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 o radius arc=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 radians in a circle.
1 radian = = 57.3
o
2
THE TRIGONOMETRIC RATIOS
rad 360 rad 180 1 rad 180 1 180 rad
4
Convert each angle in radians to degrees.
1. 2
c
114.6
o
2. 5
c
286.5
o
3. 3
c
4.
c
540
o
90
o
Convert each angle in degrees to radians.
1. 65
o
2. 200
o
3. 120
o
4. 180
o
1.13
c
3.49
c
5.
c
240
o
5. 330
o
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 tangent of the acute angle A are defined hypotenuse c as follows.
A b side adjacent to angle A B a Side opposite angle A C cos A = Side adjacent to A = hypotenuse b c sin A = Side opposite A hypotenuse = a c tan A = Side opposite A Side adjacent to A = 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 2 D E
ABC triangle is right angle at A. BC=p, AD is perpendicular at BC, DE perpendicular at AC, angle B = Q, prove that : DE=psin 2 Qcos Q
A B
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
y r
side opposite hypotenuse csc
A
r y
hypotenuse side opposite cos
A
x r
side adjacent hypotenuse sec
A
r x
hypotenuse side adjacent tan
A
y x
side opposite side adjacent cot
A
x y
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 sin
A
7 25 , cos
A
24 25 , tan
A
7 24
Trigonometric Function Values of Special Angles
• Angles that deserve special study are 30 º , 45 º , and 60 º .
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 O 45 O A X OA=OB OA 2 + OB 2 OA 2 + OA 2 = OC 2 = r 2 2OA 2 OA 2 = 1 = ½ OA = = OB How about 30 0 and 90 0 , 60 0
• •
Cofunction Identities
In a right triangle ABC, with right angle C, the acute angles A and B are complementary.
sin
A
a c
cos
B
tan
A
a b
cot
B
sec
A
c b
csc
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 cos
A
sin( 90
A
)
A
) csc
A
sec( 90 sec
A
csc( 90
A
)
A
) tan
A
cot( 90 cot
A
tan( 90
A
)
A
) If
A
is an acute angle measured in radians, then sin
A
cos
A
cos sin 2 2
A A
csc
A
sec
A
sec 2 csc 2
A A
tan
A
cot
A
cot 2 tan 2
A A
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
(a) 218 º Find the reference angle for each angle.
(b) 5 6
Solution
(a) = 218 º – 180 º = 38 º (b) 5 6 6
Special Angles as Reference Angles Example
Find the values of the trigonometric functions for 210 º .
Solution
210 º – 180 º The reference angle for 210 º = 30 º .
is Choose point
P
on the terminal side so that the distance from the origin to
P
2. A 30 º - 60 º right triangle is formed.
is sin 210 1 2 csc 210 2 cos 210 sec 210 3 2 2 3 3 tan 210 3 3 cot 210 3
Finding Trigonometric Function Values Using Reference Angles Example
Find the exact value of each expression.
(a) cos(–240 º ) (b) tan 675 º •
Solution
(a) –240 º is coterminal with 120 º . The reference angle is 180 º – 120 º = 60 º . Since –240 º lies in quadrant II, the cos(–240 º ) is negative.
cos( 240 ) cos 60 Similarly, tan 675 º = tan 315 º = –tan 45 º = –1.
1 2
Finding Angle Measure
Example Find all values of [0º, 360º) and cos 2 2 , if .
is in the interval
Solution
Since cosine is negative, either quadrant II or III. Since So the reference angle = 45º.
must lie in cos 2 2 , cos 1 2 2 45 .
The quadrant II angle quadrant III angle = 180º – 45º = 135º, and the = 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.
A 15 17 A 7.5
8.5
B B 4 C 8 C
A
Ex. 1: Finding Trig Ratios
Large sin A = opposite hypotenuse 8 17 ≈ 0.4706 Small 4 8.5
≈ 0.4706 cosA = adjacent hypotenuse 15 17 ≈ 0.8824 7.5
8.5
≈ 0.8824 tanA = opposite adjacent 8 15 ≈ 0.5333 4 7.5
≈ 0.5333 17 15 B B 8.5
8 A 7.5
C
Trig ratios are often expressed as decimal approximations.
C 4
Ex. 2: Finding Trig Ratios
sin S = opposite hypotenuse cosS = adjacent hypotenuse tanS = opposite adjacent S 5 13 ≈ 0.3846 12 13 ≈ 0.9231 5 12 ≈ 0.4167 R opposite 5 T 13 hypotenuse 12 adjacent S
Ex. 2: Finding Trig Ratios—Find the sine, the cosine, and the tangent of the indicated angle.
sin S = opposite hypotenuse cosS = adjacent hypotenuse tanS = opposite adjacent R 12 13 ≈ 0.9231 5 13 ≈ 0.3846 12 ≈ 5 2.4
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
tan 59 ° = opposite adjacent tan 59 ° =
h
45 45 tan 59 ° = h 45 (1.6643) ≈ h 75.9 ≈ h Write the ratio Substitute values Multiply each side by 45 Use a calculator or table to find tan 59 ° 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.
30 ° d 76 ft
Now the math
sin 30 ° =
opposite
hypotenuse sin 30 ° = 76
d
Write the ratio for sine of 30 ° Substitute values.
d sin 30 ° = 76 Multiply each side by d.
30 ° d
d =
76 sin 30 ° 76
d =
0.5 Divide each side by sin 30 ° Substitute 0.5 for sin 30 ° d = 152 Simplify A person travels 152 feet on the escalator stairs.
76 ft