Trig Functions II - March 7th

8/7/2019 Trig Functions II - March 7th

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Another System of Angle Measure

Mrs. Thompson 2011


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Slid s w d b iss C ll.

rs. T

ps

2011


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Defi iti f R DIAN:

A r di (r d) is t e eas re f t e central

angle w se sides intersect anarc w se

lengt is equal t t e circles radius.

rs. T

ps

n 2011


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T e circumference C fa circle wit radius r

is calculated using t eformula .

Since t e sides ofa central angleofoneradian intercept anarc w osemeasure is

equal to t e radius, it follows t at a

complete rotation will produceanangleof 2

radians (approximatel 6.28rad).

rs. T ompson 2011


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T is is t e relations ip used to

c angefromone s stemofanglemeasure toanot er...

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T e sides ofa central anglemeasuring

radians intercepts anarc w ose lengt L is

equal to times t e radius.

L=r

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Page 229

#3, 4b, 6, 13, 14, 15

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#3.

#4.

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#6.

#13.

#14.

#15.

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Slide S ow made b Miss Connell

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Trigangles is more closel related toan

angleof rotation. T e vertexof t e trigangle is located at t eorigin. One sideof t eangle is called t e

initial sideand it ist e positive orizontal

axis (positivexaxis).

T eot er side is called t eterminal side, its position isobtained b rotating t e initial

sideabout t eorigin

A is a positive

angleand B

is anegative

angleMrs. T ompson 2011


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T e terminal side toobtainangleA is rotated

Counter Clockwiseand Angle B is obtained b

rotating t e terminal side Clockwise.

Explanation On Board...

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Next Class

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Here is t eunit Circle we ave been dealing wit

for t e past two classes...

An important note:Positive angles =

Positive arc measures

Negative angles =

Negative arc measures

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The Unit Circle is an extension of Trig

Ratios:

SOH CAH TOA is only useful for acute

angles where the unit circle extends the

trig ratio properties to all angles.

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(x, )

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Find t e Coordinates of t eangles (rad).

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ArcMeasure

Quadrant

0


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Page #239#s 8, 9, 10, 12, 15, 17,

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#8.

#9.

#10.

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#12.

#15.a) Grap



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Using t eUnit Circle, lets find t e sine

values:

X 0 /6 /4 /3 /2 3/2 2

Sinx


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Using t eUnit Circle, lets find t e sine

values:

X 0 /6 /4 /3 /2 3/2 2

Sinx 0 1

2

2

2

3

2

1 0 -1 0

0.5 0.7071 0.866


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What to do with these Coordinates????Sine Function Introduced

Sin

0 0

/2 1

0

3/2 -1

2 0



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Sin

0 0

/2 1

0

3/2 -1

2 0As you can see the function

oscillates between 1 and -1...

Why do you think that is?

T is is a periodicfunction...



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A periodic function is agrap t at consists of

repeating patterns.

Acycle is t e simplest pattern t at repeats

itself toform t e curve.T eperiod (p)corresponds to t e distance

between t eendpoints ofa c cle.

T efrequency (f) is equal to t e reciprocal

of t e period. f= 1/p



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Trigonometric Functions are said to be

periodic as t ey repeat after a certain time

(period).

1 cycle 1 cycle


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T e period corresponds to t e distance

between t eendpoints ofa cycle:

PERIOD FREQUENCY

(seconds/cycle) (cycles/second)

p =x2-x1 F= 1 = 1

= (0) (-8) p 8

= 8 seconds 1 cycleevery 8s

to complete

a cycle

We will now beable to studyall six trigonometric functions.

1 cycle 1 cycle


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Rule: f(x) = sin (x)

Grap :

yDomain: Zeros =

yRange: [-1, 1] Sign = +

yMaximum = 1

yMinimum = -1 -

SINUSODAL CURVE

T ePERIODfor a

basic sinefunction is 2

T e curve

Consists ofa

pattern t at

repeats itself

indefinitely.

n



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Ruleof Basic Function (Periodic Function

called a sinusoidal curve)

f(x) = sinx

f


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P

( ,K)

|a| f

Properties

dom f = IR

ranf = [-1,1]

Initial value= 0

Zeros

X = {n} n Z

Extremes

min f= -1

max f= 1

Sign

Positive:[0 + 2n, + 2n] n Z

Negative: [ + 2n, 2 + 2n] n Z


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P

( ,K)

|a| f

Variation

Increasing:[0 + 2n, + 2n] n Z

Decreasing:[ + 2n, 2 + 2n] n Z

Period = 2

Frequency= 1/ 2


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P

( ,K)

|a| f

Amplitude is the positive distance from a peak to

the centre of the curve

A= max of function min of function

2

A= (1) (-1)

2

= 2/2 = 1

Note: consider the vertex to be (h,K)=

(0,0) The Basic Ordered Pairs

(0,0), (/2,1), (,0), (3/2, -1), (2,0)


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Growt Interval

DecayInterval

T eAmplitudeA is found usingOr it is simply t enumber w ic t e

function is oscillating between.

T efrequency is found using t eformulaWhereP is the period it takesfor thefunction to completeonefull oscillation.

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Rule

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Vertical Scale Change (affects theamplitude)

A= |a|

f1(x) =2 sin x f2(x) = 0.5 sin x


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Reflectionover x-axis (ifa is negative)

f1(x) = -3 sinx f2(x) = -0.75 sinx


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a + a -

f2(x) = -3 sinxf1(x) = 2 sinx


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Horizontal Scale Change (affects the period of )

P= 2 =2 = P =2 = 1

|b| |2| 2

1 .

|b|

f1(x) = sin (2x) f2(x) = sin (2x)


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Reflectionover they-axis (if b is negative)

P= 2 =4

|-0.5|

f(x) = sin (-0.5x)


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hand k

Horizontal scale change (h) -- phase shift

Vertical scale change (k)

f1(x) = sinx

f2(x) = sin (x-) 1

2


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y2 = 2 sin (1/2(x))

y3 = -5 sin (2(x))

y4 = -5 sin (-2(x))

y5 = 8 sin (1/2(x-)) + 2

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Ifa, b are the same sign

Ifa, b are different signs


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AmplitudeA= |a|

Period p = 2

|b|

Vertex (between min/max)

(h,k) where h is of xmax and xmin

K is of yvalues of max/min

Domain f= IR

Range f= [k-A, k+A],

where A= |a|

Initial Value f= 0


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Zeros

1. None if |K| > |a|

2.

Usuallyx1 + np and x2 +np


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Extremes

Minf= k-A

Maxf= k+A

Sign Changingat the zeros (periodically)

Variation

Increasingfrommin max (periodically)

Decreasingfrommax min (periodically

Inverse not afunction more to come!Mrs. Thompson 2011

Usually+ or - =[x1 + pn, x2+pn]+ or - =[x2 + pn, x3+pn]

=[xmin + pn, xmax+pn]

=[xmax + pn, xmin+pn]


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1. Find VERTEX (h,k) 2. Find thePERIOD = 2

|b|

3. Find the AMPLITUDE = |a|

4. Choose CURVE a, b same sign

a, b different signs

5. GRAPH 5 distinct points.

REFER TO SLIDE WITH THREE STEPSMAKE 5


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1. Identify point (h, k) which serves as the beginningof thecycle.

2. Draw a rectangular border around the cycle. The lengthofthe rectangle is equal to the cycles period p. Theheight

of the rectangle is equal to2A whereA = |a| is theamplitude.

1. Graph the cycle, taking the signs ofa and b intoaccount.Mrs. Thompson 2011


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1. V (, 1)

2. P = 2 = 4

3. A= |3| = 3

4. a and b are +

5. (, 1)(2, 4)

(3, 1)

(4, -2)

(5, 1)


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Handout given last class

March 25th

TakeUp


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Zeros: solving the trigfunction whereone

sideof theequation is equal to zero!Example:

f(x) = -45sin (t 0.25) +15

f(x) = 0

0 = -45 sin (t 0.25)+15

-15 = -45 sin (t-0.25)

-15/-45 = sin (t-0.25)

1/3 = sin (t-0.25)1/3 = sin (if = (t-0.25))

Notice thereare two possibleanswers!

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sin = 1/3

Lets refer back to theunit circle

sin = sin (-)

why?

So back to the solution:

sin = 1/3

= sin -(1/3)1 0.3398 OR 2 - 0.3398

2 2.8018

12

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Now back to the context of the problem

1= (t1 0.25) 2 = (t2 - 0.25)

0.3398 = (t1 - 0.25) 2.8018 = (t2 0.25)

0.3398/ =(t1 0.25) 2.8018/ =(t2- 0.25)

t1 = 0.3582 t2 = 1.1418

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POSITIVE OVER THE INTERVAL

NEGATIVE OVER THE INTERVAL

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1. Let thefunction ruleequal zero

2. Isolate sin b(x-h)

3. Substitute b(x-h) for

4. Find 1 and 2 from theequation (positiveand negative.)

5. Find the solutions of = b(x h)

(Zeros)

6. Write theGeneral Form using the period.

(x + np, n Z)


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Zeros: solving the trigfunction whereone

sideof theequation is equal to zero!Example:

f(x) = 3 sin(x 1) +3

f(x) = 0

0 = 3 sin(x 1) +3

-3 = 3 sin( x - 1)

-1 = sin(x - 1)

If =(x - 1)Then -1 = sin

Notice thereare two possibleanswers!


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Variation: increasing [1/2 + 2n, 3/2 + 2n] n E Z


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X = {1/6 + 2n, 5/6 + 2n} n E Z

Extremes: Min f = -1/2 Max f = 3/2

Sign: (-) [1/6 + 2n, 5/6 + 2n] n E Z

(+) [-7/6 + 2n, 1/6 + 2n] n E Z

g [ , ]

decreasing [-1/2 + 2n, 1/2 + 2n] n E Z


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Rule f(x) = a sin b(x-h) +k

Graph SinusoidalDomain

Range

Period p= 2/

Zero An infinitenumber of zeros. Ifx1 and x2 are consecutive zeros

and p is the period, then the

zeros are (x1 + np) and (x2 + np)wheren

Extremes Maximum: k + A and

Minimum: k A

Sign Related to the zeros

ariation Periodically increasingand

decreasing

Inverse Is not afunction

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Finding the Rule of the Sine Function

1. Find a

(A= |a|)

2. Find b

(P= 2/|b|)3. Find the closest positive center (h,k) to the

y-axis

4. Rule:f(x) =a sin b(x-h) + k


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Ex Jan is doing researchon the phenomenaof

vibrations. She compiles a series of results andobtains thefollowinggraphon the computer

screen:

What rule represents this function?

1

3

2

-

2

- 3

2


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Properties:

Pg 253 #6,9,10abd, 12

Graphing

Pg 253 #4,5,7

Zeros Pg 254 #10c, 11, 15y1y2y3 ,17

Rule

Pg 256 #18,19,20


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#2. a) 5 b) [-2, -3/2], [-/2,/2] and

[3/2,2]

c) [-, 0] U[,2]

#3. a) True b) False c) True

#4. a) p=2 A=2

#5a) Graph #6.

#7. Functions fand ghave the samegraph

#8. 2n

f g h

Period 2 2/3 4

Amplitude 4 1 10

Minimum -4 1 -15Maximum 4 3 5

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DIDNT USE SLIDE


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#11. a) x1=-/6+2n x2=-5/6+2n

b) x1=5/8+n x2=2/8+n

c) x10.5+2n

d) none

#12. a) b) 2 c) t(-1, ) d) min=0max= 2

#13. a) 1s b) after 0.75s c) graph

d) 1)2s 2) [12, 48]

e) 12cm #19. a) y=2 sin 2x b) y=0.5sin 4x/3-1

c)y=2 sin (x-/2)+3 d) y=3 sin /6(x+2)+2

Mrs. Thompson 2011

DIDNT USE SLIDE


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#21. a) y=20 sin 2(x-0.25)+60 x is time inhours

b) y=8 sin /6(x-3)+60 x= time inhours

#25. dom= y

y

y

y

y

y y

y

y

y

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DIDNT USE SLIDE


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BASIC

TRANSFORMED

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DIDNT USE SLIDE


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y

x

0 /2 3/2 2

Cos 1 0 -1 0 1

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0 /2 3/2 2

tan 0 Undefined 0 Undefined 0

Trig Functions II - March 7th

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Transcript of Trig Functions II - March 7th