Download x, y

The World’s Largest Ferris Wheel
• In London, British Airways built a ferris wheel which
measures 450 feet in diameter. It turns continuously at a
constant rate, completing a single rotation every 30 minutes.
• The values of f(t), your height above the ground t minutes
after boarding the wheel, are shown in the table below:
t, min
0
f(t), feet 0
7.5
15
22.5 30
225 450 225
0
37.5 45
52.5 60
225 450 225
• The graph of f(t) is shown on the next slide.
0
A function f is periodic if f(t+c) = f(t) for all t in the domain of f .
The smallest positive constant c for which this relationship holds
is called the period of f. A part of the graph of f as t varies over
an interval of one period length is called a cycle.
The midline of a periodic function is the horizontal line midway
between the function’s maximum and minimum values. The
amplitude is the vertical distance between the function’s
maximum (or minimum) and the midline.
Period = 30
Amplitude = 225
Midline: y = 225
The Unit Circle
• The unit circle is the circle of radius one that is centered at
the origin. The equation of the unit circle is x2 + y2 = 1.
• Angles (denoted by θ in the figure below) can be used to
locate points on the unit circle. These angles are measured
counterclockwise from the positive x-axis.
y (0,1)
P = (x,y)
θ
(-1,0)
(0,-1)
(1,0)
x
Definition of Sine and Cosine
• Suppose P = (x,y) is the point on the unit circle specified
by the angle θ. We define the functions , cosine of θ, or
cos θ , and the sine of θ , or sin θ , by the formulas
cos θ  x and
sin θ  y.
See figure below.
y
P  (x, y)
1
sin θ  y
θ
cos θ  x
x
Values of the Sine and Cosine Functions
• Find the values of:
cos 90, sin 90, cos 180, sin 180, cos 210, sin 210.
• In each case, find the point on the unit circle determined by
the given angle. The coordinates of this point are the
values we seek.
• We have that cos 90  0 and sin 90  1. Similarly,
cos180    1 and sin 180   0. However, in order to
evaluate cos 210 and sin 210, we may use a calculator.
We find that cos 210    0.866 and sin 210   0.5 .
Conventions for Working with Angles
• We measure angles with respect to the horizontal, not the
vertical, so that 0º describes the 3 o’clock position.
• Positive angles are measured in the counterclockwise
direction, negative angles in the clockwise direction.
• Large angles (greater than 360º or less than –360º) wrap
around a circle more than once.
Definition of Sine and Cosine in Right Triangles
• Suppose P = (x,y) is the point on the unit circle specified by
the angle θ. We define the functions, cosine of θ, or cos θ ,
and the sine of θ , or sin θ , by either pair of formulas:
cos θ  x and
Side Adjacent
cos θ  Hypotenuse
and
sin θ  y,
Side Opposite
sin θ  Hypotenuse .
y
P  (x, y)
1
Hypotenuse
(-1,0)
sin θ  y
θ
cos θ  x
θ
x
Side Adjacent
Side Opposite
Example: sine and cosine of 45º
• Consider an isosceles right triangle with sides equal 1.
1
1
• From elementary geometry, we know that the acute angles
of this triangle both equal 45º, and by Pythagoras’ theorem
the length of the hypotenuse is 2.
• Therefore, using the right triangle definition, we have
1
2
sin 45 

,
2
2

1
2
cos 45 

.
2
2

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:
θ degrees
sin θ
cos θ
0
0
1
30
1/2
45
2/2
60
3/2
90
1
3/2
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 150  1/2 and cos 150   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. See Page 314 of the text. A reference angle is
always between 0 and 90.
• Example. Find the reference angle of 150 .
y
P  ( 3 / 2, 1 / 2)
reference angle  30
θ  150 
x
For any angle :
The Unit Circle
(x,y) = (cos , sin )
( 12 ,
(
(
3
2
2
2
,
, 12 )
2
)
2
3
)
2
y (0,1)
90
120 
( 12 ,
3
)
2
60
135 
(
(
(
,
2
2
1
)
2
,
,
2
)
2
45
150 
3
2
(
30
(1,0) 180
3
2
2
2
, 12 )
0 x
360 (1,0)
330 
210 
315 
225 
2
)
2
( 12 ,
300 
240 
3
)
2
270
(0,1)
( 12 ,
(
3
)
2
2
2
(
3
2
,
, 12 )
2
)
2
Graph of sine as a function of angle t
• Domain of sine is all real numbers
• Range of sine is  1  y  1.
• Sine is an odd function—it has symmetry about the origin
• The period of sine is 360º.
Graph of cosine as a function of angle t
• Domain of cosine is all real numbers
• Range of cosine is  1  y  1.
• Cosine is an even function—it has symmetry about the y-axis
• The period of cosine is 360º.
Coordinates of a point on a circle of radius r.
• The coordinates (x,y) of the point Q in the figure below are
given by:
x  r  cos θ
and
y  r  sin θ .
Q  ( x , y)
(cos θ , sin θ )
1
θ
r
Height on the ferris wheel as a function of angle
• The ferris wheel described earlier has a radius of 225 feet.
Find your height above the ground as a function of the
angle t measured from the 3 o’clock position.
y
P = (x, y)
225sin t
t
x
225
Height  225  225sin t
Graph of ferris wheel function
• The graph of the function giving your height, y = f(t), in
feet, as a function of the angle t, measured in degrees from
the 3 o’clock position, is shown next.
Period = 360º
f(t)=225+225sin t
Amplitude  225 ft.
Midline y = 225 ft.
Definition of the tangent function
• 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, or by:
Side Opposite
tan θ 
.
Side Adjacent
Line has slope
(0, 1)
y
 tanθ.
x
P  (x, y)
Side Opposite
y
θ
x
θ
(1, 0)
Side Adjacent
Values and graph of the tangent function
• The values of the tangent function for “special” angles are:
t degrees
0
30
45
60
tan t
0
1/ 3
1
3
90
undef
• The function tan t is periodic with period 180º, and its graph is
next. It has a vertical asymptote when t is an odd multiple of 90º.
Solving Right Triangles
• One of the angles of a right triangle is 90°. Thus, one part is
already known and the only additional information necessary is
either two sides or an acute angle and a side. With this
additional information, it is possible to solve for the remaining
parts.
• Making a carefully labeled sketch of the triangle is important in
solving these problems.
• Problem 25, page 296. The top of a 200-foot tower is to be
anchored by cables that make an angle of 30° with the ground.
How long must the cables be? How far from the base of the
tower should the anchors be placed?
h
200 ft
sin 30° = 200/h => h = 200/sin 30° = 400 ft
30°
x
tan 30° = 200/x => x = 200/tan 30° = 346.4 ft
Right Triangles: Inverse Trigonometric Functions
• If we know the sides of a right triangle, what are the angles?
• Example. Consider the 3-4-5 right triangle. Value of Ө?
5
3
Ө
y
Ө=36.87º
4
y = sin Ө
Ө
Definitions of Inverse Trigonometric Functions
• arcsin x = sin–1x = the angle in a right triangle whose sine is x
• arccos x = cos–1x = the angle in a right triangle whose cosine is x
• arctan x = tan-1x = the angle in a right triangle whose tangent is x
• This means that for an acute angle Ө in a right triangle,
sin Ө = x means Ө = sin–1x
cos Ө = x means Ө = cos–1x
tan Ө = x means Ө = tan–1x
Example problem for inverse trig functions
• The grade of a road is 5.8%. What angle does the road
make with the horizontal?
5.8 ft
Ө
100 ft
• We have tan Ө = 5.8/100 = 0.058. Using a calculator in
degree mode, we have
Ө = tan–1(0.058) = 3.319º.
Solving Non-right Triangles
• Suppose the triangle to be solved does not contain a right angle.
If we are given three parts of the triangle, not all angles, then we
can solve for the remaining parts. The solution exists and is
unique except in one case--discussed in a later slide.
• Usually we label the angles of the triangle as A, B, C and the
sides opposite these angles as a, b, and c, respectively.
b
C
A
a
B
c
• Law of Cosines:
c 2  a 2  b 2  2ab  cos C.
• Law of Sines:
sin A sin B sin C


.
a
b
c
An example using the Law of Cosines
• A person leaves her home and walks 5 miles due east and
then 3 miles northeast. How far away from home (as the
crow flies) is she? As seen from her home, what is the
angle  between the easterly direction and the direction to
her destination?
Destination
N
x
Home
θ
3
135° 45°
5
• By applying the Law of Cosines, we can solve for x:
x 2  5 2  32  2  5  3  cos 135  34  30  ( 2 / 2)  55.2 .
So x  55.2  7.43 miles.
• A second application of the Law of Cosines gives:
5 2  (7.43) 2  2  (5)  (7.43)  cos θ  32 , so
θ  arccos(0.9583)  16.6 .
An example using the Law of Sines
• Problem 35, page 306. Two fire stations are located 56.7
miles apart, at points A and B. There is a forest fire at
point C. If CAB = 54° and CBA = 58°, which fire
station is closer? How much closer?
C
b
68°
a
58° B
A 54°
c = 56.7 miles
56.7
a
56.7  sin 54

, so a 
 49.474 miles.
sin 68 sin 54
sin 68
56.7
b
56.7  sin 58

, so b 
 51.861 miles.
sin 68 sin 58
sin 68
• The fire station at point B is closer to the fire by
51.861 – 49.474 = 2.387 miles.
The ambiguous case for solving a triangle
• Suppose we are given two sides of a triangle and the angle
opposite one of them, and we are asked to find the
remaining parts of the triangle. Depending on the data
given, there may be: two possible triangles, one possible
triangle, or no possible triangle.
• Example. CAB = 45°, b = 1, and various values of a:
C
b=1
45°
A
a=3/10
No triangle exists.
C
b=1
45°
A
C
a=2/2
One triangle exists.
B
C
b=1
45°
A
a=8/10
Two triangles exist.
b=1
45°
A
a=11/10
B
Summary for Trigonometry in Circles and Triangles
• Periodic function, period, midline, and amplitude were defined.
• Sine, cosine, and tangent were defined in terms of the unit circle
and in terms of right triangles (SOHCAHTOA).
• Do not use SOHCAHTOA to solve non-right triangles!
• If (x,y) is a point on a circle of radius r specified by angle θ, then
x  r  cos θ and y  r  sin θ .
• Values of sine, cosine, and tangent for “special” angles were
given.
• Reference angles can be used to find values of sine, cosine, and
tangent for angles which are not in the first quadrant.
• The sine and cosine functions are periodic with period 360º
and their domains, ranges, and graphical symmetries were given.
• The function tan t is periodic with period 180º and it has vertical
asymptotes when t is an odd multiple of 90º.
Summary for Trigonometry in Circles and Triangles, cont’d.
• The definitions of the trig functions and the inverses of the
trig functions were used in solving right triangles.
• The law of cosines was given, and it is a generalization of
Pythagoras’ theorem. Also, the law of sines was given.
• The use of the law of cosines and the law of sines in
solving non-right triangles was discussed.