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TRIGONOMETRIC FUNCTIONS Chapter 5

TRIGONOMETRIC FUNCTIONS Chapter 5. Section 5.1 Angles and Degree Measure Learn how to convert decimal degree measures to degrees, minutes, and seconds

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TRIGONOMETRIC FUNCTIONS

Chapter 5

Section 5.1

Angles and Degree Measure Learn how to convert decimal degree measures to

degrees, minutes, and seconds. Find the number of degrees in a given number of

rotations Identify angles that are coterminal with a given angle

5.1 Angles and Degree Measure

Vertex Endpoint of an angle

Initial Side The ray of the angle that is fixed

Terminal Side The second ray that rotates to form the angle

Standard Position An angle with its vertex at the origin and

its initial side along the positive x-axis

5.1 Angles and Degree Measure

5.1 Angles and Degree Measure

Degree Most common unit to measure an angle

Minutes A degree is subdivided into 60 equal parts called minutes (1’)

Seconds A minute is subdivided into 60 equal parts called seconds

(1”)Example 1:

Change 15.735 degrees to degrees, minutes, and seconds. 15.735o=15o+(0.735*60)’ =15o+44.1’ =15o+44’+(0.1*60)” =15o+44’+6”

5.1 Angles and Degree Measure

Example 2: Change 329.125 degrees to degrees, minutes, and

seconds. 329o+7’+30”

Example 3: Change 39o+5’+34” to degrees

39o+5’+34”=39o+5’*(1 o/60’) +34” (1 o/3600’) 39.093o

Example 4: Change 35o+12’+7” to degrees

35.202o

5.1 Angles and Degree Measure

Quadrantal Angle If the terminal side of an angle that is in standard

position coincides with one of the axes How many quadrantal angles are there?

4 What are their measures?

90 degrees 180 degrees 270 degrees 360 degrees

5.1 Angles and Degree Measure

Give the angle measure represented by the rotation about the axis 5.5 Rotations clockwise

Which way do we rotate to go clockwise? Negative, clockwise rotations ALWAYS have negative measures

5.5 * -360 = -1980 degrees 3.3 Rotations counterclockwise

Which way do we rotate to go counterclockwise? Positive, counterclockwise rotations ALWAYS have positive

measures 3.3*360 = 1188 degrees

9.5 Rotations clockwise -3420 degrees

6.75 Rotations counterclockwise 2430 degrees

5.1 Angles and Degree Measure

Coterminal Angles If α is the degree measure of an angle, then all angles

measuring α+ 360k degrees, where k is an integer; are coterminal with α.

Two angles in standard position that have the same initial side

Identify all angles that are coterminal with a 45 degree angle. Find one positive and one negative angle that are also coterminal. All angles having a measure of 45o+ 360ko

Positive Angle 45o+ 360*(1)o=405o

Negative Angle 45o+ 360*(-2)o=-675o

5.1 Angles and Degree Measure

Identify all angles that are coterminal with a 294 degree angle. Find one positive and one negative angle that are also coterminal. All angles having a measure of 294o+ 360ko

Positive Angle 294o+ 360*(1)o=654o

Negative Angle 294o+ 360*(-1)o=-66o

5.1 Angles and Degree Measure

If an angle with 775 degrees is in standard position, determine a coterminal angle that is between 0 and 360 degrees. State the quadrant in which the terminal side lies. Find the number of rotations about the axis by dividing 775 by

360 = 2.15278 Since we need an angle between 0 and 360 degrees, what

rotation number should we use? Subtract 2 and use .15278 α= .15278*360 Could also take the number and continue to subtract 360

until you get a number between 0 and 360. 55o

What quadrant does the terminal side fall in? Quadrant 1

5.1 Angles and Degree Measure

If an angle with -777 degrees is in standard position, determine a coterminal angle that is between 0 and 360 degrees. State the quadrant in which the terminal side lies. 303o

What quadrant does the terminal side fall in? Quadrant 4

5.1 Angles and Degree Measure

Reference Angle The acute angle formed by the terminal side of the

given angle and the x-axisReference Angle Rule

For any angle α, 0o< α <360o, its reference angle α’ is defined by: α, when the terminal side is in Quadrant I. 180o- α, when the terminal side is in Quadrant II. α – 180o, when the terminal side is in Quadrant III. 360o- α, when the terminal side is in Quadrant IV.

5.1 Angles and Degree Measure

Find the measure of the reference angle for an angle with a measurement of 120 degrees What Quadrant does this angle’s terminal side fall in?

Between 90o and 180o

Quadrant II We use which formula? 180o- α 180o-120o=60o

Find the measure of the reference angle for an angle with a measurement of -135degrees First we have to find a positive coterminal angle

360o-135o = 225o

What quadrant does this angle’s terminal side fall in? Between 180o and 270o

Quadrant III α – 180o

225o-180o=45o

5.1 Angles and Degree Measure

Find the measure of the reference angle for an angle with a measurement of 312 degrees 48o

Find the measure of the reference angle for an angle with a measurement of -195 degrees 15o

Section 5.2

Trigonometric Ratios in Right Triangles Learn how to find the values of trigonometric ratios

for actue angles of right triangles

5.2 Trigonometric Ratios in Right Triangles

What is a Right Triangle? A triangle with a 90 degree angle in it. How can we classify the other two angles in the right

triangle? Must be acute Are also complementary

What are the parts of a Right Triangle?< 1 + < 2 = 90o

Side C is the hypotenuseSides A and B are the legs

1

2

5.2 Trigonometric Ratios in Right Triangles

When looking at one specific acute angle in a triangle, we can classify the legs by: Adjacent Side: the leg that is a side of the acute angle Opposite Side: the leg that is the side opposite the

angle Looking at Triangle ABC, what are the adjacent and

opposite sides for < B?BHypotenuse

Adjacent

Opposite

5.2 Trigonometric Ratios in Right Triangles

Trigonometric Ratios The ratios of the sides of right triangles based on a specific

acute angle within the right triangleEasy way to remember Sin, Cosine, and Tangent

Ratios SOHCAHTOA

Sine Ratio of the side opposite Θ and the hypotenuse

Cosine Ratio of the side adjacent Θ and the hypotenuse

Tangent Ratio of the side opposite Θ and the side adjacent to Θ

5.2 Trigonometric Ratios in Right Triangles

5.2 Trigonometric Ratios in Right Triangles

Find the values of sine, cosine, and tangent for <B. What do we need to do first?

Find the third side’s length How do we do that?

Pythagorean Theorem, A2 + B2 = C2

C=3*(85)1/2

Which angle is our Θ? Sin Θ

Opposite/hypotenuse 18/33

Cos Θ Adjacent/hypotenuse 3*(85)1/2/33

Tan Θ Opposite/Adjacent 18/3*(85)1/2

18 m

33 m

C

B

A

Θ

3*(85)1/2 m

5.2 Trigonometric Ratios in Right Triangles

Find the values of sine, cosine, and tangent for <A. Find the third side’s length

C=17 m Which angle is our Θ? Sin Θ

8/17 Cos Θ

15/17 Tan Θ

8/15

8 m

15 m

C

B AΘ

17 m

5.2 Trigonometric Ratios in Right Triangles

Cosecant Opposite of sine, cscΘ = 1/sin Θ Hypotenuse/side opposite

Secant Opposite of cos, secΘ = 1/cosΘ Hypotenuse/side adjacent

Cotangent Opposite of tan, cotΘ = 1/tanΘ Side Adjacent/ Side opposite

5.2 Trigonometric Ratios in Right Triangles

If cos Θ = ¾, what is sec Θ? sec Θ = 1/cos Θ =1/(3/4) =4/3

If sin Θ=0.8, what is csc Θ? =1.25

If csc Θ = 1.345 find sin Θ? sin Θ = 1/csc Θ =1/1.345 =.7435

If cot Θ = 6/5, what is tan Θ? =5/6

5.2 Trigonometric Ratios in Right Triangles

Find the values of the six trigonometric ratios for <E. First find third side using pythagorean theorem (58)1/2 m Cos E = 3* (58)1/2 /58 Sin E= 7* (58)1/2 /58 Tan E=7/3 Sec E= (58)1/2 /3 Csc E= (58)1/2 /7 Cot E=3/7

7 m

3 m

D

F EΘ

(58)1/2 m

5.2 Continued

Special Triangles 30O-60O-90O

45O-45O-90O

What are the special relationships we know about these triangles? 30O-60O-90O

5.2 Continued

What are the special relationships we know about these triangles? 45O-45O-90O

5.2 Continued

Trigonometric Ratios for 30o, 60o, 90o

5.2 Continued

Cofunctions Trigonometric functions that are equal when their

arguments are complementary angles, such as sine and cosine, tangent and cotangent, and secant and cosecant.

Sin Θ = Cos (90O- Θ) Cos Θ = Sin (90O- Θ) Tan Θ=Cot (90O- Θ) Cot Θ=Tan (90O- Θ) Sec Θ=Csc (90O- Θ) Csc Θ=Sec (90O- Θ)

Section 5.3 Trigonometric Functions on the Unit Circle

Find the values of the six trigonometric functions using the unit circle

Find the values of the six trigonometric functions of an angle in standard position given a point on its terminal side

Trigonometric Functions on the Unit Circle

Unit Circle A circle with a radius of 1 Usually with the center on the

origin on the coordinate system Symmetric with respect to the

x-axis, y-axis, and the origin

Trigonometric Functions on the Unit Circle

Unit Circle Consider an angle between 0O and 90O in standard

position Let P(x,y) be where the angle intersects with the unit

circle Draw a perpendicular segment from intersection point

back down to the positive x axis Creates a right triangle Find the sin Θ and cos Θ

Sin Θ = y; Cos Θ = x

Trigonometric Functions on the Unit Circle

Sine and Cosine on the Unit Circle If the terminal side of an angle Θ in standard position

intersects the unit circle at P(x,y), then cos Θ = x and sin Θ=y.

Trigonometric Functions on the Unit Circle

Trigonometric Functions on the Unit Circle

Circular Functions Functions defined using the unit circle Ie Sin and Cosine How can we define the other cosine functions on the

unit circle? Tan Θ = y/x Csc Θ =1/y Sec Θ =1/x Cot Θ = x/y

Trigonometric Functions on the Unit Circle

Use the unit circle to find each value Cos (-180O)

Which way do you go for negative angles? Clockwise

What is the ordered pair of the intersection of this angle on the unit circle? (-1,0) Cos Θ = x-axis Cos (-180O) = -1

Sec(90O) Where is the angle located ?

Terminal Side is on positive y axis Where is the intersection on the unit circle?

Intersection at (0,1) Sec Θ = 1/x Sec(90O)=1/0 = undefined

Trigonometric Functions on the Unit Circle

Use the unit circle to find each value Sin(-90O)

-1 Cot(270O)

0

Trigonometric Functions on the Unit Circle

Use the unit circle to find the values of all six trigonometric functions for a 210 degree angle. What is the intersection with the unit circle?

(-√(3) /2, -1/2) Sin Θ = -1/2 Cos Θ =- √(3) /2 Tan Θ = √(3) /3 Csc Θ =-2 Sec Θ =-2 √(3) /3 Cot Θ = √(3)

Unit Circle Quiz

Use the unit circle to find each value Tan 3600

Cos 450

Sin(-600) Csc(-2100 ) Sec(2250 )

Name the six trigonometry functions for the angles below 1500

4200

Trigonometric Functions on the Unit Circle

What if the angle doesn’t fall within the unit circle? What if the length is greater or less than 1? Use length R instead of 1 in unit circle and R =

(x2+y2)1/2

Trigonometric Functions on the Unit Circle

What are the six trigonometric values using length r? Sin Θ =y/r Cos Θ =x/r Tan Θ =y/x Csc Θ =r/y Sec Θ =r/x Cot Θ =x/y

Trigonometric Functions on the Unit Circle

Find the six trigonometric functions for angle Θ in standard position if a point with the coordinates (-15, 20) lies on its terminal side. Draw Figure on xy axis What are the side measures?

Leg = -3 Leg = 4 Hypotenuse = 5

Where is the theta located? At the point they gave us

What are the six functions? Sin Θ =4/5 Cos Θ =-3/5 Tan Θ=-4/3 Csc Θ =5/4 Sec Θ =-5/3 Cot Θ =-3/4

Review Quiz 5.3

Suppose Θ is an angle in standard position whose terminal side lies in Quadrant III. If sin Θ = -4/5, find the values of the remaining five trigonometric functions of Θ? What do we need to do first?

Draw figure Next?

Find Missing Side +/-3, use negative 3 because Quadrant III

Find 5 remaining trig functions. Cos Θ = -3/5 Tan Θ =4/3 Csc Θ =-5/4 Sec Θ =-5/3 Cot Θ =3/4

5.4 Applying Trigonometric Functions

Use trigonometry to find the measures of the sides of right triangles

Make sure calculators are in degrees and not radians Press Mode Scroll down to third line and arrow left one Hit enter Quit Quick Check Cos(90) = 0

Applying Trigonometric Functions

In triangle PRQ, P = 35o and r = 14. Find Q. First draw figure How is Q related to Θ?

Adjacent Side What function should we use?

Cosine Cos P = q/r Cos 35=q/14 14*cos(35)=Q Q is about 11.5

Applying Trigonometric Functions

Angles of Elevation The angle between a horizontal line and the line of sight

from an observer to an object at a higher level Angles of Depression

The angle between a horizontal line and thel ine of sight from the observer to an object at a lower level

5.5 Solving Right Triangles

Evaluate inverse trig functions.Find missing angle measurements.Solve right triangles.

Solving Right Triangles

Inverse of a Trigonometric Function The arcsine, arccosine, and arctangent relations with

their corresponding trigonometric functions Arcsine

Sin x = √3/2 Can be written as x = arcsine √3/2 or x = sin-1 √3/2 Read this as x is an angle whose sine is √3/2

Same for other two trig functions Arccosine Arctangent

Solving Right Triangles

Sin x = √3/2 X is an angle with sine √3/2 X = arcsine √3/2 or x = sin-1 √3/2 60 o, 120 o, or any coterminal angles with these

Tan x = 1 45 o,225o

Sin x = -1/2 210 o, 330 o

Solving Right Triangles

Evaluate and assume angles are in Quadrant I Cos(arcsin 2/3)

Let B = arcsin 2/3 Sin B = 2/3 Draw figure in Quadrant I Find X = √5/3 Try in calculator

Tan(cos-1 4/5) 3/4

Solving Right Triangles

Solve the Triangle with A = 33o and B = 5.8 First fill in known amounts Find Angle B

Two acute angles in a right triangle are complementary

B =90-33o = 57o degrees Find side A

Tan A = a/b Tan 33o = a / 5.8 5.8 Tan 33o = a a = 3.767

Find side C Cos A = b/c Cos 33o = 5.8/c C=5.8/cos 33o

c=6.916

B

CA

Solving Right Triangles

Solve the Triangle with K = 40o and k = 26 < L = 50o l = 31.0 j = 40.4 K

LJ

Section 5.6The Law of Sines

Solve triangles by using the Law of Sines if the measures of two angles and one side are given

Find the area of a triangle if the measures of sides and the included angle or the measures of two angles and a side are given.

The Law of Sines

Let Triangle ABC by any triangle with a, b, and c representing the measures of the sides opposite the angles with measures A, B, and C, respectively. Then the following is true:

a = b = c SinA SinB SinC

The Law of Sines

Solve triangle ABC if A = 33o, B = 105o, and b = 37.9 Find < C by subtracting the other two angle measures

from 180 C= 42o

a = 37.9 Sin33o Sin105o A = 21.37o

c = 37.9 Sin42o Sin105o C=26.25o

The Law of Sines

Area of a Triangle Let triangle ABC be any triangle with a, b, and c

representing the measures of the sides opposite the angles with measurements A, B, C, respectively. Then the area K can be determined using: K = ½ * b* c* sinA K = ½ * a* c* sinB K = ½ * b* a* sinC

The Law of Sines

Find the area of triangle ABC if a = 4.7, c = 12.4, and B = 47o 20’ First convert B to decimal degrees B = 47.33o

K = ½ * a* c* sinB K = ½ * 4.7*12.4*sin47.33o

K = 21.4 units2

Find the area of triangle ABC if b = 21.2, c = 16.5, and A = 25o

K = ½ * b* c* sinA K = 73.9 units2

The Law of Sines

Area of a Triangle Let triangle ABC be any triangle with a, b, and c

representing the measures of the sides opposite the angles with measurements A, B, C, respectively. Then the area K can be determined using: K = ½*a2*sinB*sinC

SinA K = ½*b2*sinA*sinC

SinB K = ½*c2*sinB*sinA

SinC

The Law of Sines

Find the area of triangle DEF if d = 13.9, D = 34.4o, and E = 14.8o

Find the measure of <F first, 180-34.4-14.8 <F = 130.8o

K = ½*d2*sinE*sinF SinD

K = ½*13.92*sin 14.8o *sin 130.8o

SinD 34.4o

K =33.065 units2

Section 5.8The Law of Cosines

Let Triangle ABC by any triangle with a, b, and c representing the measures of the sides opposite the angles with measures A, B, and C, respectively. Then the following is true: a2=b2 + c2 -2*b*c*cosA b2=a2 + c2 -2*a*c*cosB c2=b2 + a2 -2*b*a*cosC

The Law of Cosines

Solve the triangle with the following information:A = 120o,b=9, c=5

Solve for Side a a2=b2 + c2 -2*b*c*cosA a2=92 + 52 – 2*9*5*cos(120o) a2=426.97 a=12.3 Now use Law of Sines to find <B 12.3 = 9 Sin120o sinB B=arcsin (9*sin120o) 12.3B= 39.3o

C= 180o – 120o-39.3o

C=20.7o

The Law of Cosines

Try a triangle with A = 39.4o,b = 12, c = 14 B=58.2o

C = 82.4o

a = 9.0Try a triangle with a =19, b = 24.3, c = 21.8

A = 48.3o

B = 72.7o

C = 59o

The Law of Cosines

Hero’s FormulaIf the measures of the sides of a triangle are

a, b, and c, then the area, K, of the triangle is found as follows: K = √s*(s-a)*(s-b)*(s-c)

Where S = ½ (a+b+c)

The Law of Cosines

Find the area of the triangle with the following information:

a=72 cmb=83 cmc=95 cm

K = √s*(s-a)*(s-b)*(s-c) Where S = ½ (a+b+c) S=125 K=2889.2 cm2