A.maths 緊急~ 10分

2007-01-14 6:24 am
1.Prove the identity

[ 1 - sec(π/2 -θ) - cot(2π - θ ) ] /[1+sec(3π/2+θ) + cot ( π - θ ) ] = cot θ / 1+csc θ

請以簡單的方法去列明step~ thz

2.If sec^2(π - θ ) / { 3cot^2 ( π /2 +θ) -7 } =1 and π/2 < θ < π ,find the value of cosθ .

Ans : -1/√5

請列埋step~~~ thz

回答 (1)

2007-01-14 5:21 pm
✔ 最佳答案
1.
 L.H.S.
= [1 - sec(π/2 - θ) - cot(2π - θ)] / [1 + sec(3π/2 + θ) + cot(π - θ)]
 1 - 1/cos(π/2 - θ) - cos(2π - θ)/sin(2π - θ)
= ---------------------------------------------------------------
 1 + 1/cos(3π/2 + θ) + cos(π - θ)/cos(π - θ)
 1 - 1/sinθ - cosθ/(-sinθ)   sinθ
= ------------------------------------ × ------------
 1 + 1/sinθ + (-cosθ)/sinθ   sinθ
 sinθ - 1 + cosθ   sinθ + 1 + cosθ
= ------------------------ × ----------------------------
 sinθ + 1 - cosθ   sinθ + 1 + cosθ
 (sinθ + cosθ)² - 1²
= ---------------------------
 (sinθ + 1)² - cos²θ
 sin²θ + 2sinθcosθ + cosθ² - 1
= ---------------------------------------------
  sin²θ + 2sinθ + 1 - cos²θ
  2sinθcosθ + 1 - 1
= --------------------------------
 sin²θ + 2sinθ + sin²θ
  2sinθcosθ
= ------------------------
 2sinθ(sinθ + 1)
  cosθ   1/sinθ
= -------------×------------
 sinθ + 1  1/sinθ
  cotθ
= ---------------
 1 + cscθ

2.
  sec²(π - θ)
--------------------------- = 1
3cot²(π/2 + θ) - 7

  sec²(π - θ) = 3cot²(π/2 + θ) - 7
1 / [cos(π - θ)]² = 3[cos(π/2 + θ) / sin(π/2 + θ)]² - 7
  1 / (-cosθ)² = 3(-sinθ / cosθ)² - 7
   1 / cos²θ = 3(sin²θ / cos²θ) - 7
       1 = 3sin²θ - 7cos²θ
       1 = 3(1 - cos²θ) - 7cos²θ
       1 = 3 - 3cos²θ - 7cos²θ
    10cos²θ = 2
     cos²θ = 1/5
      cos = 1/√5 (rejected as π/2 < θ < π) or cos = -1/√5


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