chapter 17 17-1 trigonometric functions in triangles
TRANSCRIPT
Chapter 17
17-1 Trigonometric Functions in Triangles
Special Right TrianglesCONCEPT
SUMMARY
30-60-90 Triangle 45-45-90 Triangle
2xx
3x 2x
The hypotenuse is the times the leg!2
The hypotenuse is 2 times the short leg!
3The long leg is the times the short leg
Special Right Triangles
2x
2x
52x
25x
28x
2hyp
leg
2221
x
3x
2xx
3x
2xx
10 is the short leg since it is opposite the 30° Angle
Hypotenuse = 2(Short Leg)
y = 2(10)
y = 20
310x
LegShort3LegLong
Exact Trig ValuesCONCEPT
SUMMARY
2xx
3x 2x
Trig Values
1sin30
2 3
sin 602
3cos30
2
1cos60
2
3tan30
3 tan 60 3
2sin 45
2
2cos 45
2
tan 45 1
Find the value of each variable
Find x first
372tan
x
x72tan3
2331.9x
Find y next
y3
72cos
3)72(cos y
72cos3y 708.9
Exact Value
Approximation
Find the value of each variable
Find x first
x8
40cos
8)40(cos x
40cos8x
Find y next
840tan
y
)40(tan8y 7128.6
Exact Value
Approximation
4433.10
A B
6631.4)25(tan10 x
0338.1125cos
10 y
x = 12(cos 63) = 5.4479
y = 12(sin 63) = 10.6921
Find the value of each variable
SHORT-RESPONSE TEST ITEM A wheelchair ramp is 3 meters long and inclines at Find the height of the ramp to the nearest tenth centimeter.
Answer: The height of the ramp is about 0.314 meters,
Multiply each side by 3.
Simplify.
Y
W
Method 2The horizontal line from the top of the platform to which the wheelchair ramp extends and the segment from the ground to the platform are perpendicular. So, and are complementary angles. Therefore,
Y
W
Answer: The height of the ramp is about 0.314 meters,
Multiply each side by 3.
Simplify.
SHORT-RESPONSE TEST ITEM A roller coaster car is at one of its highest points. It drops at a angle for 320 feet. How high was the roller coaster car to the nearest foot before it began its fall?
Answer: The roller coaster car was about 285 feet above the ground.
Evaluate the six trigonometric functions of the angle shown in the right triangle.
SOLUTION
The sides opposite and adjacent to the angle are given. To find the length of the hypotenuse, use the Pythagorean Theorem.
13
17
8
If , find the other 5 trig functions.
SOLUTION
Draw a triangle such that one angle has the given cosine value use the Pythagorean theorem
8cos
17
2 217 8 225 15
15
1517
1715
178
158
815
817
HW #17-1Pg 732 1-23
Chapter 17
17-2 More fun with Trigonometric Functions
ANGLES IN STANDARD POSITION
GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS
r
(x, y)
Let be an angle in standard position and (x, y) be any point (except the origin) on the terminal side of . The six trigonometric functions of are defined as follows.
0
00
0
GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS
0
sin =0yr
r
y
y
r
csc = , y 00 ry
GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS
0r
cos =0xr
x
x
r
sec = , x 00rx
GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS
0xy
x
tan = , x 00yx
y
cot = , y 00xy
GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS
Pythagorean theorem gives
r = x 2 +y
2.
r
(x, y)
0
Let be an angle in standard position and let P(x, y) be a point on the terminal side of . Using the Pythagorean Theorem, the distance r from the origin to P is given by . The trigonometric functions of an angle in standard position may be defined as follows.
2 2r x y
0
Let (3, – 4) be a point on the terminal side of an angle instandard position. Evaluate the six trigonometric functions of .
0
0
SOLUTION
Use the Pythagorean theorem to find the value of r.
r = x 2 + y
2
= 3 2 + (– 4 )
2
= 25
= 5
Evaluating Trigonometric Functions Given a Point
r
(3, – 4)
Using x = 3, y = – 4, and r = 5,you can write the following:
45
sin = = – yr0
cos = =xr
350
tan = = – 0yx
43
csc = = – ry
54
0
sec = =rx
53
0
cot = = – 0xy
34
Evaluating Trigonometric Functions Given a Point
0r
(3, – 4)
The values of trigonometric functions of angles greater than90° (or less than 0°) can be found using corresponding acuteangles called reference angles.
Let be an angle in standard position. Its reference angle is the acute angle (read theta prime) formed by the terminal side of and the x-axis.
00'
0
90 < < 180;0
2
< <00
Radians: = 0–0'
0'
= 180Degrees: 0' – 0
0
0'
– 1800=Degrees: 0'
Radians: = 0 – 0'
180 < < 270;
32
< <
0
0
0
0'
– 0360=Degrees: 0'
Radians: = 0–20'
270 < < 360;
232
< <
0
0
Evaluating Trigonometric Functions Given a Point
EVALUATING TRIGONOMETRIC FUNCTIONSCONCEPT
SUMMARY
Use these steps to evaluate a trigonometric function ofany angle .0
2
Use the quadrant in which lies to determine thesign of the trigonometric function value of .
00
3
1 Find the reference angle .0'
Evaluate the trigonometric function for angle .0'
EVALUATING TRIGONOMETRIC FUNCTIONSCONCEPT
SUMMARY
Evaluating Trigonometric Functions Given a Point
Signs of Function Values
Quadrant IQuadrant II
Quadrant III Quadrant IV
sin , csc : +0 0
cos , sec : –0 0
tan , cot : –0 0
sin , csc : +0 0
cos , sec : +0 0
tan , cot : +0 0
sin , csc : –0 0
cos , sec : –0 0
tan , cot : +0 0
sin , csc : –0 0
cos , sec : +0 0
tan , cot : –0 0
Using Reference Angles to Evaluate Trigonometric Functions
Evaluate tan (– 210).
SOLUTION
The angle – 210 is coterminal with 150°.
The tangent function is negative in Quadrant II,so you can write:
tan (– 210) = – tan 30 = – 33
0' = 30
0 = – 210
The reference angle is = 180 – 150 = 30.0'
1
2
2
2
2
2 2,
2 2
1
2
2
2
2
1
2
2
2
2
2 2,
2 2
2 2,
2 2
1
2
2
2
2
2 2,
2 2
2 2,
2 2
2 2,
2 2
2 2,
2 2
2 2,
2 2
1 1
2
3
2
3 1,
2 2
11
2
3
2
3 1,
2 2
1
1
2
3
2
3 1,
2 2
1
1
2
3
2
3 1,
2 2
2 2,
2 2
2 2,
2 2
2 2,
2 2
2 2,
2 2
1
1
2
3
2
3 1,
2 2
3 1,
2 2
3 1,
2 2
3 1,
2 2
1 3,
2 2
1
1
2
3
2
1
1
2
3
2
1 3,
2 2
1 3,
2 2
1
1
2
3
2
1 3,
2 2
2 2,
2 2
2 2,
2 2
2 2,
2 2
2 2,
2 2
3 1,
2 2
3 1,
2 2
3 1,
2 2
3 1,
2 2
1 3,
2 2
1 3,
2 2
1 3,
2 2
1 3,
2 2
30
45
60
2
3,
2
1
0
3
3
1
3
undef
3
1
3
3
0
3
3
1
3 undef
3
3
1
3
HW #17-2Pg 739-740 1-61 Odd, 62-63
Chapter 17
17-3 Radians, Cofunctions, and Problem solving
Theorem 17-1
The radian measure of a rotation is the ratio of the distance s traveled by a point at a radius r from the center of rotation to the length of the radius.
sr
2 2,
2 2
2 2,
2 2
2 2,
2 2
2 2
,2 2
3 1,
2 2
3 1,
2 2
3 1,
2 2
3 1
,2 2
1 3,
2 2
1 3,
2 2
1 3,
2 2
1 3
,2 2
0,1
0, 1
1,0 1,0
One Radian is the measure of an angle in standard position whose terminal side intercepts an arc of
length r.
The arc length s of a sector with radius r and central angle is given by the formula: s = r
A
B
Linear Speed
Angular Speed
Linear Speed
A child is spinning a rock at the end of a 2-foot rope at the rate of 180 revolutions per minute (rpm). Find the linear speed of the rock when it is released.
The rock is moving around a circle of radius r = 2 feet.
The angular speed of the rock in radians is:
The linear speed of the rock is:
The linear speed of the rock when it is released is 2262 ft/min 25.7 mi/hr.
Linear Speed on Earth Earth rotates on an axis through its poles. The distance from the axis to a location on Earth 40° north latitude is about 3033.5 miles. Therefore, a location on Earth at 40° north latitude is spinning on a circle of radius 3033.5 miles. Compute the linear speed on the surface of Earth at 40° north latitude.
3033.5 79412
v rw mph
409
s
.08 4.58radians
Co-Function Identities
cosb
Ac
sinb
Bc
90A B 90A B
sin cosB A
sin cos(90 )B B
The sine of an angle is the cosine of the complement and the cosine of an angle is the sine of the complement
The same is true of each trig function and its co-function
HW #17-3 Pg 746-747 1-47 Odd, 48-49
Chapter 17
17-4 Finding Function Values
2
3,
2
1
2
3,
2
1
Let's label the unit
circle with values of
the tangent. (Remember this is just y/x)
0
3
3
1
3
undef3
1
3
3
0
3
3
1
3undef
3
3
1
3
= 45°
What is the measure of this angle?
You could measure in the positive direction
= - 360° + 45°You could measure in the positive direction and go around another rotation which would be another 360°
= 360° + 45° = 405°
You could measure in the negative direction
There are many ways to express the given angle. Whichever way you express it, it is still a Quadrant I angle since the terminal side is in Quadrant I.
= - 315°
If the angle is not exactly to the next degree it can be expressed as a decimal (most common in math) or in degrees, minutes and seconds (common in surveying and some navigation).
1 degree = 60 minutes 1 minute = 60 seconds
= 25°48'30" degrees
minutesseconds
To convert to decimal form use conversion fractions. These are fractions where the numerator = denominator but two different units. Put unit on top you want to convert to and put unit on bottom you want to get rid of.
Let's convert the seconds to minutes
30" "60
'1 = 0.5'
1 degree = 60 minutes 1 minute = 60 seconds
= 25°48'30"
Now let's use another conversion fraction to get rid of minutes.
48.5' '60
1 = .808°
= 25°48.5' = 25.808°
(1,0)
(cos ,sin )
What is the length of this segment?
HW #17-4Pg 753 1-73 Odd, 74-78
Chapter 17
17-6 Trig Functions and Relationships
2 2,
2 2
2 2,
2 2
2 2,
2 2
2 2
,2 2
3 1,
2 2
3 1,
2 2
3 1,
2 2
3 1
,2 2
1 3,
2 2
1 3,
2 2
1 3,
2 2
1 3
,2 2
0,1
0, 1
1,0 1,0
(1,0)
(cos ,sin )
Let’s consider the length of this
segment?
HW #17-6Pg 767-768 1-33 Odd, 34-37
Chapter 17
17-8 Algebraic Manipulations
Row 2, 4, 6
Row 1. 3. 5
Row 2, 4, 6
Row 1. 3. 5
Row 1, 2, 3. 4, 5, 6
HW #17-8Pg 772-773 1-47 Odd, 48-51
Test Review
Determine the quadrant in which the terminal side of the angle lies
Find one positive angle and one negative angle coterminal with the given angle.
Rewrite each degree measure in radians and each radian measure in degrees.
Find the arc length of a sector with the given radius r and central angle
Evaluate the trigonometric function without using a calculator.
Evaluate the trigonometric function without using a calculator.
Find the values of the other five trigonometric functions of .
Verify the identity.
Verify the identity.
HW #R-17Pg 776-778 1-46