Question

在三角形ABC中求证sin^2A=sin^2B+sin^2C-2sinBsinCcosA

Answer 1

令三个角的对边分别为a、b、c，则：
b/a=sinB/sinA,c/a=sinC/sinA
a^2=b^2+c^2-2bccosA
1=(b/a)^2+(c/a)^2-2*b/a*c/a*cosA
1=(sinB/sinA)^2+(sinC/sinA)^2-2sinB/sinA*sinC/sinA*cosA
(sinA)^2=(sinB)^2+(sinC)^2-2sinB*sinC*cosA

Answer 2

这只是余弦定理的变化形式而已：
a^2=b^2+c^2-2bcCosA
根据正弦定理：a/sinA=b/sinB=c/sinC=2R
(2RsinA)^2=(2RsinB)^2+(2RsinC)^2-2(2RsinB)(2RsinC)CosA
公式两边同时消除(2R)^2
sin^2A=sin^2B+sin^2C-2sinBsinCcosA