在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1

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在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1

在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1
在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1

在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1
由题意:1-sin^2A=cos^2A
sin^2B+cos^2C+2sinAsinBcos(A+B)=
=sin^2B+cos^2C-2sinAsinBcosC
=sin^2B +cosC(cosC-2sinAsinB)
=sin^2B -cosC[cos(A+B)+2sinAsinB]
=sin^2B-cosC[cosAcosB-sinAsinB+2sinAsinB]
=sin^2B-cosC[cosAcosB+sinAsinB]
=sin^2B +cos(A+B)(cosAcosB+sinAsinB)
=sin^2B+(cosAcosB-sinAsinB)(cosAcosB+sinAsinB)
=sin^2B +cos^2Acos^2B -sin^2Asin^2B
=sin^2B(1-sin^2A)
=sin^2Bcos^2A +cos^2Bcos^2A
=cos^2A
所以sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1

原式
=(1-cos2A+1-cos2B)/2+cos^2C-2sinAsinBcosC
=1-cos(A+B)cos(A-B)+cos^2C-2sinAsinBcosC (和差化积)
=1+cosC[cos(A-B)+cosC]-2sinAsinBcosC
=1+cosC[cos(A-B)-cos(A+B)]-2sinAsinBcosC
=1+cosC*2sinAsinB-2sinAsinBcos
=1
这类题目主要就是降幂和差化积的应用,多做题目熟练就会了

就是不断地和差化积:sin^2A+sin^2B+cos^2C=1-1/2(cos2A+cos2B)+cos^2C=1-cos(A+B)cos(A-B)-cosCcos(A+B)=1+cosCsinAsinB.证毕。