✔ 最佳答案
Q1)
cosθ = 1/k
sin²θ = 1 - cos²θ
sin²θ = 1 - (1/k)²
sin²θ = (k² -1)/k²
因 0° < θ < 90°,sinθ > 0
所以 sinθ = [√(k² - 1)]/k
tan(θ - 270°)
= tan(360° + θ - 270°)
= tan(θ + 90°)
= - 1/tanθ
= - cosθ / sinθ
= -(1/k) / {[√(k² -1)]/k}
= -1/√(k² -1)
= [√(k² - 1)]/(k² -1)
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Q2)
cos³x = cosx
cos³x - cosx = 0
cosx(cos²x - 1) = 0
cosx(cosx + 1)(cosx - 1) = 0
cosx = 0 或 cosx = 1 或 cosx = -1
x = 90°, 270° 或 0°, 360° 或 x = 180°
所以 x = 0°, 90°, 180°, 270°, 360°
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Q3)
情況一:2 ÷ (3 + sin²θ) 的極大值
當 0° ≤ θ ≤ 90° 時:
0 ≤ sinθ ≤ 1
0 ≤ sin²θ ≤ 1
3 + 0 ≤ 3 + sin²θ≤ 3 + 1
3 ≤ 3 + sin²θ ≤ 4
2 ÷ 4 ≤ 2 ÷ (3 + sin²θ)≤ 2 ÷ 3
1/2 ≤ 2 ÷ (3 + sin²θ)≤ 2/3
所求的極大值 = 2/3
情況二:(2 ÷ 3) + sin²θ 的極大值
當 0° ≤ θ ≤ 90° 時:
0 ≤ sinθ ≤ 1
0 ≤ sin²θ ≤ 1
(2 ÷ 3) + 0 ≤ (2 ÷ 3) + sin²θ≤ (2 ÷ 3) + 1
2/3 ≤ (2 ÷ 3) + sin²θ≤ 5/3
所求的極大值 = 5/3
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Q4)
tanx = 2 sinx
sinx/cosx = 2 sinx
(sinx/cosx) cosx = 2 sinx cosx
sinx = 2 sinx cosx
sinx - 2 sinx cosx = 0
sinx(1 - 2 cosx) = 0
sinx = 0 或 cosx = 1/2
x = 0°, 180°, 360° 或 x = 60°, (360 - 60)°
x = 0°, 60°, 180°, 300°, 360°