在三角形ABC中,8sin^2[(B+C)/2]-2cos2A=7 (1)求A大小(2)若a=√3,b+c=3,求b,c

A+B+C=180 2A=360-(2B+2C) cos2A=cos(2B+2C)=2[cos(B+C)]^2-1 因为cos2X=1-2(sinX)^2 所以(sinX)^2=(1-cos2X)/2 所以sin^2[(B+C)/2]=[1-cos(B+C)]/2 所以8sin^2[(B+C)/2]-2cos2A=7 4-4cos(B+C)-4[cos(B+C)]^2+2=7 4[...