Transcript Chapter 4 Trigonometric Graphs

Trigonometric Graphs
y
1.5
Period = 3600
Period
Amplitude = 1
1
Amplitude
0.5
– 60
60
120
180
240
300
x
– 0.5
– 1
– 1.5
y  sin x
The horizontal extent of the basic pattern is called the period.
Half of the vertical extent is called the amplitude.
y
1.5
Period
Period = 3600
1
Amplitude
0.5
Amplitude = 1
– 60
60
120
180
– 0.5
– 1
– 1.5
y  cos x
240
300
x
Period
y
20
15
Period = 1800
10
Amplitude = undefined
5
– 60
60
120
180
240
300
x
– 5
– 10
– 15
– 20
y  tan x
The dotted vertical lines are known as asymptotes.
approaches but never touches them.
The graph
1. Find the period and amplitude of the function y  3sin 4 x  2
y
1
90
180
270
360
x
– 1
– 2
– 3
– 4
– 5
360
0

90
Period
=
The graph repeats 4 times over
4
Vertical extent of graph (height) = 6 units. Amplitude = 3.
3600.
For y  a sin bx0 and y  a cos bx0
Amplitude = a
3600
Period =
b
For y  a tan bx0
Amplitude = can not be measured
1800
Period =
b
Sketching Trigonometric Graphs
The strategy for sketching trig graphs of the form
y = a sin bx  c or y = a cos bx  c is :
1. Start with a simple y  sin bx or y  cos bx
2. Put in the scale to match the amplitude a.
3. Slide the graph vertically to match the constant  c.
90
180
270
360
1. Sketch
the graph of y  2sin 3x  1 for 0  x  3600
Step 1.
y
Step 2.
y  sin 3x
90
180
270
360
x
– 3
1
2
3
–
1
2
90
180
270
1. 321123360
Sketch
the graph of y  2sin 3x  1 for 0  x  3600
Step 1.
y
3
Step 2.
Step 3.
y  2sin 3x
2
1
90
– 1
– 2
– 3
180
270
360 x
– 3
1
2
3
–
1
2
90
180
270
1. 321123360
Sketch
the graph of y  2sin 3x  1 for 0  x  3600
Step 1.
y
3
Step 2.
Step 3.
y  2sin 3x  1
2
1
90
– 1
– 2
– 3
180
270
360 x
Radians
Degrees are not the only units used to measure angles.
It is often, and usually, useful to measure angles in radians.
The angle subtended at the
centre of a circle by an arc
equal in length to the radius
is 1 radian.
r
1 radian
r
r
r
1 radian
r
r
Circumference = 2 r
Hence the radius will fit the circumference 2 times.
So there are 2 radians in a complete turn.
3600  2 radians
1800   radians
1. Convert 1200 into radians.
1800  
1200  x
(cross multiply starting with the ‘x’ term)
180 x  120
2
120
x

180 3
2
  radians.
3
2
This is usually written as
radians.
3
2. Convert
1800  
5
x0 
9
x 
5
radians to degrees.
9
(cross multiply starting with the ‘x’ term)
180  5
9
20
x
180  5
9 1
20  5

1
 1000
Special angles and triangles
It is useful and necessary to know exact values of common, and
some not so common, trigonometric ratios.
If you remember only 2 of these you can work the others out easily.
This saves trying to remember them all. Although by the end of
the course you will probably have remembered them all as we
use them quite a lot in higher maths.
The two ratios that you MUST remember are:
1
sin 30 
2
0
(remember SOH)
1
60
tan 450  1
(remember TOA)
2
0
1
450
2
450
300
3
1
This gives us exact trig ratios from all 4 quadrants.
The two ratios that you MUST remember are:

1
sin 
6 2
(remember SOH)
1

3
tan

4
1
(remember TOA)
2
1

4
2


6
3
4
1
This gives us exact trig ratios from all 4 quadrants.
1. What is the exact value of
(a) sin 3000
(b) cos(-135)0
(C) tan
3000
600

4
450
11
4
3
4
-1350
sin 300   sin 60
0
3

2
0
cos(135)   cos 45
0
0
1

2
11

tan
  tan
4
4
 1
2. Calculate the exact length of the side marked x in the triangle.
x
600
10 m
x
sin 60 
10
0
x  10sin 600
5
3
 10 
2
5 3m
10 3

2 1
Solving Problems using exact Values
An oil tanker is sailing north. At 0115 hours a lighthouse is due east
of the tanker. By 0230 hours the lighthouse is 10km from the tanker
on a bearing of 1500. Calculate the speed of the tanker.
N
1500
T2
x
cos30 
10
x  10cos300
0
300
x
T1
10 3

2
 5 3 km
10 km
Distance
Speed =
Time
10 3

2
5
4
10 3 4


2
5
Lighthouse
5
Time = 1 hour and 15 minutes  hours
4
 4 3 km/h
Algebraic solution of equations
1. Solve for x, 2sin x  3  0 for 0  x  3600
2sin x  3  0
2sin x   3
3
sin x  
2
Acute value of x.
 3
0
x  sin 

60

2


1
x  180  60 and 360  60
 2400 ,3000
S
A
T

C
x is in quadrant 3 and 4

2. Solve tan 2 x  3 for 0  x  3600
tan 2 x  3
tan x   3
S

A
T

C
x is in quadrants 1, 2, 3 and 4

x  tan 1 3  600
x  60, 180  60, 180  60 and 360  60
 600 , 1200 , 2400 ,3000
3. Solve 2sin 4 x  3  0 for 0  x  1800
2sin 4 x  3  0
2sin 4 x   3
3
sin 4 x  
2
Acute value of 4x.
 3
0
4 x  sin 

60

 2 
1
S
A
T

C

Remember there are 4 repeats of the sine
graph giving 8 solutions over 3600.
4 x  240, 300, 600, 660, 960, 1020
x  600 , 750 , 1500 , 1650.
x is in quadrant 3 and 4
4. Solve 3sin 2 x  4sin x  1  0 for 0  x  3600
3sin 2 x  4sin x  1  0
(3sin x  1)(sin x  1)  0
3sin x  1  0

S
3sin x  1
1
sin x 
3
T
A
C
sin x  1  0
sin x  1
1
x  sin 1  
3
x  19.50 , 160.50
x  19.50 , 900 , 160.50
x  900
5. Solve 5cos x  2  0.72 for 0  x  2
5cos x  2  0.72
5cos x  1.28
cos x  0.256
x  cos 1 0.256  1.312 radians
x  1.312 rads, 2  1.312 rads
x  1.312rads, 4.971 rads
S
A
T
C


Algebraic Solution of Compound Angle
Equations
Algebraic solutions of trigonometric equations can be extended to
problems involving compound angles such as (3x + 45)0 etc.
1. Solve for x, 4 tan(3 x  45)0  2.5, 0  x  1800
4 tan(3x  45)0  2.5
5
tan(3 x  45)  
8
Acute value of (3x+45).
(3x  45)  320

S
A
T
C

Since 0  x  1800 ,
450  3 x  45  5850
3 x  45  180  32 and 360  32
 1480 , 3280, 5080.
3 x  1030 , 2830 , 4630.
x  34.330 , 94.330 , 154.330.


2. Solve for x, 2sin  2 x    1, 0  x  2
3


S
A
 1

sin  2 x   
3 2

C
T
Since 0  x  2 ,


11
  2x  
3
3
3
   5 13 17

 2x    , 6 , 6 , 6 .
3 6

2x 
x
 7 15 19
2
,
6
,
,
6
6
 7 5 19
,
4 12
,
4
,
12
.
.