急! F4 Basic Trigonometry 13q4

2012-05-11 7:25 am
請詳細步驟教我計以下幾條 :
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回答 (1)

2012-05-11 8:14 am
✔ 最佳答案
13.
(a)
240° = 180° + 60°
240° = 270° - 30°

sin240°
= sin(180° + 60°)
=-sin60°
= -(√3)/2

sin240°
= sin(270° - 30°)
= -cos30°
= -(√3)/2

(b)
120° = 180° - 60°
120° = 90° + 30°

sin120°
= sin(180° - 60°)
=sin60°
= (√3)/2

sin120°
= sin(90° + 30°)
= cos30°
= (√3)/2


14.
(b)
(1 - tan²θ) / (1 + tan²θ) +sin²θ
= [1 - (sinθ/cosθ)²] / [1 + (sinθ/cosθ)²]+ sin²θ
= [(cos²θ - sin²θ)/cos²θ] / [(cos²θ+ sin²θ)/cos²θ] + sin²θ
= (cos²θ - sin²θ) / (cos²θ + sin²θ)+ sin²θ
= (cos²θ - sin²θ) + sin²θ
= cos²θ


15.
(b)
cos(180° + θ)•cos(360° - θ) + 1
= -cosθ•cosθ + 1
= 1 - cos²θ
= sin²θ


[tanθ - sin(-θ)] / [1 -cos(180° - θ)]
= (tanθ + sinθ) / (1 + cosθ)
= [(sinθ/cosθ) + sinθ] / (1 + cosθ)
= [(sinθ + sinθcosθ)/cosθ] / (1 + cosθ)
= [(sinθ + sinθcosθ)] / [cosθ(1 +cosθ)]
= [sinθ(1 + cosθ)] / [cosθ(1 + cosθ)]
= sinθ / cosθ
= tanθ


17.
sin(270° - θ)•cos(90° + θ) / tan(180° + θ)
= -cosθ•(-sinθ) / tanθ
= sinθ•cosθ / (sinθ/cosθ)
= cos²θ


18.
cos(90° - θ)•sin(360° - θ) + cos(180° - θ)•sin(90°- θ)
= sinθ•(-sinθ) + (-cosθ)•cosθ
= -sin²θ - cos²θ
= -(sin²θ + cos²θ)
= -1
參考: fooks


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