sin^2[(B+C)/2]=[1-cos(B+C)]/2

在△ABC中,求证；sin^(A/2)+sin^(B/2)+sin^(C/2)=1-2sin(A/2)sin(B/2)sin(C/2） 设二次函数f(x)=x^2+bx+c(b,c∈r),已知不论α,β为何实数恒有f(sinα)≥0和f(2+co设二次函数f(x)=x^2+bx+c(b,c∈R),已知不论α,β为何实数恒有f(sinα)≥0和f(2+cosβ)≤0.(1)求证:b+c=-1(2)求证：c≥3.(3)若f(sinα)的 证明cosA＋cosB＋cosC=1+4sin(A/2)sin(B/2)sin(C/2)证明：cosA＋cosB＋cosC=1+4sin(A/2)sin(B/2)sin(C/2)尽量详细一点选做cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2) cos(180-B-C)+cosB+cosC=1+2sin(A/2)[2sin(B/2)sin(C/2)] cos(180-B-C)+cosB+cosC 在三角形ABC中sin^A+sin^B=2sin^C,则角C为? 3 在三角形ABC中,已知（a2+b2）sin（A-B）=（a2-b2）sin(A+B) 求证：ABC是等腰或直角三角形(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B) sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B)) sin^A*2c 求证(a^2+b^2-c^2)/(b^2+c^2-a^2)=(sin(A+B)+sin(A-B))/(sin(A+B)-sin(A-B)) 已知(sin A)^2+(sin B)^2+（sin C)^2=1,且A,B,C均为锐角,求cos Acos Bcos C的最大值 三角形ABC中证明 COSA+COSB+COSC=1+4SIN(A/2)*SIN(B/2)*SIN(C/2) 三角形ABC中,已知（sin^2 A-sin^2 B-sin^2 C)/(sinB sinC)=1 求A? 三角形ABC中证明 COSA+COSB+COSC=1+4SIN(A/2)*SIN(B/2)*SIN(C/2) Rt三角形中 角c等于九十度 求证 sin^2A?tanA+sin^2B?tanB=1-2sin^2A?sin^2B/cosA?cosB 在△ABC中,求证sin²A+sin²B+sin²C=2(1+cosAcosBcosC） △ABC中,sin^2A+sin^2C-sin^2B+sinAsinC=0(1)求B（2）求(a+c)/b的取值范围 sin^2A+sin^2B=sin^2C求证△ABC的形状 sin^2 A=sin^2 B+sinBsinC+sin^2 C 求A角 证明 SIN²A+SIN²B-SIN²C=2SINASINBCOSC 非线性方程解析解-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos(b)-y0*sin(a)*co 已知:sin^2A/sin^2B+cos^2Acos^2C=1,求证:tan^2Acot^2B=sin^2C