[數學]代數題2

[數學]代數題2

a*cosx+b*sinx=c
a*cosy+b*siny=c  (abc≠0  x-y/2≠kπ)
prove that
a/cos(x+y/2)=b/sin(x+y/2)=c/cos(x-y/2)

kai

a*cosx+b*sinx=c
a*cosy+b*siny=c

a*(cosx+cosy)+b*(sinx+siny)=2c

a*2*cos[(x+y)/2]*cos[(x-y)/2]+b*2*sin[(x+y)/2]*cos[(x-y)/2]=2c

a*cos[(x+y)/2] +b*sin[(x+y)/2]=c/cos[(x-y)/2]

p

a*cos[(x+y)/2] +b*sin[(x+y)/2]=c/cos[(x-y)/2]---------(1)

a*cosx+b*sinx=c
a*cosy+b*siny=c

a*(cosx-cosy)+b*(sinx-siny)=0

-a*2*sin[(x+y)/2]*sin[(x-y)/2]+b*2*cos[(x+y)/2]*sin[(x-y)/2]=0
a*sin[(x+y)/2] =b*cos[(x+y)/2]
a =b*cos[(x+y)/2]/ sin[(x+y)/2]  b=a*sin[(x+y)/2]/cos[(x+y)/2] 代入(1)

b*cos^2 [(x+y)/2]/sin[(x+y)/2] +a*sin^2 [(x+y)/2]/cos[(x+y)/2]=c/cos[(x-y)/2]-----(2)

(1)+(2)
a*sin^2 [(x+y)/2]/ cos[(x+y)/2]+ a*cos[(x+y)/2]+
b*cos^2 [(x+y)/2]/ sin[(x+y)/2]+ b*sin[(x+y)/2]=
2c/cos[(x-y)/2]

a*sin^2 [(x+y)/2]/ cos[(x+y)/2]+ a*cos^2 [(x+y)/2] / cos[(x+y)/2]+
b*cos^2 [(x+y)/2]/ sin[(x+y)/2]+ b*sin^2 [(x+y)/2] / sin[(x+y)/2]=
2c/cos[(x-y)/2]

a/cos[(x+y)/2]+ b/sin[(x+y)/2]=2c/cos[(x-y)/2]

p

kai

a*sin[(x+y)/2] =b*cos[(x+y)/2]
a/cos[(x+y)/2] =b/sin[(x+y)/2]
a/cos[(x+y)/2]+ b/sin[(x+y)/2]=2c/cos[(x-y)/2]
2a/cos[(x+y)/2] =2c/cos[(x-y)/2]

a/cos[(x+y)/2] =b/sin[(x+y)/2]=c/cos[(x-y)/2]

p

(cosx,sinx)  (cosy,siny)在上面

kai

p

kai 寫到:

(cosx,sinx)  (cosy,siny)在上面

y-sinx/x-cosx=siny-sinx/cosy-cosx ------(1)
cos(x+y/2)x+sin(x+y/2)y=cos(x-y/2) ---(2)

p

1. siny-sinx/cosy-cosx =-cos(x+y/2)/sin(x+y/2)
cosx+cos(x+y/2)+sinxsin(x+y/2)=cos(x-y/2)
2. 因為兩個代表同一個直線

kai

siny-sinx/cosy-cosx =-cos(x+y/2)/sin(x+y/2)

y-sinx/x-cosx==-cos(x+y/2)/sin(x+y/2)

cosx+cos(x+y/2)+sinxsin(x+y/2)=cos(x-y/2)

p

p

1. cos間多打一個加號
2.  過同樣兩點

kai

y-sinx/x-cosx==-cos(x+y/2)/sin(x+y/2)

cosxcos(x+y/2)+sinxsin(x+y/2)=cos(x-y/2)

cos(x+y/2)x+sin(x+y/2)y=cos(x-y/2)

p