已知A,B,C是三角形ABC的三个内角,且y=cot A+(2sin A)/【cos A+cos(B-C)】,求证交换任
1个回答
因为
y=cot A+(2sin A)/【cos A+cos(B-C)】
=cot A+(2sin A)/【cos (B-C)-cos(B+C)】
=cot A+(2sin A)/(2sinBsinC)
=cosA/sinA+sinA/(sinBsinC)
=[cosAsinBsinC+sinA*sinA]/(sinAsinBsinC)
=[cosAsinBsinC+1-cosA*cosA]/(sinAsinBsinC)
=[1+cosA(sinBsinC+cosA)]/(sinAsinBsinC)
={1+cosA[sinBsinC-cos(B+C)]}/(sinAsinBsinC)
=(1+cosAcosBcosC)/(sinAsinBsinC)
所以
由ABC的对称性可知,求证交换任意两角,y不变.
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