在三角形ABC中,已知4cos^A/－cos2(B＋C）＝7/2.求角A?注意：那^代表平方

因,在△ABC中,-cos2(B+C)=-[2cos^2(B+C)-1]
cos^2(B+C)=[cos(B+C)]^2=(-cosA)^2=cos^2A.
故-cos2(B+C)=-[2cos^2(B+C)-1]=-[2cos^2A-1]
故,4cos^2A/[-cos2(B+C)=4cos^2A/[-(2cos^2A-1)]=7/2
8cos^2A=7*[-(2cos^2A-1)]
=-14cos^2A+7
22cos^2A=7
cos^2A=7/22
cosA=±(根号154)/22
即,A=±arccos[(根号154)/22]
故,角A=55.66度,或角A=124.34度.
cos2(B＋C）=－cos2(180-A)=－cos(360-2A)=－cos2A
4cos^A/－cos2(B＋C）=(4cos^A-2+2)/－cos2A=2（cos2A+1）/－cos2A=7/2
cos2A=4/11,2cos^A-1=cos2A
A=arccos√330/12

－cos2(B＋C）=－cos2(180-A)=－cos(360-2A)=－cos2A
4cos^A/－cos2(B＋C）=(4cos^A-2+2)/－cos2A=2（cos2A+1）/－cos2A=7/2
cos2A=4/11,2cos^A-1=cos2A
A=arccos√330/12

因，在△ABC中，-cos2(B+C)=-[2cos^2(B+C)-1]
cos^2(B+C)=[cos(B+C)]^2=(-cosA)^2=cos^2A.
故-cos2(B+C)=-[2cos^2(B+C)-1]=-[2cos^2A-1]
故，4cos^2A/[-cos2(B+C)=4cos^2A/[-(2cos^2A-1)]=7/2
8cos^2A=7*[-(2...