Transcript File

Chapter 8
Unit Circle and Radian Measures
Opening problem
1. Consider an equilateral triangle with sides 2
cm long. Altitude [AN] bisects side [BC] and
the vertical angle BAC.
A
Can you see from this figure that sin 30o =1/2?
Use your calculator to find the values of sin 30o
, sin 150o , sin 390o , sin 1110o and sin(-330o).
2
30o 30o
60o
What do you notice? Can you explain why this
result occurs even though the angles are not
between 0o and 90o ?
B
60o
N
1
C
2. Degree, minute, second:
One degree :
1o = 1/360 of a revolution
One minute :
1’ = 1/60 of a degree
One second:
1’’ = 1/60 of a minute
3. 1 radian (1c) ≈ 57.3o
Radian is an abbreviation for
“radial angle”
Degree-radian conversion
x

180 
Degree
Radian
x
180 

5. Convert to radians, in terms of 
a. 60o
b. 80o
c. 315o
6. Convert to degrees:
a. /5
b. 3 /4
c. 5 /6
Parts of a circle
minor arc
radius
sector
center
chord
segment
major arc
*minor if it involves less than half the circle
*major if it involves more than half the circle.
Arc Length
X
l
O
q
Y
r
For q in radians, arc length
l = qr
For q in degrees, arc length l = q/360 x 2r
O
q
r
For q in radians, area of a sector
A = ½ qr2
For q in degrees, area of a sector
A= q/360 x r2
9. Use radians to find the arc length and the area of a
sector of a circle of
Radius 9 cm and angle 7/4
Arc Length: l = qr
Area of a sector: A = ½ qr2
10. A sector has an angle of 1.19 radians and an area of 20.8
cm2. Find its radius and its perimeter.
Area of a sector: A = ½ qr2
Perimeter: P =
11. A sector has an angle of 107.9o and an arc length of 5.92
m. Find its:
For q in degrees, arc length l = q/360 x 2r
A= q/360 x r2
a. radius
b area.
12. Find, in radians, the angle of a sector of: Radius 4.3 m
and arc length 2.95 m
Arc Length: l = qr
13. (number 8 in Exercise 8B)
14. A nautical mile (nmi) is the distance on the Earth’s surface
that subtends an angle of 1 minute (or1/60 of a degree) of the
Great Circle arc measured from the center of the Earth. A knot
is a speed of 1 nautical mile per hour.
a. Given that the radius of the Earth is 6370 km, show that 1 nmi
is approximately equal to 1.853 km.
b. Calculate how long it would take a plane to fly from Perth to
Adelaide (a distance of 2130 km) if the plane can fly at 480
knots.
14. A nautical mile (nmi) is the distance on the Earth’s surface that
subtends an angle of 1 minute (or1/60 of a degree) of the Great Circle
arc measured from the center of the Earth. A knot is a speed of 1
nautical mile per hour.
a. Given that the radius of the Earth is 6370 km, show that 1 nmi is
approximately equal to 1.853 km.
((1/60)/360) x 2  (6370) = 1.853 km
b. Calculate how long it would take a plane to fly from Perth to
Adelaide (a distance of 2130 km) if the plane can fly at 480 knots.
(2130/1.853) /480 = time
2.4 hours = time
The unit circle is the circle with center (0, 0) and radius 1
unit.
(0, 1)
(-1, 0)
(0, 0)
(0, -1)
(1, 0)
x2 + y2 = r2 is the equation of a circle with center (0, 0) and
radius r.
The equation of the unit circle is x2 + y2 = 1.
q is positive for anticlockwise rotations and negative for
clockwise rotations.
+
P
-
Definition of sine and cosine
SohCahToa
P(cos q, sin q)
q
cos q is the x-coordinate of P
sin q is the y-coordinate of P
Pythagorean identity of sine and cosine
cos2 q + sin2 q = 1.
Domain and range of sine and cosine of a
unit circle.
For all points on the unit circle, -1 < x < 1 and -1 < y < 1.
So, -1 < cos q < 1 and -1 < sin q < 1 for all q.
(0, 1)
(-1, 0)
(0, 0)
(0, -1)
(1, 0)
Definition of tangent
tan q 
sin q
cos q
22. PERIODICITY OF TRIGONOMETRIC
RATIOS
For q in radians and k є Z ,
cos (q + 2k) = cos q and sin (q + 2k) = sin q.
For q in radians and k є Z , tan(q + k) = tan q.
24.
sin (180 – q) = sin q
and
cos (180 – q) = -cos q
25. Find all possible values of cos q for sin q = 2/3. Illustrate
your answer.
25. Find all possible values of cos q for sin q = 2/3. Illustrate
your answer.
sin q = opp/hyp
3
2
3
q
q
-√5
2
0
√5
Use Pythagorean Theorem to solve for the 3rd side.
cos q = √5/3 or -√5/3
26. If sin q = -3/4 and  < q < 3/2, find cos q and tan q
without using a calculator.
26. If sin q = -3/4 and  < q < 3/2, find cos q and tan q
without using a calculator.
the domain for q is in the between  and 3/2 puts us in the
third quadrant.
cos q = -√7/4
-√7
-3
4
tan q = -3/-√7
= (3√7) / 7
27. If tan q = -2 and 3/2 < q < 2, find sin q and cos q.
27. If tan q = -2 and 3/2 < q < 2, find sin q and cos q.
cos q = √5/5
sin q = -2√5/5
1
√5
-2
28. If q is a multiple of /2, the coordinates of the points on
the unit circle involve 0 and +1.
29. If q is a multiple of /4, but not a multiple of /2, the
coordinates involve +√2 / 2.
30. If q is a multiple of /6, but not a multiple of /2, the
coordinates involve + ½ and + √3 / 2.
31. Use a unit circle diagram to find sin q, cos q and tan q
for q equal to:
a. /4
b. 5/4
c. 7/4
d. 
e. -3/4
31. Use a unit circle diagram to find sin q, cos q and tan q
for q equal to:
a.
a. /4
b. 5/4
c. 7/4
d. 
e. -3/4
2
2
,
2
b. 
, 1
2
2
2
, 
2
c. 
2
, 1
2
2
,
2
, 1
2
d . 0 ,  1, 0
e. 
2
2
, 
2
2
, 1
32. Without using a calculator, evaluate:
a. sin2 60o
b. sin 30o cos 60o
c. 4sin 60o cos 30o
32. Without using a calculator, evaluate:
a. sin2 60o
a . sin 60  (sin
2
o
o
60 )(sin
3
60 ) 
o
2
b. sin 30o cos 60o
b . sin 30 cos 60 
o
o
 3  3 

3
c . 4 sin 60 cos 30  4 
 2  2 



o
c. 4sin 60o cos 30o
1 1
1
 
2 2
4
Complete the rest and report out.
o

3
2

3
4
33. Equation of a straight line.
If a straight line makes an angle of q with the positive x-axis
then its gradient is m = tan q