Transcript No Slide Title

Radians
• We have previously measured angles in degrees, but now we
introduce a new way to measure angles--in radians.
• The arc length spanned, or
cut off, by an angle is shown next:
y
θ
Arc length spanned by θ
x
• An angle of 1 radian is defined to be the angle, in the
counterclockwise direction, at the center of a unit circle which
spans an arc of length 1. An angle of 1 radian is
approximately equal to 57.3.
• If an angle is referred to without units, radians is assumed.
• The radian measure of a positive angle is the length of arc
spanned by the angle in a unit circle. For a negative angle,
the radian measure is the negative of the arc length.
• To convert from degrees to radians,
multiply the degree measure by

180
.
• To convert from radians to degrees,
multiply the radian measure by
180

.
• Examples. - /2 radians  - 90 and 1  0.01745 radians.
• Problem. Convert 4 radians to degrees. In which quadrant
is this angle?
Arc Length
• The arc length, s, spanned in a circle of radius r by an
angle of θ radians is given by s  r θ .
y
r
Arc length  r θ
θ radians
x
• Problem. What length of arc is cut off by an angle of
120 degrees on a circle of radius 12 cm?
Solution. First convert to radians, then multiply by 12 to
get an arc length of 8 cm.
Effect of tire wear on mileage
• The odometer in your car measures the mileage travelled.
• The odometer uses the angle θ that the axle turns to
compute the mileage s in the formula s  r θ .
• Question. With worn tires, are the actual miles travelled
more or less than the odometer miles?
Values of Sine and Cosine
• The values for sine and cosine in the following table of
“special” angles (see pg 314 of text) should be memorized:
sin θ
0
/6
1/2
cos θ
1
3/2
θ radians
0
/4
/3
/2
2/2 3/2 1
2/2
1/ 2
0
• You should be able to use reference angles (see next slide) to
find the values of sine and cosine for angles which are not in
the first quadrant. For example,
sin 5/6  1/2 and cos 5/6   3 / 2.
Reference Angles
• For an angle θ corresponding to the point P on the unit
circle, the reference angle of θ is the angle between the
line joining P to the origin and the nearest part of the
x-axis. A reference angle is always between 0 and 90,
that is, between 0 and /2. (See page 346 of the textbook.)
• Example. Find the reference angle of 5 / 6.
y
P  ( 3 / 2, 1 / 2)
reference angle  /6
θ  5/6
x
For any angle :
The Unit Circle
(x,y) = (cos , sin )
( 12 ,
(
2
2
,
3
2
, 12 )
(1,0)
5
6

2
2
3
2
)
2
3
3
)
2
( 12 ,
90


60 
135
45 
150
(
(
2
2
,
7
6
2
)
2
2
)
2
3
2
(

6
, 12 )
0 0
x
360 2 (1,0)
180
, 12 )
,
4
30 

3
2
2
2
(
3
120
4
(
y (0,1)
3
)
2
210
330
315
225
5
4
( 12 ,
300
240
4
3
3
)
2
270
(0,1)
3
2
7
4
5
3
( 12 ,
11
6
(
3
)
2
(
2
2
3
2
,
, 12 )
2
)
2
Graphs of sin x and cos x, where x is in radians
• The function sin x is periodic with period 2 .
• The function cos x is periodic with period 2 .
Generation of Sine Curve Using Unit Circle
Q1
Red arc length is π/3.
Q2
Graph of sin x, where x is in radians
• One cycle of the function sin x.
Graph of cos x, where x is in radians
• One cycle of the function cos x.
Graph of ferris wheel function
• The graph of the function giving your height, y = f(x), in
feet, as a function of the angle x, measured in radians from
the 3 o’clock position, is shown next.
Period = 2π
f(x)=225+225sin x
Amplitude 225 ft.
Midline y = 225 ft.
Sinusoidal Functions
• We consider functions which can be expressed in the form:
y  A  sin B(t  h)  k
and
y  A  cos B(t  h)  k,
where A, B, h, and k are constants and t is measured in
radians. The graphs of these functions resemble those of
sine and cosine. They may start with sine or cosine and
then shift, flip, or stretch the graph. Which of these
transformations are to be applied is determined by the
values of the constants A, B, h, and k as follows:
• |A| is the amplitude
• 2 / | B | is the period
• h is the horizontal shift
• y = k is the midline
• |B| is the number of cycles completed in 0  t  2 .
• | B | /2 is number of cyclesper unit time- -the frequency
One Cycle of y  A  sin B(t - h)  k
• Assume A, B, h, and k are positive.
y
2
B
kA
h
k
A
kA
h
h

B
2
h
B
t
The London Eye Ferris wheel function as a sinusoidal function
• Problem. Use the sinusoidal function to represent your
height f(t) above ground while riding the Ferris wheel.
Solution. The diameter of the Ferris wheel is 450 feet so
the midline is k = 225 and the amplitude, A, is also 225.
The period of the Ferris wheel is 30 minutes, so
2 
B
 .
30 15
The graph is shifted to the right by t = 7.5 minutes since
we reach y = 225 (the 3 o’clock position) when t = 7.5.
Thus, the horizontal shift is h = 7.5, and the formula for
f(t) is:

f(t)  225  sin (t  7.5)  225,
15
where t is in minutes and height is in feet.
Height of capsule on London Eye Ferris wheel
f(t)  225sin

15
( t  7.5)  225
Phase shift
• Assume that a sinusoidal function has been given. The
phase shift  is defined as   hB. Then since
h
hB



,
(2/B) 2
2
it follows that the ratio of the phase shift to 2 equals the
fraction of a period by which the sinusoidal function is
shifted.
• For sinusoidal functions written in the form
y  A  sin(Bt   ) and y  A  cos(Bt   ),
 is thephaseshift.
Phase shift, continued
• Phase shift is significant because we often want to know if
two waves reinforce or cancel each other. For two waves
of the same period, a phase shift of 0 or 2 tells us that the
two waves reinforce each other while a phase shift of 
tells us that the two waves cancel.
• Example. The graph of cos (3t  /4) is the same as the
graph of cos (3t) but shifted  /12 units to the left, which
is ( /4)/(2) = 1/8 of the period of cos(3t).
cos(3t +/4)
cos(3t)
Sine with a phase shift of π/2
• plot({sin(t),sin(t+Pi/2)},t=-Pi..Pi,color=black);
• The shifted sine curve is the cosine. Likewise, we can
shift the cosine curve right by π/2 to get the sine.
Interpreting the shifted sine curve
• From the previous slide, we have
cos t  sin(t 
• Upon replacing t by –t, we have
cos t  cos(  t)  sin(  t 

2

2
).
)  sin(

2
 t),
using the fact that cosine is an even function.
• In terms of angles of a right triangle, the latter result is:

cos  sin(   ), or
φ
2
c
a
b
cos  sin  
c
Ө
b
The tangent function revisited
• Suppose P = (x, y) is the point on the unit circle specified
by the angle θ. We define the tangent of θ , or tan θ, by
tan θ = y/x. The graphical interpretation of tan θ is as a
slope. In the figure below, the slope m of the line passing
from the origin through P is given by m  tanθ .
y
Line has slope  t an θ
x
P  (x, y) and we have
sin θ
t an θ 
.
y
cos θ
(0, 1)
θ
x
(1, 0)
Values and graph of the tangent function
• The values of the tangent function for “special” angles are:
x radians
0
/6
tan x
0
1/ 3
/4
/3
1
3
/2
undef
• The function tan x is periodic with period  , and its graph is next.
It has a vertical asymptote when x is an odd multiple of  / 2.
Reciprocals of the trig functions
• The reciprocals of the trig functions are given special
names. Where the denominators are not equal to zero, we
define:
1
secant t sec t 
cos t
1
cosecant t csc t 
sin t
1
cos t
cotangentt  cot t

tan t sin t
• The graphs of these new functions can be easily obtained
from the graphs of the functions of which they are the
reciprocals.
Summary of Trigonometric Relationships
• Sine and Cosine functions
sin t  cos(t  2 )  cos(2  t)   sin(t)
cos t  sin(t  2 )  sin(2  t)  cos(t)
• Pythagorean Identity
cos2 t  sin 2 t  1
• Tangent and Cotangent
sin t
1

cos t cot t
cos t
1
cot t 

sin t t an t
• Secant and Cosecant
t an t
1
cos t
1
csc t 
sin t
sec t 
Inverse Trigonometric Functions
• Solving equations involving trigonometric functions can
often be reduced (after some algebraic manipulation) to an
equation similar to the following example:
Find a value of t radians satisfying cos t = 0.4 .
• In order to find a value of t as in the above example, we can
use a calculator which has support for inverse cosine. Such a
calculator in radian mode allows us to find the required value
of t by pressing the button labeled as shown below:
cos-1
• Pressing this button yields cos-1(0.4) 1.16, where
cos(1.16)  0.4 .
The Inverse Cosine Function.
• The inverse cosine function, also called the arccosine
function, is denoted by cos-1y or arccos y. We define
t  cos -1 y provided that y  cos t and 0  t   .
In other words, if t = arccos y, then t is the angle between
0 and  whose cosine is y. The inverse cosine has
domain  1  y  1 and range 0  t   .
• In conjunction with the inverse cosine, we may have to use
reference angles to find answers which are not in 0  t   .
• Example. cos-1(0.4)  1.16 gives us an answer in the first
quadrant. Suppose we want an answer in the fourth
quadrant. We simply subtract the reference angle from 2
to obtain t  5.12 .
The Inverse Sine Function.
• The inverse sine function, also called the arcsine function,
is denoted by sin-1y or arcsin y. We define
t  sin -1 y provided that y  sin t and   / 2  t   / 2.
The inverse sine has domain  1  y  1 and range
  / 2  t   / 2.
• In conjunction with the inverse sine, we may have to use
reference angles to find answers which are not in its range.
• Example. sin-1(0.707) =  / 4 gives us an answer in the
first quadrant. Suppose we want an answer in the second
quadrant. We simply subtract the reference angle from 
to obtain t = 3 / 4.
The Inverse Tangent Function.
• The inverse tangent function, also called the arctangent
function, is denoted by tan-1y or arctan y. We define
t  tan-1 y provided that y  tan t and   / 2  t   / 2.
The inverse tangent has domain    y   and range
  / 2  t   / 2.
• In conjunction with the inverse tangent, we may have to
use reference angles to find answers which are not in its
range.
• Example. tan-1(1.732) =  / 3 gives us an answer in the
first quadrant. Suppose we want an answer in the third
quadrant. We simply add the reference angle to  to
obtain t = 4 / 3.
• While riding the London Eye Ferris wheel, at which times
during the first turn is your height = 400 feet? As shown in
the text, we must find two solutions for t between 0 and 30
in the equation

 175
sin  (t  7.5)  
.
 15
 225
The first solution is given by the arcsin. We solve for t in
the equation


175
t   arcsin
 0.891.
15
2
225
This yields t = 11.75 minutes. The second solution
corresponds to another angle on the circle with the same
reference angle, 0.891, and a positive value of the sine.
This is in the second quadrant, so we have

15
t

2
   0.891, which yields t  18.25 min.
Summary for the Trigonometric Functions
• Radian measure of an angle was defined and  radians  180.
• The arc length, s, spanned by an angle of θ radians in a circle of
radius r is s  r θ.
• Values of sine and cosine for “special” angles in radians were
given.
• Reference angles can be used to find values of sine and cosine
for angles in radians which are not in the first quadrant.
• The sine and cosine functions are periodic with period 2 and
their graphs are similar but shifted horizontally.
• For sinusoidal functions, we related amplitude, period, hor. shift,
and midline to the parameters of the function.
• Phase shift can be used to determine the fraction of a full period
by which a sinusoidal graph is shifted.
• The values of tangent for special angles in radians were given.
Summary for the Trigonometric Functions, cont’d.
• The function tan x is periodic with period  and it has
vertical asymptotes when x is an odd multiple of  / 2 .
• Three other trig functions were defined as reciprocals:
secant, cosecant, and cotangent.
• A number of trigonometric relationships were given.
• The definitions of inverse functions for cosine, sine, and
tangent were extended and their domains and ranges were
given.
• Use of the inverse trig functions in solving problems was
illustrated.