Question

在三角形ABC中,内角A.B.C的对边分别为a,b,c,已知B=C,2b=(根号3)a. ----求cos(2A+45

在三角形ABC中,内角A.B.C的对边分别为a,b,c,已知B=C,2b=(根号3)a.
----求cos(2A+45度)

Answer 1

解：B=C ；2b=√3a；∴ b = √3/2 a
sin(A/2) = (a/2) / b = √3/3
cos(A/2)= √（1-1/3）= √橡核6/3
sinA = 2sin(A/2)cos(A/2) = 2√2/3
cosA = √（则如备1-8/9）=1/3
sin2A = 2sinAcosA = 4√2/9
cos2A = -√（1-32/81）孙毁= - 7/9
cos（2A+45°）= cos2Acos45°-sin2Asin45° = √2/2（-7/9-4√2/9）= -(8-7√2)/18

Answer 2

B=C 说明颂如是等腰三角形 b=c

2b=(根号3)a 得出野并启 b*b=3/4a*a
由a*a=*b*b+c*c-2*b*c*cosA 得出
cosA=1/3
cos（2A+45°）= cos2Acos45°-sin2Asin45° = √2/2（-7/9-4√蔽旦2/9）= -(8-7√2)/18

Answer 3

解：B=C ；咐雀2b=√3a；∴ b = √3/2 a
sin(A/2) = (a/2) / b = √3/3
cos(A/2)= √（1-1/3）= √6/3
sinA = 2sin(A/2)cos(A/2) = 2√2/3
cosA = √（1-8/9）=1/3
sin2A = 2sinAcosA = 4√2/9
cos2A = -√（1-32/银启81）= - 7/9
cos（2A+45°）= cos2Acos45°-sin2Asin45° = √2/2（-7/衡搏早9-4√2/9）= -(8-7√2)/18
是正解

Answer 4

太简单
解B=C 说明是镇基等腰三御兆谨角形 b=c
2b=(根号3)a 得出 b*b=3/4a*a
由a*a=*b*b+c*c-2*b*c*cosA 得出
cosA=1/猜雹3
cos（2A+45°）= cos2Acos45°-sin2Asin45° = √2/2（-7/9-4√2/9）= -(8-7√2)/18
就这点