Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 1 Test 2 Lecture Notes
Section 6.4 I. Inverse Functions Definition f
1
f f
f x
x for all x in the domain of f.
1
x for all x in the domain of f 1 .
x
domain(f) = range ( f 1 ) range(f) = domain ( f 1 )
A. ArcSine Function
sin
1
x if and only if sin( ) = x.
Domain: 1 x 1
Notice that this is the range of sin(x)
Range:
the restricted domain of sin(x).
2
2
Example Find the exact value of: a) sin
1
b) sin
1
c) sin
1
1
2
3
2
2
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 2 Test 2 Lecture Notes
B. ArcCosine Function cos
1
x if and only if cos( ) = x.
Domain: 1 x 1
Notice that this is the range of cos(x)
Range: 0
the restricted domain of cos(x).
Example Find the exact value of: a) cos
1
2
2
b) cos 1 1 c) cos 1 0.75
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 3 Test 2 Lecture Notes
C. ArcTangent Function tan
1
x if and only if tan( ) = x.
Domain: x R Range:
2
Example Find the exact value of: a) tan
1
b) tan
1
c) tan
1
d) tan
1
0 1
3 4
2
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 4 Test 2 Lecture Notes
II. Calculating inverses from Triangles Example: A laser beam is to be directed through a small hole in the center of a circle of radius 10 feet. The origin of the beam is 35 feet from the circle. At what angle of elevation should the beam be aimed to ensure that it goes through the hole?
Ex: A 680 foot rope anchors a hot-air balloon.
a) Express the angle as a function of the height of the balloon. b) Find the angle if the balloon is 500 feet high.
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 5 Test 2 Lecture Notes
Example: A carpenter is preparing to put a roof on a garage that is 20 feet by 40 feet. A steel support beam 46 feet in length is positioned in the center of the garage. To support the roof, another beam will be attached to the top of the center beam. At what angle of elevation (pitch of the roof) is the new beam? Express your answer using Degrees/Minutes/Seconds notation.
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 6 Test 2 Lecture Notes
III. Inverse Properties of Trigonometric Functions
1.
sin sin
sin
2)
1
1
x
1
x
x
cos
tan tan
tan
x
sin
cos cos cos
3)
1
1
1
tan
x
x
Check: Is x
___, ___
Check: Is
___, ___
Check: Is x
___, ___
Check: Is
___, ___
Check: Is x _____ Check: Is
___, ___
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 7 Test 2 Lecture Notes
Examples Type I – Answer is a ratio because sin, cos, or tan is on the outside. a) cos cos 1 b) tan tan
1
5
Type II – Answer is an angle because sin 1 , cos 1 , tan 1 is on one outside. The angle one you have memorized on the unit circle.
a) tan tan
b) sin
1
1
sin
5
5 3
Type III – Answer is an angle because sin 1 , cos 1 , tan 1 is on one outside. The angle not one you have memorized on the unit circle. a) tan 1 (tan(
b) sin
1
sin
c) cos cos 1
d)
sin
1
sin
18
))
3 5
11 8
4 7
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 8 Test 2 Lecture Notes
V. Solving Equations Example: Find the exact solution of each equation for all angles between [00 ,1800 ] or else [0, ] , as specified. Type I: Angles you have memorized. a) cos( x)
2 2
(radians)
b) sin( x)
3 2 (degrees)
c) tan( x)
1 (radians)
Type II: Angles you don’t have memorized. a) cos( x) 0.7 (degrees) b) sin x
1 4 (degrees)
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 9 Test 2 Lecture Notes
IV. Steps to find the exact value of expressions involving a combination of inverse functions: 1. Let θ = inside expression. 2. Draw a picture if needed. MAKE SURE YOU ARE IN THE RIGHT QUADRANT. 3. Solve for θ 4. Plug θ or θ’s value into the expression. 5. Find the function value.
Example: Find the exact value of a) sin cos 1 1 2 b) csc cos
1
c) cot cos
1
d) tan
1
cos
1
e)
tan 2 sin 7
3
3
2
3
3 6
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 10 Test 2 Lecture Notes
6.5 Solving Oblique Triangles Oblique Triangle – A triangle with no right angle. To Solve an oblique triangle means to find the side lengths and angles. To solve, you must know: o Case I: two angles and the length of any side (SAA or ASA)
o
Case II: Two sides and the angle opposite of one of them (SSA)
o
Case III: Three sides (SSS)
o
Case IV: Two sides and their include angle (SAS)
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 11 Test 2 Lecture Notes Law of Sines
sin A a
sin B b
sin C c
1800 Property: A + B + C = 1800
To solve Case I: SAA and ASA Use 1800 property to find the third angle. Use the Law of Sines to find the two sides.
Example The bell tower of the cathedral in Pisa, Italy leans 5.6 0 from the vertical. A tourist stands 105 m from its base, with the tower leaning directly towards her. She measures the angle of elevation to the top of the tower to be 29.2 0 . Find the length of the tower to the nearest meter.
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 12 Test 2 Lecture Notes Example: The highest bridge in the world is the bridge over the Royal Gorge of the Arkansas River in Colorado. Sightings to the same point at water level directly under the bridge are taken from each side of the 880 foot long bridge. How high is the bridge?
Example Observers at P and Q are located on the side of a hill that is inclined at 320 to the horizontal, as shown. The observer at P determines the angle of elevation to a hot air balloon to be 620 . At the same instant the observer at Q measures the angle of elevation to the balloon to be 710 . If P is 60 m downhill from Q, find the distance from Q to the balloon.
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 13 Test 2 Lecture Notes To solve Case II: SSA There are three possible triangles that can occur when A is an acute angle.
There are two possible triangles that can occur when A is an obtuse angle.
Steps to solve SSA triangles 1. Use the law of sines to find the second angle. Check: If sin(A) is not equal to a number in the interval [-1, 1], then there is no solution.
Otherwise, there can be one to two angles that angle A equals:
2. Use the 1800 property to determine the third angle. 3. Use the law of sines to find the side lengths.
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 14 Test 2 Lecture Notes Example: No solution case (SSA) a = 15cm
b = 25cm
Example: One solution case (SSA) a = 22”
b = 12”
A = 420
A = 850
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 15 Test 2 Lecture Notes Example: A short-wave radio antenna is supported by two guy wires, 165 ft and 180 ft long. Each wire is attached to the top of the antenna and anchored to the ground, at two anchor points on opposite sides of the antenna. The shorter wire makes an angle of 67 0 with the ground. How far apart are the anchor points?
Recall: Two parallel lines cut by a transversal have congruent alternate interior and alternate exterior angles.
Example: A communications tower is located at the top of a steep hill. The angle of inclination of the hill is 580 . A guy wire is to be attached to the top of the tower and to the ground, 100 meters downhill from the base of the tower. The angle required for the guy wire.
in the figure is determined to be 120 . Find the length of cable
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 16 Test 2 Lecture Notes Example: Two solutions case (SSA) a =12m
b = 31m
A = 20.50
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 17 Test 2 Lecture Notes Section 6.6 Recall ---To solve, you must know: o Case I: two angles and the length of any side (SAA or ASA) o Case II: Two sides and the angle opposite of one of them (SSA) o Case III: Three sides (SAS) o Case IV: Two sides and their include angle (SSS)
Law of Cosines
a2
b2
c2
2bc cos A
“The square of one side of a triangle is equal to the sum of the squares of the other two sides minus twice their product times the cosine of the included angle.”
This also implies an ALTERNATIVE FORM.
cos A
b2
c2 a2 2bc
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 18 Test 2 Lecture Notes Case III: Two sides and their include angle (SAS) 1. Use the law of cosines to find the missing side. 2. Use the law of sines to find the smallest angle (opposite the shortest side). 3. Use the 180 0 property to find the third angle. Recall: distance = rate*time Example: Two straight roads diverge at an angle of 65 0 . Two cars leave the intersection at 2:00PM, one traveling at 50 mi/hr and the other at 30 mi/hr. How far apart are the cars at 2:30pm?
Example: A pilot flies in a straight path for 1 hr 30 min. she then makes a course correction, heading 10 0 to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 625 mi/hr, how far is she from her starting position?
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 19 Test 2 Lecture Notes Bearings:
Example Two boats leave the same port at the same time. One travels at a speed of 30 mi/h in the direction N 50 0 E and the other travels at a speed of 26 mi/h in a direction of S 70 0 E . Use the law of cosines to determine how far apart the boats are after one hour.
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 20 Test 2 Lecture Notes Recall: Two parallel lines cut by a transversal have congruent alternate interior angles and congruent alternate exterior angles.
Example A fisherman leaves his home and heads in the direction N 70 0 W . He travels 30 mi and reaches Egg Island. The next day he sails N10 0 E for 50 mi, reaching Forrest Island. a) Find the distance between the fisherman’s home port and Forrest Island. b) Find the bearing from Forrest Island back to his home port.
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 21 Test 2 Lecture Notes Case IV: Three sides (SSS) 1. Use the law of cosines to find the largest angle. 2. Use the law of sines to find the second angle. 3. Use the 180 degree property to find the third angle. Example: Solve the triangle with a = 8ft, b = 19 ft, and c = 14 ft.
Example: An airboat travels 60 miles due east, then adjusts its course northward. After traveling 80 miles in that direction, the ship is 139 miles from its point of departure. Describe the bearing from point of departure to where the boat stopped.
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 22 Test 2 Lecture Notes Area of a SSS Triangle:
K
s s a s b s c
where s
1 a b c 2
Example Find the area of a triangle with lengths a = 43m, b = 53m, and c = 72m.
Example You want to buy a triangular lot measuring 510 yards by 840 yards by 1170 yards. The price of the land is $2000 per acre. How much does the land cost? (1acre = 4840 yd 2 ).
Sections 6.4, 6.5, 6.6, 7.3, 7.4, 7.5, 7.6 23 Test 2 Lecture Notes
Section 7.3
