F.4 Trigonometry

2013-04-27 6:20 am
1. Solve -sin2x° = 2/(5-3sin2x°) for 0 <= x <360

2. It is given that sin θ = 2/3 and cosθ <0
Without using a calculator, find the values of the following expressions.
i) -sin(-θ) - cos (θ-270°) ii) sin(180°+θ) + sin^2 (θ-270°)
________________________
tan(-540°-θ)

3. Solve the following equation for 0 <= x <=360
a) -2 tan (3x+6°) = 1
b) √2 cos 2x - 4 cos (90°-2x) = 0

3 answer: a/ 49.1°, 109.1°, 169.1°, 229.1°, 289.1°, 349.1°

b) 9.7°, 99.7° , 189.7°, 279.7°


Thank you !!!

回答 (1)

2013-04-27 8:34 am
✔ 最佳答案
1.
0 ≤ x ≤ 360
Then 0 ≤ 2x ≤ 720

- sin 2x° = 2 / (5 - 3 sin 2x°)
- sin 2x° (5 - 3 sin 2x°) = 2
3 sin² 2x° - 5 sin 2x - 2 = 0
(3 sin 2x + 1) (sin 2x - 2) = 0
sin 2x = -1/3 or sin 2x = 2 (rejected)
2x = 180+19.47, 360-19.47, 540+19.47, 720-19.47
x = 99.7, 170.3, 279.7, 350.3


2.
(i)
sin θ = 2/3
cos θ = -√[1 - (2/3)²] = -(√5)/3

- sin (-θ) - cos (θ - 270°)
= - (-sin θ) - cos (270° - θ)
= sin θ + sin θ
= 2 sin θ
= 2 x (2/3)
= 4/3

(ii)
[sin (180° + θ) + sin² (θ - 270°)] / tan (-540° - θ)
= [- sin θ + (- sin (270° - θ))²] / tan -(360° + 180° + θ)
= [- sin θ + cos²θ)] / [- tan θ]
= [- (2/3) + (-(√5)/3)²] / [- sin θ / cos θ]
= [- (2/3) + (5/9)] / [- (2/3) / (-√5)/3]
= - (1/9) / (2/√5)
= -(1/9) x [(√5)/2]
= -(√5)/18


3.
(a)
0° ≤ x ≤ 360°
6° ≤ (3x + 6°) ≤ 1086°

-2 tan (3x + 6°) = 1
tan (3x + 6°) = -1/2
3x + 6° = (180-26.57)°, (360-26.57)°, (540-26.57)°, (720-26.57)°, (900-26.57)°,(1080-26.57)°
x = 49.1°, 109.1°, 169.1°, 229.1°, 289.1°, 349.1°

(b)
0° ≤ x ≤ 360°
0° ≤ 2x ≤ 720°

√2 cos 2x - 4 cos (90° - 2x) = 0
√2 cos 2x = 4 cos (90° - 2x)
√2 cos 2x = 4 sin 2x)
sin 2x / cos 2x = (√2)/4
tan 2x = (√2)/4
2x = 19.47°, (180+19.47)°, (360+19.47)°, (540+19.47)°
x = 9.7°, 99.7°, 189.7°, 279.7°
參考: 賣女孩的火柴


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