a maths

假設a是下列x的二次方程的公共根

x^2-4xsiny-2=0------------(1)

x^2-4xcosy+2=0-----------(2)

a)證明a=1/(cosy-siny)

因為a是公共根,

a^2-4asiny-2=0------------(3)

a^2-4acosy+2=0-----------(4)

(3) - (4)

-4asiny - 2 - (-4acosy + 2) = 0

-4asiny + 4acosy - 4 = 0

a(cosy - siny) - 1 = 0

a = 1/(cosy - siny)


b)由此,證明sin^2y=1/4

因為a = 1/(cosy - siny) (已證)

x^2-4xsiny-2=0

[ 1/(cosy - siny) ]^2 - 4[ 1/(cosy - siny) ]siny - 2 = 0

1/(cosy - siny)^2 - 4siny/(cosy - siny) - 2 = 0

1 - 4siny(cosy - siny) - 2(cosy - siny)^2 = 0

1 - 4sinycosy + 4sin^2y + 4sinycosy = 0

1 + 4sin^2y = 0

sin^2y = 1/4

2007-02-09 20:12:28 補充：
sorry,there is typing mistakeshere is the corrections1 - 4sinycosy 4sin^2y - 2 4sinycosy = 0-1 4sin^2y = 0sin^2y = 1/4

1/(cosy - siny)

x^2-4xsiny-2=0------------(1)
x^2-4xcosy+2=0----------(2)
siny-cosy=-1
1/(cosy-siny)=1