求证,cos(A/2)+cos(B/2)+cos(C/2)=4cos((π-A)/4)*cos((π-B)/4)*cos
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求证,cos(A/2)+cos(B/2)+cos(C/2)=4cos((π-A)/4)*cos((π-B)/4)*cos((π-C)/4)
A+B+C=180度.谢谢高一的
A+B+C=180度.谢谢高一的
4cos((π-A)/4)*cos((π-B)/4)*cos((π-C)/4)
=4cos((B+C)/4)*cos((A+B)/4)*cos((A+C)/4)
=2[cos((A+C+2B)/4)+cos((A-C)/4)]*cos((A+C)/4)
=2cos((A+C+2B)/4)*cos((A+C)/4)+2cos((A-C)/4)*cos((A+C)/4)
=[cos((A+B+C)/2)+cos(B/2)]+cos(A/2)+cos(C/2)
=cos(A/2)+cos(B/2)+cos(C/2)
得证
=4cos((B+C)/4)*cos((A+B)/4)*cos((A+C)/4)
=2[cos((A+C+2B)/4)+cos((A-C)/4)]*cos((A+C)/4)
=2cos((A+C+2B)/4)*cos((A+C)/4)+2cos((A-C)/4)*cos((A+C)/4)
=[cos((A+B+C)/2)+cos(B/2)]+cos(A/2)+cos(C/2)
=cos(A/2)+cos(B/2)+cos(C/2)
得证
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