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三角函数万能公式证明 由余弦定理：a^2+b^2-c^2-2abcosC=0 正弦定理：a/sinA=b/sinB=c/sinC=2R 得 (sinA)^2+(sinB)^2-(sinC)^2-2sinAsinBcosC=0 转化 1-(cosA)^2+1-(cosB)^2-[1-(cosC)^2]-2sinAsinBcosC=0 即 (cosA)^2+(cosB)^2-(cosC)^2+2sinAsinBcosC-1=0 又 cos(C)=-cos(A+B)=sinAsinB-cosAcosB 得 (cosA)^2+(cosB)^2-(cosC)^2+2cosC[cos(C)+cosAcosB]-1=0 (cosA)^2+(cosB)^2+(cosC)^2=1-2cosAcosBcosC 得证 (8)(sinA)^2+(sinB)^2+(sinC)^2=2+2cosAcosBcosC (责任编辑：admin) |