Question

三角形ABC中,若C=90度 则三边(a+b)/c=2^(1/2) cos((A-B)/2)

Answer 1

由二倍角公式可知,cos(A-B)=[cos((A-B)/2)]^2-[sin((A-B)/2]^2=2[cos((A-B)/2)]^2-1,所以2[cos((A-B)/2)]^2=cos(A-B)+1=cosAcosB+sinAsinB+1.

三角形ABC中C=90度,则三角形为直角三角形.cosA=b/c,cosB=a/c,sinA=a/c,sinB=b/c,代入上式得2[cos((A-B)/2)]^2=ab/c^2+ab/c^2+1=2ab/c^2+1=(2ab+c^2)/c^2=(2ab+a^2+b^2)/c^2=(a+b)^2/c^2,两边同时开方得2^(1/2) cos((A-B)/2)=(a+b)/c,证毕.