急! 三角學09Q2

2012-06-21 7:27 pm
請詳細步驟教我計以下幾條 :
(不要網址回答)


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回答 (1)

2012-06-21 8:28 pm
✔ 最佳答案
8a.
sin(θ-270°)
= -sin(270°-θ)
= -( -cosθ)
= cosθ

8b.
sinθcos(90°-θ)-sin(θ-270°)cos(180°-θ)
= sin²θ + cos²θ (因 cos(90°-θ) = sinθ, cos(180°-θ) = -cosθ )
= 1
------------
9a
1/(1+cosθ) + 1/(1-cosθ)
= [(1-cosθ)+(1+cosθ)]/[(1+cosθ)(1-cosθ)]
= 2 / (1-cos²θ)
= 2/sin²θ

9b.
1/(1+cosθ) + 1/(1-cosθ)
= 2/sin²θ
= 2/(1/3)²
= 18
----------------
10.
留意tan(90°-θ) = 1/tanθ,

tanθ + 1/tanθ = 3
(tanθ + 1/tanθ)² = 3²
tan²θ + 2 + 1/tan²θ = 9
tan²θ + 1/tan²θ = 7
------------------
11.
sinθ + cosθ = k
(sinθ + cosθ)² = k²
sin²θ +2sinθcosθ+cos²θ = k²
1+ 2sinθcosθ = k²
sinθcosθ = (k²-1)/2


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