求证:sin(x-y)/(sinx-siny)=cos[(x-y)/2]/cos[(x+y)/2](cosy-cosx)/(sinx-siny)=tan[(x+y)/2]

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求证:sin(x-y)/(sinx-siny)=cos[(x-y)/2]/cos[(x+y)/2](cosy-cosx)/(sinx-siny)=tan[(x+y)/2]求证:sin(x-y)/(si

求证:sin(x-y)/(sinx-siny)=cos[(x-y)/2]/cos[(x+y)/2](cosy-cosx)/(sinx-siny)=tan[(x+y)/2]
求证:sin(x-y)/(sinx-siny)=cos[(x-y)/2]/cos[(x+y)/2]
(cosy-cosx)/(sinx-siny)=tan[(x+y)/2]

求证:sin(x-y)/(sinx-siny)=cos[(x-y)/2]/cos[(x+y)/2](cosy-cosx)/(sinx-siny)=tan[(x+y)/2]
你可以把分母 sinx - siny 用和差化积 化成 2sin((x-y)/2) cos((x+y)/2)
这样答案就很显然了