Lecture 27 Sections 4.3 and 4.4

Section 4.3

Right Triangle Trigonometry

SOH CAH TOA

Cofunctions of complementary angles

Trigonometric identities

Section 4.4 Trigonometric Functions of Any

Angle

Reference angle

Angle of elevation and angle of depression

L27 - 1

Def. Let θ be an acute angle of a right triangle. Thesix trigonometric functions of the angle θ are definedas follows.

sin θ = csc θ =

cos θ = sec θ =

tan θ = cot θ =

L27 - 2

HYPOTENUSE OPPOSITE

10ADJACENT

AL1

OPP HYPSOHTpp

ADJ HYPCAHHYP ADJ

0 Toa ADIADJ OPP

SOH CAH TO Ap s

ex. If cos θ =3

4, sketch a right triangle with acute

angle θ, and find the other values of the five trigono-metric functions of θ.

Checkpoint: Lecture 27, problem 1

L27 - 3

COS 0 3 ADIdHYP ft HYP4 t

OPP PYTHAGOREANTHEOREM

42 32 t COPPIvj 0 16 9 t Copp I3g 7 PP5 Opp ItADJ

Opp 57

Sino ftp.pp E Csce HotpIp ftp.r4ff4coso

Iseco HffgI

z4tanooafj FI cot o Appt Ff Ffg 3

Theorem: Cofunctions of complementary anglesare equal. That is,

sin(90◦ − θ) = cos θ cos(90◦ − θ) = sin θ

tan(90◦ − θ) = cot θ cot(90◦ − θ) = tan θ

sec(90◦ − θ) = csc θ csc(90◦ − θ) = sec θ

a

b

c

θ

ex. Evaulate:

1) sin 36◦ − cos 54◦

2)tan 56◦

cot 34◦

Checkpoint: Lecture 27, problem 2

L27 - 4

EX COS 700 sin zoo icos 0 _ADI ibiBK sink200 sin 900700 Goo HYP if7051700 sin 900 o j OPP.I.be

HYP ICI

coso s.in 9oo o

90 54 36

sin 900 540 costs40 cos 540 cos 49 0

tan 900 340 Cot 340otL34 Cott 440

Trigonometric Identities

1. Reciprocal Identities

sin θ =1

csc θcsc θ =

1

sin θ

cos θ =1

sec θsec θ =

1

cos θ

tan θ =1

cot θcot θ =

1

tan θ

2. Quotient Identities

tan θ =sin θ

cos θcot θ =

cos θ

sin θ

3. Pythagorean Identities

sin2 θ + cos2 θ = 1

1 + tan2 θ = sec2 θ

1 + cot2 θ = csc2 θ

L27 - 5

MUST KNOW THESE

a

L

sin 20 050 1

sin 20 050 1 ot l

atan 01 1 see 0

I t 050 ITino sin ltcotzo cs.co

ex. Let θ be an acute angle such that sin θ =1

3.

Use trigonometric identities to find

1) cos θ

2) tan θ

Checkpoint: Lecture 27, problem 3

L27 - 6

ACUTEANGLEsin 20 050 1 2 I ONLY

I cos 9 LPOSITIVEVALUES

f tcoso I LOSE too9cos't I tf FI3

COULD USE tan I see ZI3

OR QUICKER

tano siisfozfyj f.frEra FT

Lecture 27, Part II: Section 4.4

Trigonometric Functions of Any Angle

Def. Let θ be an angle in standard position and let(x, y) be a point on the terminal side. Ifr =

√

x2 + y2 is the distance from the origin to thepoint (x, y), then

sin θ = csc θ =

cos θ = sec θ =

tan θ = cot θ =

θ

(x,y)

NOTE: If x = 0, tan θ and sec θ are undefined. Ify = 0, cot θ and csc θ are undefined.

L27 - 7

NOT NECESSARILY A UNIT CIRCLE WHEREr I

y eF y

t er x

1 xx y

RADIUS HYPOTENUSEi rY on I

ex. Given point (−5, 12) on the terminal side of theangle θ, find the six trigonometric function values ofthe angle θ.

ex. If cos θ =3

5and θ lies in Quadrant IV, find the

values of all the trigonometric functions.

Checkpoint: Lecture 27, problem 4

L27 - 8

I 5,12f 2 Of 12

144 TOI 13 l

5Sino _r1 if cKO F

ELATESE tr

050 1 1 see 1 SOHCAHTOA Tr 13 TO1tano I L cot o yI

55 12

cos 0 1 3 Sec fr o

3Sino In csco Iy to 4

r i 4J i

tano f 4g Coto Ky ZITS

NEGATIVEB C QHI

Tty2 52

y2 25 9

y2 16

y 4

Signs of the Trigonometric Functions

ll ine

angent osine

A

T C

S

ex. Find the quadrant in which θ lies if tan θ < 0and sin θ < 0

ex. Given sec θ = 5 and sin θ < 0, find csc θ.

L27 - 9

ALL STUDENTSTAKE CALCULUS

1 it t t

l Ct

tano IY 15 NEGATIVE X IS POSITIVE

QI

1 Sec 0 55,2 I x L r 5

ALSO Y IS NEGATIVE SINCE sin 0 LO

WE ARE IN QII

O I

4 r 525 Hy Z

J 24 42Ys Tty I

rzy y 4 76co I off EE r 4 256

Reference Angles

Def. Let θ be an angle in standard position. Thereference angle is the acute angle θ′ formed bythe terminal side of θ and the x-axis.

θθ

θθ

L27 - 10

i iii i

q0 Oi n e0 0 01 01 1800

01 1800 O0 1 0 _IT0 11 0

w lo YO't Ii ri

0 11 0 01 0121101211 0O IT _O

OR 0 01 3600

0 180 to O 360 O

0 1800 0

ex. Find the reference angle for

1) θ =5π

3

2) θ = 870◦

Checkpoint: Lecture 27, problem 5

Evaluating Trigonometric Functions for Any

Angle

1. Find the reference angle θ′ associated with theangle θ.

2. Determine the sign of the trigonometric functionof θ by noting the quadrant in which θ lies.

3. The value of the trigonometric function of θ isthe same, except possibly for sign, as the value of thetrigonometric function of θ′.

L27 - 11

997O ZIT IT

31061 5115 O 61T3

3 IT3

O II3

FIND SMALLESTPOSITIVECOTERMINAL ANGLE THAT IS LESSTHAN360 1800 0 08700 3600 5100 y 01 180 O5100 3600 1500 O 01 1800 150001 300

ex. Use the reference angle to find

1) sin 240◦

2) cot 495◦

3) sin

(

16π

3

)

4) sec(

−π

4

)

L27 - 12

Sin2400 sin 600

o ioooi oi s.fi III240 1800101

01 600l i I

C it 4950 3600 1350

f35 Lot 4957 cot 13504501

cot 450

0 01 1800 EFI ray01 1800 0 1800 1350 450 JI I

2

6ft 217 1631 6,1 1031 Sin lbj sin 415

Ift 21T loft 6,1 431 Sin EsB

01 415 it 2It

4Iz LI27,1 817 7,1 secftfj seyttyj

seytfj fz fz.FIEiEzrz

O'IIIII oh.at E pti

ex. If tan θ =2

3and θ is in Quadrant III, find cos θ.

Method 1: Use trigonometric identities.

Method 2: Use the reference angle.

Checkpoint: Lecture 27, problem 6

L27 - 13

ta n O t I see20

Z t I see Zoe

TecateIg sect

secco i tf FI3

cos o seto3

tan D OPIADJ

3

iii so Affp fff 3

I 25 t l 35 h13 h Z h 13T

Applications

depressionangle of

elevationangle of

object

object

observer

observer

ex. A giant redwood tree casts a shadow 600 feetlong. Find the height of the tree if the angle ofelevation of the sun is 30◦.

L27 - 14

tan 300_Off 00TREE h

h 600 tan 300

1300 E IE t 4 60013K 600ftSHADOW h 20053 FT

fan 300 1IzIz Z

B

f E3TI33

ex. From a point on the ground 500 feet from thebase of a building, an observer finds that the angleof elevation to the top of the building is 30◦ and thatthe angle of elevation to the top of a flagpole atopthe building is 32◦. Find the height of the buildingand the length of the flagpole.

L27 - 15

If START WITH SMALLER b

BUILDING 0 IPod tan 300

500

BIGGER D

Opp h 500 BjHgtoh tan 320h 5O0B

f th 500 tan 320 3

f 500 tan 1320 hI STBSTITUTE h

f 5OOtanl3z 5O0z3