一.设△ABC的角A.B.C的对边长为a.b.c,且3b^2+3c^2-3a^2=四倍根号二倍的bc.求①sinA的值②

一,①cosA=(b^2+c^2-a^2)/2bc
3b^2+3c^2-3a^2=（4根号2）bc
(b^2+c^2-a^2)/2bc=（2根号2）/3
cosA=（2根号2）/3
sinA=1/3
②2sin(A+π/4)sin(B+C+π/4)/（1-cos2A)
=2sin(A+π/4)sin(π-A+π/4)/（2sin^2A)
=sin(A+π/4)sin(π-A+π/4)/（sin^2A)
=sin(A+π/4)sin(A-π/4)/（sin^2A)
=1/2(cos^2A-sin^2A)/sin^2A
=1/2(8/9-1/9)/1/9
=7/2
二,（1）cos2A=3/5,cos2A=1-2（sinA）^2=3/5,sinA=（根号5）/5,cosA=（2根号5）/5,sinB=（根号10）/10,cosB=（3根号10）/10,sin(A+B)=sinAcosB+cosAsinB=(根号2)/2,因为2A为锐角,所以0

①cosA=(b^2+c^2-a^2)/2bc
3b^2+3c^2-3a^2=四倍根号二倍的bc
(b^2+c^2-a^2)/2bc=2根号2/3
cosA=2根号2/3
sinA=1/3
②2sin(A+π/4)sin(B+C+π/4)除以（1-cos2A)
=2sin(A+π/4)sin(π-A+π/4)除以（2sin^2A)
=sin(...

直接在卷子上面写百度一下 你就知道 绝对满分