已知：acosx+bsinx=c,acosy+bsiny=c(abc不等于零，(y-x)/2不等于k派（就是３．１４１５９．．．），k为整数）证：a/cos((x+y)/2)=b/sin((x+y)/2)=c/cos((x-y)/2)
热心网友
解: acosx+bsinx=c,。。。。(1) acosy+bsiny=c。。。。。(2)(1)-(2)得:a×(cosx-cosy)+b×(sinx-siny)=0和差化积 -2a ×sin［(x+y/)2］×sin［(x-y/)2］+2b×cos［(x+y/)2］× sin ［(x-y/)2］=0sin［(x-y/)2］×｛2b×cos［(x+y/)2］-2a ×sin［(x+y/)2｝=0∵(y-x)/2不等于π ∴sin［(x-y/)2］≠0∴2b×cos［(x+y/)2］-2a ×sin［(x+y/)2=0 ∴ a/cos［(x+y)/2］=b/sin［(x+y)/2］(1)-(2)得:a×(cosx+cosy)+b×(sinx+siny)=2c2a×cos［(x+y)/2］×cos［(x-y)/2］+2b×sin［(x+y)/2］×cos［(x-y)/2］=2c 即a×cos［(x+y)/2］=b×sin［(x+y)/2］=c/cos［(x+y)/2］∵a/cos［(x+y)/2］=b/sin［(x+y)/2］ ∴a=b×cos［(x+y)/2］/sin［(x+y)/2］式左=b×｛cos［(x+y)/2］｝＾2/sin［(x+y)/2］+b×｛sin［(x+y)/2］｝＾2/sin［(x+y)/2］=b×1/sin［(x+y)/2］=式右=c/cos［(x+y)/2］∴a/cos((x+y)/2)=b/sin((x+y)/2)=c/cos((x-y)/2) 。