00sin

radians2

0 radians

radians2

3

radians

radians2

0

90

180

270

360or

(0, 1)

02

cotundefinedis2

tan

undefinedis2

sec02

cos

12

csc12

sin

(-1, 0)

undefinediscot0tan

1sec1cos

undefinediscsc0sin

(0, -1)

02

3cotundefinedis

2

3tan

undefinedis2

3sec0

2

3cos

12

3csc1

2

3sin

Now let’s cut each quadrant in half,

which basically gives us 8 equal

sections.

0

4

4

2

4

4

4

6

4

3

4

5

4

7

4

8We can again count around the

circle, but this time we will count in

terms of

radians.

4

.4

8,

4

7,

4

6,

4

5,

4

4,

4

3,

4

2,

4

1 and

4

2

2

2

3

Then reduce appropriately.

45

90

135

180

225

270

315

360

2

45

45 The lengths of the legs of the 45 – 45 – 90

triangle are equal to each other because their

corresponding angles are equal.

If we let each leg have a length of 1, then we find

the hypotenuse to be using the Pythagorean

theorem.

2

1

1

2

You should memorize this triangle or at

least be able to construct it. These angles

will be used frequently.

Next we will look at two special triangles: the 45 – 45 – 90 triangle and the 30 – 60 – 90 triangle.

These triangles will allow us to easily find the trig functions of the special angles, 45 , 30 , and 60 .

(Knowing this derivation is not necessary,

but knowing the ratio of sides and angles is.)

45

45

1

1

2

145cot145tan

245sec2

245cos

245csc2

2

2

145sin

Using the definition of the trigonometric functions as the ratios of the sides of a right triangle,

we can now list all six trig functions for a angle.45

For the 30 – 60 – 90 triangle, we will construct an equilateral triangle (a triangle with 3 equal angles of

each, which guarantees 3 equal sides).60

If we let each side be a length of 2, then cutting

the triangle in half will give us a right triangle

with a base of 1 and a hypotenuse of 2. This

smaller triangle now has angles of 30, 60, and

90 .

We find the length of the other leg to be using

the Pythagorean theorem.

3

3

60

1

2

30

You should memorize this triangle or at least

be able to construct it. These angles, also,

will be used frequently.(Knowing this derivation is not necessary,

but knowing the ratio of sides and angles is.)

Again, using the definition of the trigonometric functions as the ratios of the sides of a right triangle, we

can now list all the trig functions for a 30 angle and a 60 angle.

330cot3

3

3

130tan

3

32

3

230sec

2

330cos

230csc2

130sin

3

3

3

160cot360tan

260sec2

160cos

3

32

3

260csc

2

360sin

60

30

1

32

All ISine II

III

Tangent

IV

Cosine

I. All

II. Students

III. Take

IV. Calculus

Positive Values for Trigonometric Functions

Note: Because they are reciprocals, the sign of cosecant matches the sign of sine,

the sign of secant matches the sign of cosine, and the sign of cotangent matches

the sign of tangent. Always.


EXAMPLE 4 Determining the Quadrant in Which an Angle Lies

Solution

If tan θ > 0 and cos θ < 0, in which quadrant does θ lie?

Because tan θ > 0, θ lies either in quadrant I or in quadrant III. However, cos θ > 0 for θ in quadrant I; so θ must lie in quadrant III.


DEFINITION OF A REFERENCE ANGLE

Let be an angle in standard position that is not a quadrantal angle.

The reference angle for is the positive acute angle (“theta prime”) formed by the terminal side of and the x-axis.


TRIGONOMETRIC FUNCTION VALUES OF COTERMINAL ANGLES

360sinsin n n 2sinsin

360coscos n

These equations hold for any integer n.

in degrees in radians

n 2coscos



Quadrant I Quadrant II



Quadrant III Quadrant IV


EXAMPLE 6 Identifying Reference Angles

Find the reference angle for each angle .

a. = 250º b. = c. =

5.75Solution

a. Because 250º lies in quadrant III, = º. So = 250º 180º = 70º.

• Because lies in quadrant II, = π .

So = π


EXAMPLE 6 Identifying Reference Angles

Solution continued

c. Since no degree symbol appears in θ = 5.75,

has radian measure. Now ≈ 4.71 and

2π ≈ 6.28. So lies in quadrant IV and

= 2π . So

= 2π – ≈ 6.28 – 5.75 = 0.53.


USING REFERENCE ANGLES TO FIND TRIGONOMETRIC FUNCTION VALUES

Step 1 Assuming that > 360º or θ < 0°, find a coterminal angle for with degree measure between 0º and 360º. Otherwise, go to Step 2.

Step 2 Find the reference angle for the angle resulting from Step 1. Write the trigonometric function of .


USING REFERENCE ANGLES TO FIND TRIGONOMETRIC FUNCTION VALUES

Step 3 Choose the correct sign for the trigonometric function value of θ based on the quadrant in which it lies. Write the given trigonometric function of θ in terms of the same trigonometric function of θ with the appropriate sign.


EXAMPLE 8 Using the Reference Angle to Find Values of Trigonometric Functions

30º360º

3ta

330º

n tan 30º3

Find the exact value of each expression.

Step 1 0º < 330º < 360º; find its reference angle.

Step 2 330º is in Q IV; its reference angle is

a.Solution

.



3tan t 3330 0ºnº a

3

Solution continued

Step 3 In Q IV, tan θ is negative, so

b. Step 1

11

6

is between 0 and 2π coterminal with

.

.



1159 2 3sec sec

66 6sec

3

Solution continued

Step 3 In Q IV, sec θ > 0; so

Step 2

2

2 3sec sec

6 3

6 6

11

is in Q IV; its reference angle is

.

.

y

x

330

3

60

1

2

30 AS

T C

3330cot3

3

3

1330tan

3

32

3

2330sec

2

3330cos

2330csc2

1330sin

Example 1 Continued: The six trig functions of 330 are:

30

Your reference angle is 30 .

y

x

Example 1: Find the six trig functions of 330 .

First draw the 330 degree angle.

3

60

1

2

30

AS

T C

330

30

(Answers on next page.)

Your reference angle is 30 .

y

x

Example 2: Find the six trig functions of .

3

60

1

2

30

3

4

First determine the location of .

3

4

3

3

2

3

3

3

3

3

4

3

(Answers on next page.)

Your reference angle is .3

3

60

1

2

30

AS

T C

Example 2: Find the six trig functions of . 3

4

y

x

3

3

2

3

4

3

3

3

3

1

3

4cot3

3

4tan

23

4sec

2

1

3

4cos

3

32

3

2

3

4csc

2

3

3

4sin

Your reference angle is .

3

0 radians

Example 3: Find the exact value of cos .

4

5

We will first draw the angle to determine the quadrant.

4

5

4

4

2

4

3

4

4

AS

T C

45

45

1

12

4

Your reference angle is .4

0 radians

Problem 3: Find the sin .

AS

T C

6

5

6

6

2

6

36

4

6

5

is the reference angle.6

6

0 radians

Problem 7: Find the exact value of cos .

We will first draw the angle to determine the quadrant.

AS

T CNote that the reference angle is .

4

4

13

4

4

2

4

4

4

64

5 4

7

4

8

4

94

10

4

114

3We see that the angle is located in the

3rd quadrant and the cos is negative in the

3rd quadrant.

4

13

4

12

4

13

4


EXAMPLE 5 Evaluating Trigonometric Functions

Solution

3tan

3

2 2

y

x

Since tan θ > 0 and cos θ < 0, θ lies in Quadrant III; both x and y must be negative.


EXAMPLE 5 Evaluating Trigonometric Functions

Solution continued

With , , and we can find

sin

132

and s c .

3

e

rx y

3

1

3 13si

3n

13r

y

13 13se

2c

2x

r

,

Technology

Lecture 14 section 5.3 trig fcts of any angle