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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)
#ed_op#DIV#ed_cl#1. cos間多打一個加號#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2.&nbsp; 過同樣兩點#ed_op#/DIV#ed_cl#

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)
#ed_op#DIV#ed_cl##ed_op#FONT size=2#ed_cl#1. siny-sinx/cosy-cosx =-cos(x+y/2)/sin(x+y/2)#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT size=2#ed_cl#cosx+cos(x+y/2)+sinxsin(x+y/2)=cos(x-y/2)#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT size=2#ed_cl#2. 因為兩個代表同一個直線#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl#
kai 寫到:

(cosx,sinx)&nbsp; (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)

#ed_op#DIV#ed_cl#直線ax+by=c#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(cosx,sinx)&nbsp; (cosy,siny)在上面#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故又可寫成y-sinx/x-cosx=siny-sinx/cosy-cosx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#即cos(x+y/2)x+sin(x+y/2)y=cos(x-y/2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#得證#ed_op#/DIV#ed_cl#

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]
#ed_op#DIV#ed_cl#是三個相等耶#ed_op#/DIV#ed_cl#
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]

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]

### [數學]代數題2

#ed_op#DIV#ed_cl#a*cosx+b*sinx=c#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a*cosy+b*siny=c&nbsp; (abc≠0&nbsp; x-y/2≠kπ)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#prove that#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a/cos(x+y/2)=b/sin(x+y/2)=c/cos(x-y/2)#ed_op#/DIV#ed_cl#