举报
zbsaza
1,3b²+3c²-3a²=4√2bc
a²=b²+c²-2*(2/3√2)bc
cosA=2/3√2
因为A是三角形ABC的内角,sinA只取正
sinA=√(1-cos²A) =(√3)/3
2,网上找的
已知在△ABC中,cosA=2√2/3,
所以1-cos2A=2-2cos²A=2/9
因为2sin(A+π/4)sin(B+C+π/4)
=cos(B+C-A)-cos(A+B+C+π/2)
=cos(π-A-A)-cos(π+π/2)
=cos(π-2A)-cos(3π/2)
=-cos(2A)-0
=-cos2A
所以[2sin(A+π/4)sin(B+C+π/4)]/(1-cos2A)
=-cos2A/(1-cos2A)
=1-1/(1-cos2A)
=1-1/(2/9)
=-7/2