在△ABC中,求证;sin^(A/2)+sin^(B/2)+sin^(C/2)=1-2sin(A/2)sin(B/2)sin(C/2)

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在△ABC中,求证;sin^(A/2)+sin^(B/2)+sin^(C/2)=1-2sin(A/2)sin(B/2)sin(C/2)

在△ABC中,求证;sin^(A/2)+sin^(B/2)+sin^(C/2)=1-2sin(A/2)sin(B/2)sin(C/2)
在△ABC中,求证;sin^(A/2)+sin^(B/2)+sin^(C/2)=1-2sin(A/2)sin(B/2)sin(C/2)

在△ABC中,求证;sin^(A/2)+sin^(B/2)+sin^(C/2)=1-2sin(A/2)sin(B/2)sin(C/2)
由倍角公式:
(sin A/2)^2=(1-cosA)/2,(sin B/2)^2=(1-cosB)/2,(sin C/2)^2=(1-cosC)/2.所以等式左边=3/2-1/2(cosA+cosB+cosC).又因为
1-2sin(A/2)sin(B/2)sin(C/2) (积化和差公式)
=1-[cos (A-B)/2 - cos (A+B)/2]sin(C/2)
=1-[cos (A-B)/2 - sin C/2]sin(C/2)
=1-cos (A-B)/2*sin(C/2)+(sin C/2)^2 (对第二项用积化和差公式,第三项用倍角公式)
=1-1/2[sin (A-B+C)/2-sin (A-B-C)/2]+(1-cosC)/2
=1-1/2[sin (180-2B)/2+sin (180-2A)/2]+(1-cosC)/2
=1-1/2(cosB+cosA)+(1-cosC)/2
=3/2-1/2(cosA+cosB+cosC)
因此等式两边都等于 3/2-1/2(cosA+cosB+cosC),故所证式成立.