Trig Functions II  March 7th

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Transcript of 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...
rs. T ompson 2011

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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
Mrs. T ompson 2011

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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
Mrs. T ompson 2011

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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...
Mrs. T ompson 2011

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Next Class
Mrs. T ompson 2011

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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
Mrs. T ! ompson 2011

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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.
Mrs. T " ompson 2011

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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
Mrs. T 4 ompson 2011

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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 =x2x1 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
Mrs. T 5 ompson 2011

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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.
Mrs. Thompson 2011

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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 xaxis (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 theyaxis (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
Mrs. Thompson 2011

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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= [kA, 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= kA
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 (t0.25)
15/45 = sin (t0.25)
1/3 = sin (t0.25)1/3 = sin (if = (t0.25))
Notice thereare two possibleanswers!
Mrs. Thompson 2011

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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
Mrs. Thompson 2011

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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(xh)
3. Substitute b(xh) 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(xh) +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
Mrs. Thompson 2011

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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
yaxis
4. Rule:f(x) =a sin b(xh) + 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
Mrs. Thompson 2011
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/31
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(x0.25)+60 x is time inhours
b) y=8 sin /6(x3)+60 x= time inhours
#25. dom= y
y
y
y
y
y y
y
y
y
Mrs. Thompson 2011
DIDNT USE SLIDE

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BASIC
TRANSFORMED
Mrs. Thompson 2011
DIDNT USE SLIDE

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y
x
0 /2 3/2 2
Cos 1 0 1 0 1
Mrs. Thompson 2011

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0 /2 3/2 2
tan 0 Undefined 0 Undefined 0