已知正弦定理a/sinA = b/sinB = c/sinC = 2R证明 (a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R就是推出它们全部相等.

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/24 00:41:49
已知正弦定理a/sinA = b/sinB = c/sinC = 2R证明 (a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R就是推出它们全部相等.
x){}K]tϲf=Ж_`b8 3ab}} t$j'i'kkkgM[n.LJ$ k Xo]O5?ٽi늗+|mMR>록!N"l;!u n5{:  m@t ,'d,_FAB 0Y'$! aS4f^';V=ٽiҧOx? l{6w]m:@sՁ)N!!|V جg Ov/}>e#4ZtIS 1I/

已知正弦定理a/sinA = b/sinB = c/sinC = 2R证明 (a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R就是推出它们全部相等.
已知正弦定理a/sinA = b/sinB = c/sinC = 2R
证明 (a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R
就是推出它们全部相等.

已知正弦定理a/sinA = b/sinB = c/sinC = 2R证明 (a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R就是推出它们全部相等.
证明,已知
a/sinA = b/sinB = c/sinC = 2R(1)

a=2RsinA, b=2RsinB,c=2RsinC
(a+b+c)/(sinA+sinB+sinC)=2R(sinA+sinB+sinC)/(sinA+sinB+sinC)=2R(2)
(a-b-c)/(sinA-sinB-sinC)=2R(sinA-sinB-sinC)/(sinA-sinB-sinC)=2R(3)
前2个代入后提取2R就出来了,后面3个是正弦定理已知的
所以由(1)(2)(3)得到
(a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R

根据正弦定理a / sinA = b / sin B ,推导出a / sinA = a+b / sin (A+B),根据正弦定理a / sinA = b / sin B 推导出a / sinA = a+b / sin A +sin 正弦定理不是a/sin=b/sin=c/sin吗 为什么a/sinA=2R?求 已知a/sinA=b/sinB(正弦定理)则a=?b=?sinA=?sinB=?RT 正弦定理中,sinA:sinB:sinC=a:b: 用正弦定理证明a+b/c=sin A+sin B/sin c 用正弦定理判断三角形形状b+a/a=sinB/sinB-sinA且2sinAsinB=2sin^2c请判断三角形形状 关于三角函数正弦定理问题正弦定理sinA/a=sinB/b=sinC/c=2R 怎样算出a=sinA ,b=sinB 请问由正弦定理:a/sinA=b/sinA,得:sinB=(b/a)sinA,可是a/sinA不是等于b/sinB吗 用正弦定理判断三角形形状sinA=2sinBcosc 且sin^2A=sin^2B+sin^C请判断三角形形状应是sin^2C 正弦定理既然有a/sinA=b/sinB,那么余弦定理为什么不是b/cosA=a/cosB? 正弦定理中sinA=a,sinB=b,sinC=c怎么得的啊 在三角形ABC中,设a+c=2b,A-C=60度,求sinB的值根据正弦定理 a/sinA=b/sinB=c/sinC得:a=(sinA/sinB)*b c=(sinC/sinB)*b将其带入已知条件 a+c=2b中可得sinA+sinC=2sinB根据三角函数和公式sinA+sinC=2sin[(A+C 正弦定理中 sinA+sinB=2sinC 可不可以等于是 a+b=2c? 正弦定理中的a:b:c=sinA:sinB:sinC是怎样证明的? 为什么正弦定理中a:b:c=sinA:sinB:sinC? 求解正弦定理中为什么a/cosA=a/sinA 正弦定理计算已知三角的边分别是abc,所对角为ABC,a=1b=根号3,A+B=2C,求sinA. 正弦定理 在三角形ABC中,已知A=60,a=3,则(a+b+c)/(sinA+sinB+sinC)=?怎么化解而来