Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of...
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Transcript of Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of...
![Page 1: Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:](https://reader035.fdocuments.in/reader035/viewer/2022081513/56649f435503460f94c63846/html5/thumbnails/1.jpg)
Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
Aim: How do we use the principals of trigonometry to solve problems?
Do Now:
3If sin , the values of in
2the interval 0 360 are
1) 60 and 120 3) 120 and 300
2) 120 and 200 4) 240 and 300
![Page 2: Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:](https://reader035.fdocuments.in/reader035/viewer/2022081513/56649f435503460f94c63846/html5/thumbnails/2.jpg)
Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
Find the value of w to the nearest tenth.
w
10 represents the side adjacent to the 540 angle.
w represents the side opposite the 540 angle.
Tangent ratio
Tan 540 = Leg opposite 540 Leg adjacent to 540
Tan 540 = w 10 1.37638192 = w
10
w = (10)1.37638192 = 13.7638192 . . .
10540
w 13.8
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Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
All but two of the pyramids built by the ancient Egyptians have faces inclined at 520. Suppose an archaeologist discovers the ruins of a pyramid. Most of the pyramid has eroded, but she is able to determine that the length of a side of the square base is 82m. How tall was the pyramid, assuming its faces were inclined 520?
520
h
Tan 520 = h/41
1.279941. . . = h/41
(41)1.279941. . . = h
52.4776. . . = h
52 meters h 41 m
82 m82 m
520
Height h
m.
![Page 4: Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:](https://reader035.fdocuments.in/reader035/viewer/2022081513/56649f435503460f94c63846/html5/thumbnails/4.jpg)
Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
1st Degree Trig Equations
algebraic trigonometric2 1 0x 2cos 1 0
what values of x make this
statement true?
conditional equalities
what values of make this
statement true?
2 1 0x 2cos 1 0 1 1 1 1
2 1x 2cos 1 1
2x
1cos
2
1 1cos ?
2 o60 or
3
![Page 5: Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:](https://reader035.fdocuments.in/reader035/viewer/2022081513/56649f435503460f94c63846/html5/thumbnails/5.jpg)
Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
1st Degree Trig Equations2cos 1 0
1 12cos 1
1cos
2
1 1cos 60 or
2 3
Periodic Function - period of 2
Domain = | Real numbers
Range = | 1 1
x x
y y
2
-5 5
g x = 0.5
f x = cos x QI QII QIII QIV
cosine is positive in QI and QIV
52 and 2
3 6k k
![Page 6: Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:](https://reader035.fdocuments.in/reader035/viewer/2022081513/56649f435503460f94c63846/html5/thumbnails/6.jpg)
Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
Beware!
3sin 5 7 5
sin 4
1sin 4 ?!!!
5
3sin 12
no solution
or
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Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
Model Problem
Find for values between 0 and 360o to the nearest degree for
3 tan - 5 = 75
tan 4
5
3tan 12
arctan 4 76 or 1.3258 radians
tan is positive in QI and QIII
76o and 256o
10
8
6
4
2
-2
-4
5
q x = 7
h x = 3tan x -5
QI QII QIII QIV
![Page 8: Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:](https://reader035.fdocuments.in/reader035/viewer/2022081513/56649f435503460f94c63846/html5/thumbnails/8.jpg)
Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
Model Problem
Solve for cos : cos = 3 cos + 1 for all values between 0 and 2.
cos = 3 cos + 13cos 3cos
2cos 1 1
cos2
1arccos reference angle of
2 3
cosine is negative in QII and QIII
2 in QII =
3 34
in QIII = 3 3
4
3
2
1
-1
-2
2 4 6
g x = 3cos x +1
f x = cos x
QI QII QIII QIV
![Page 9: Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:](https://reader035.fdocuments.in/reader035/viewer/2022081513/56649f435503460f94c63846/html5/thumbnails/9.jpg)
Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
Model Problem
Solve for : 5 (sin + 3) = sin + 12 for all values between 0 and 360o to the nearest minute.
3arcsin reference angle of 48.59
4
sine is negative in QIII and QIV
in QIII 180 48.59 228.59
in QIV 360 48.59 311.41
5 (sin + 3) = sin + 125sin 15 sin 12
4sin 3 3
sin4
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Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
Model Problem
Solve for : 5 (sin + 3) = sin + 12 for all values between 0 and 360o to the nearest minute.
=228.59 228 .59 60 228 35'
=311.41 311 .41 60 311 25'
20
15
10
5
-5
2 4 6
g x = sin x +12
f x = 5sin x +3
QI QII QIII QIV
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Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig.
Regents Prep
If 0 < < 360o, find all values of that satisfy the equation -4 cos = 1
1. 104o, 256o
2. 76o, 104o
3. 24o, 104o, 156o, 256o
4. 76o, 104o 156o, 256o