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trigonometry (1)

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plz step by step~thanks 圖片參考:http://upload.lsforum.net/users/public/s33401RWs18.bmp

最佳解答:

1. (tanθ)^2 / [ 1 + (tanθ)^2 ] + (cosθ)^2 = (sinθ)^2 / [ (cosθ)^2 + (sinθ)^2 ] + (cosθ)^2 = (sinθ)^2 + (cosθ)^2 = 1 2. cosθ≧1 , 4+2cosθ≧4+2(1)=6 3/(4+2cosθ) ≦ 3/6 = 1/2 So , the greatest value of 3/(4+2cosθ) is 1/2 . 3. When 0°<θ<90°, cosθ > 0 (tanθ)^2 = 4a^2b^2 / (a^2-b^2)^2 (a^2-b^2)^2(sinθ)^2 = 4a^2b^2 (cosθ)^2 (a^2-b^2)^2 [ 1 - (cosθ)^2 ] = 4a^2b^2 (cosθ)^2 (a^2 - b^2)^2 = [4a^2b^2 + (a^2-b^2)^2 ](cosθ)^2 (a^2 - b^2)^2 = [4a^2b^2+a^4-2a^2b^2+b^4](cosθ)^2 [(a^2-b^2) / (a^2+b^2) ] = (cosθ)^2 cosθ = |a^2 - b^2| / (a^2 + b^2) ( since cosθ > 0 ) 4. ∠ADC = 90° . Consider ΔADC , tanα = AD / DC = p / DC ... (1) Consider ΔABD , tanβ = p / BD = p / (DC + BC) (DC + BC )tanβ = p BC = p/tanβ - DC = p/tanβ - p/tanα ( By (1) ) 5. AC = BCcosα , Area of ΔABC = 0.5*BC*AC*sin∠ACB =0.5*BC^2*sinαcosα =0.5*4*(sin2α / 2) =sin2α BD = BC tanβ area of ΔBCD = 0.5*BC*BD = 0.5*2*2tanβ = 2tanβ So , the area of ABDC = sin2α + 2tanβ 2012-01-28 13:30:54 補充: 更正第2條 : 2. cosθ ≧ -1 , 4+2cosθ ≧ 4+2(-1) = 2 > 0 So , 3 / (4+2cosθ) ≦ 3/2 The greatest value of 3 / (4+2cosθ) is 3/2 . 2012-01-28 13:32:08 補充: cosθ (greater than or equal) -1 , 4+2cosθ (greater than or equal ) 4+2(-1) = 2 > 0 So , 3 / (4+2cosθ) (less than or equal) 3/2 The greatest value of 3 / (4+2cosθ) is 3/2 .

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