如果b+c=π (派) 证明 2(1-sinb sinc )=cos**2b+cos**2c
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC
在△ABC中,求证:sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
11.在△ABC中,求证sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
不等式题目:在△ABC中请证明sinA+sinB+sinC≤cos(A/2)+cos(B/2)+cos(C/2)
sina+sinb+sinc=0 cosa+cosb+cosc=0求证cos*2a+cos*2b+cos*2c=3|2
在三角形ABC中求证sinA+sinB+sinC=4cos(A/2)COS(B/2)COS(C/2)证到这步然后怎么证:
在三角形ABC中,角A,B,C的对边分别为a,b,c,且满足sinA:sinB:sinC=2:3:4 求cos
已知b,c∈{0,π/2},b=sin(cosb),c=cos(sinc),比较b,c大小
sinA+sinB+sinc=0 cosA+cosB+cosC=0 cos(B-C)
sinA+sinB+sinC=0; cosA+cosB+cosC=0,求cos(B-C)的值?
sina+sinb+sinc=0,cosa+cosb+cosc=0,求cos(B-C)的值?