∵sinC=2sin0.5C ×cos0.5C,cosC=cos0.5C×cos0.5C-sin0.5C×sin0.5C
∴2sin0.5C ×cos0.5C+cos0.5C×cos0.5C-sin0.5C×sin0.5C+sin0.5C=1
∵∠C0,设sin0.5C=x
则有:2x×(1-x^2)^(1/2)+1-2×x^2+x=1
解得:x=[7^(1/2)-1]/4
∴cosC=cos0.5C×cos0.5C-sin0.5C×sin0.5C=1-2sin0.5C×sin0.5C
=1-2×x^2=7^(1/2)/4
又∵a^2+b^2=4(a+b)-8 化简得到:(a-2)^2+(b-2)^2=0
∴a=b=2
根据余弦定理:c^2 = a^2 + b^2 - 2·a·b·cosC
=4+4-8×7^(1/2)/4
∴c=7^(1/2)-1好快。。哈哈,你看看对不对,早上刚做过
在△abc中,角a,b,c的对边分别为a,b,c,已知sinc +cosc = 1 -sin(c/2) (1)求sinc的值 (2)a^2 +b^2 =4(a+b)-8
在△abc中,角a,b,c的对边分别为a,b,c,已知sinc +cosc = 1 -sin(c/2) (1)求sinc的值 (2)a^2 +b^2 =4(a+b)-8
若 a^2 +b^2 =4(a+b)-8 求边c的值
若 a^2 +b^2 =4(a+b)-8 求边c的值
数学人气:871 ℃时间:2019-10-19 21:20:13
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