若sina,seca為方程2x^2-kx+3=0的2根,其中圓周<3圓周/2,試求k的值(答案要保留根式)
更新1:
2: 答案系-19√13/13
AMATHS
2007-02-14 3:00 am
1:如果sec(0字中間有一畫)= -17/8,圓周(3.1415xxxx)/2<0<圓周 試求tan0(中間有一畫)的值
若sina,seca為方程2x^2-kx+3=0的2根,其中圓周<3圓周/2,試求k的值(答案要保留根式)
若sina,seca為方程2x^2-kx+3=0的2根,其中圓周<3圓周/2,試求k的值(答案要保留根式)
回答 (2)
2007-02-16 9:06 pm
我印象中唔知邊個白痴仔就抄左呢度答案去答其他問題
而投票者竟然有一半以上投佢
而投票者竟然有一半以上投佢
2007-02-14 3:40 am
secθ = -17/8 where π/2 < θ < π
cosθ = -8/17
sinθ = 15/17 (as π/2 < θ < π)
tanθ = sinθ/cosθ = -15/8
QUESTION 2
sin a + sec a = k/2
sin a sec a = 3/2
sin a + sec a = k/2
tan a = 3/2
As π < θ < 3π/2
tan a = 3/2 gives
sin a = 3/√13
cos a = 2/√13
sin a + sec a
= sin a + (1/cos a)
= (3/√13) + [(√13)/2]
= (3√13/13) + [(√13)/2]
= [(6√13)/13]/2 + [(√13)/2]
= [(6√13)/13 + (√13)]/2
= k/2
Hence, k = (6√13)/13 + (√13)
cosθ = -8/17
sinθ = 15/17 (as π/2 < θ < π)
tanθ = sinθ/cosθ = -15/8
QUESTION 2
sin a + sec a = k/2
sin a sec a = 3/2
sin a + sec a = k/2
tan a = 3/2
As π < θ < 3π/2
tan a = 3/2 gives
sin a = 3/√13
cos a = 2/√13
sin a + sec a
= sin a + (1/cos a)
= (3/√13) + [(√13)/2]
= (3√13/13) + [(√13)/2]
= [(6√13)/13]/2 + [(√13)/2]
= [(6√13)/13 + (√13)]/2
= k/2
Hence, k = (6√13)/13 + (√13)
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