设三角形ABC的内角A、B、C的对边长分别为abc,且3b^2+3c^2-3a^2=4(根号下2)bc

题目

设三角形ABC的内角A、B、C的对边长分别为abc,且3b^2+3c^2-3a^2=4(根号下2)bc
a)求SinA的值
b)求[2sin(A+π/4)sin(B+C+π/4)]/1-cos2A的值.

答案

余弦定理CosA = (b^2+c^2-a^2)/(2bc) = 2/3 * 根号(2)所以(SinA)^2 = 1-(CosA)^2 = 1 - 8/9 = 1/9所以SinA = 1/3 因为A是 π的角,SinA>0 2B+C = π - A所以分子变为2Sin(A + π/4)Sin(π - A + π/4) = 2Sin(A + π/...