Transcript 9.5 Trigonometric Ratios

ANGLE
ANGLE
ANGULAR MEASUREMENT
Terminal side
Initial side
• Angle is defined a rotation result from initial side to
the terminal side
• An angle has “a positive” sign if its rotation is
anticlockwise
• An angle has “a negative” sign if its rotation is
clockwise
• We only talk about the magnitude of angle~ not
observe its signs
What is a radian?
radius
arc=radius
o
radius
Angle = 1 rad
Why do mathematicians use radians instead of
degrees?
How many times does the radius
divide into the circumference?
There are 2
1 radian =
radians in a circle.
= 57.3o
THE TRIGONOMETRIC RATIOS

2 rad  360

 rad  180

180
1 rad 



1 
rad
180
4
Convert each angle in
radians to degrees.
Convert each angle in
degrees to radians.
1. 2c
114.6o
1. 65o
1.13c
2. 5c
286.5o
2. 200o
3.49c
3. 3
4.
5.
c
c
c
3.
120o
90o
4.
180o
240o
5.
540o
330o
Two important formula using radians
Length of an arc using
radians
Area of a sector using radians
Trigonometric
Ratios
Objectives/Assignment
• Find the since, the cosine, and the tangent of an
acute triangle.
• Use trionometric ratios to solve real-life problems
Finding Trig Ratios
• A trigonometric ratio is a ratio of the lengths
of two sides of a right triangle. The word
trigonometry is derived from the ancient
Greek language and means measurement of
triangles. The three basic trigonometric ratios
are sin, cos, and tan
Trigonometric Ratios
• Let ∆ABC be a right
triangle. The since,
the cosine, and the hypotenusec
tangent of the acute
angle A are defined
b
A
as follows.
side adjacent to angle A
cos A =
sin A =
Side opposite A
hypotenuse
=
B
Side
a opposite
angle A
C
Side adjacent to A b
=
hypotenuse
c
a
c
tan A =
Side opposite A
a
=
Side adjacent to A b
Exercise:
• In the PQR triangle is right angled at R
happens cos P = 8/ 17…find the value of tg P
and tg Q ?
Exercises
1. PQR triangle is right angled at R & sin P cosQ = ⅗. Find out
the value tgP
tgQ
c
ABC triangle is right angle at A. BC=p,
2
AD is perpendicular at BC, DE
D
E
perpendicular at AC, angle B = Q, prove
that : DE=psin2 Qcos Q
B
A
Evaluating Trigonometric Functions
• Acute angle A is drawn in
standard position as shown.
Right-Triangle-Based Definitions of Trigonometric Functions
For any acute angle A in standard position,
sin A 
cos A 
tan A 
y
r
x
r
y
x
csc A 
r

side opposite
hypotenuse
y

side adjacent
sec A 
r

side opposite
cot A 
x
hypotenuse
side adjacent
x
y



hypotenuse
side opposite
hypotenuse
side adjacent
side adjacent
side opposite
Finding Trigonometric Function Values
of an Acute Angle in a Right Triangle
Example Find the values of sin A, cos A, and tan A
in the right triangle.
Solution
– length of side opposite angle A is 7
– length of side adjacent angle A is 24
– length of hypotenuse is 25
7
sin A  ,
25
24
cos A  ,
25
7
tan A 
24
Trigonometric Function Values of
Special Angles
• Angles that deserve special study are 30º, 45º, and 60º.
Using the figures
above, we have the
exact values of the
special angles
summarized in the
table on the right.
Let’s try to prove it !
Y
B
45O
O
A
X
OA=OB
OA2 + OB2 = OC2
OA2 + OA2 = r2
2OA2 = 1
OA2 = ½
OA =
How about 300 , 600
and 900
= OB
Cofunction Identities
•
In a right triangle ABC, with right angle C, the acute
angles A and B are complementary.
a
sin A   cos B
c
a
tan A   cot B
b
c
sec A   csc B
b
•
Since angles A and B are complementary, and
sin A = cos B, the functions sine and cosine are called
cofunctions. Similarly for secant and cosecant, and
tangent and cotangent.
Cofunction Identities
If A is an acute angle measured in degrees, then

sin A  cos(90  A)

cos A  sin( 90  A)

csc A  sec(90  A)

sec A  csc(90  A)

tan A  cot(90  A)

cot A  tan(90  A)
If A is an acute angle measured in radians, then



sin A  cos  A 
2 
 
cos A  sin   A 
2 



csc A  sec  A 
2 


sec A  csc  A 
2 
   A 
2 


cot A  tan   A 
2 
tan A  cot
Note
These identities actually apply to all angles (not just
acute angles).
Reference Angles
•
A reference angle for an angle , written  , is
the positive acute angle made by the terminal
side of angle  and the x-axis.
Example
Find the reference angle for each angle.
(a) 218º
(b) 5
6
Solution
5 
(a)   = 218º – 180º = 38º (b)     

6
6
Special Angles as Reference Angles
Example
Find the values of the trigonometric
functions for 210º.
Solution
The reference angle for 210º is
210º – 180º = 30º.
Choose point P on the terminal side so
that the distance from the origin to P is
2. A 30º - 60º right triangle is formed.
1
3
3


sin 210  
cos 210  
tan 210 
2
2
3
2 3



csc 210  2 sec 210  
cot 210  3
3

Finding Trigonometric Function Values Using
Reference Angles
Example Find the exact value of each expression.
(a) cos(–240º)
(b) tan 675º
Solution
(a) –240º is coterminal with 120º.
The reference angle is
180º – 120º = 60º. Since –240º
lies in quadrant II, the
cos(–240º) is negative.
1
cos( 240 )   cos 60  
2
Similarly,
tan 675º = tan 315º
= –tan 45º = –1.

•

Finding Angle Measure
Example Find all values of , if  is in the interval
[0º, 360º) and cos   22 .
Solution Since cosine is negative,  must lie in
either quadrant II or III. Since
So the reference angle   = 45º.
cos  22 , cos1 22  45.
The quadrant II angle  = 180º – 45º = 135º, and the
quadrant III angle  = 180º + 45º = 225º.
Ex. 1: Finding Trig Ratios
• Compare the sine, the
cosine, and the tangent
ratios for A in each
triangle beside.
• By the Similarity
Theorem, the triangles
are similar. Their
corresponding sides are
in proportion which
implies that the
trigonometric ratios for
A in each triangle are
the same.
B
17
8
A
C
15
B
8.5
4
A
7.5
C
Ex. 1: Finding Trig Ratios
Large
sin A =
cosA =
tanA =
opposite
hypotenuse
adjacent
hypotenuse
opposite
adjacent
Small
8 ≈ 0.4706
17
4 ≈ 0.4706
8.5
7.5 ≈ 0.8824
8.5
15 ≈ 0.8824
17
8 ≈ 0.5333
15
4 ≈ 0.5333
7.5
B
B
17
8.5
4
8
A
A
15
7.5
C Trig ratios are often
expressed as decimal
approximations.
C
Ex. 2: Finding Trig Ratios
S
sin S =
cosS =
tanS =
opposite
hypotenuse
adjacent
hypotenuse
opposite
adjacent
5 ≈ 0.3846
13
12 ≈ 0.9231
13
5 ≈ 0.4167
12
R
opposite
5
13 hypotenuse
T
12
adjacent
S
Ex. 2: Finding Trig Ratios—Find the sine, the
cosine, and the tangent of the indicated angle.
R
sin S =
cosS =
tanS =
opposite
hypotenuse
adjacent
hypotenuse
opposite
adjacent
12 ≈ 0.9231
13
5 ≈ 0.3846
13
12 ≈ 2.4
5
R
adjacent
5
13 hypotenuse
T
12
opposite
S
Notes:
• If you look back, you will notice that the
sine or the cosine of an acute triangles is
always less than 1. The reason is that
these trigonometric ratios involve the
ratio of a leg of a right triangle to the
hypotenuse. The length of a leg or a right
triangle is always less than the length of
its hypotenuse, so the ratio of these
lengths is always less than one.
Using Trigonometric Ratios in Real-life
• Suppose you stand and look up at a point in
the distance. Maybe you are looking up at the
top of a tree as in Example 6. The angle that
your line of sight makes with a line drawn
horizontally is called angle of elevation.
Ex. 6: Indirect Measurement
• You are measuring the height of
a Sitka spruce tree in Alaska.
You stand 45 feet from the base
of the tree. You measure the
angle of elevation from a point
on the ground to the top of the
top of the tree to be 59°. To
estimate the height of the tree,
you can write a trigonometric
ratio that involves the height h
and the known length of 45 feet.
The math
opposite
tan 59° =
Write the ratio
adjacent
h
tan 59° =
Substitute values
45
45 tan 59° = h
Multiply each side by 45
45 (1.6643) ≈ h
Use a calculator or table to find tan 59°
75.9 ≈ h
Simplify
The tree is about 76 feet tall.
Ex. 7: Estimating Distance
• Escalators. The escalator at
the Wilshire/Vermont Metro
Rail Station in Los Angeles
rises 76 feet at a 30° angle.
To find the distance d a
person travels on the
escalator stairs, you can write
a trigonometric ratio that
involves the hypotenuse and
the known leg of 76 feet.
d
30°
76 ft
Now the math
opposite
sin 30° =
hypotenuse
76
sin 30° =
d
Write the ratio for
sine of 30°
30°
Substitute values.
d
d sin 30° = 76
76
d=
Divide each side by sin 30°
sin 30°
76
d=
Multiply each side by d.
0.5
d = 152
Substitute 0.5 for sin 30°
Simplify
A person travels 152 feet on the escalator stairs.
76 ft