在三角形ABC中,内角A,B,C的对边分别为a,b,c,已知a=2,c=根号2,cosA=-根号2/4求sinC和b的值,求cos(2A+派/3)的值

Question

在三角形ABC中,内角A,B,C的对边分别为a,b,c,已知a=2,c=根号2,cosA=-根号2/4
求sinC和b的值,求cos(2A+派/3)的值

Answer 1

∵cosA=-√2/4
∴sinA=√14/4
由正弦定理,有
a/sinA=c/sinC
则 sinC=c*sinA/a
=√2×(√14/4)÷2
=√7/4
cosC=3/4
∵sinB=sin[π-(A+C)]=sin(A+C)
∴sinB=sinA*cosC+cosA*sinC
=(√14/4)×(3/4)+(-√2/4)×(√7/4)
=√14/8
故 b=a*sinB/sinA
=2×(√14/8)÷(√14/4)
=1
∵cos(A+π/6)=cosA*cos(π/6)-sinA*sin(π/6)
=(-√2/4)×(√3/2)-(√14/4)×(1/2)
=-(√6+√14)/8
∴cos(2A+π/3)=2cos²(A+π/6)-1
=2×[-(√6+√14)/8]²-1
=(√21-3)/8