sinB+sinC=sinA(cosB+cosC) 2sin[(B+C)/2]coc[(B-C)/2]=4sin(A/2)cos(A/2)cos[(B+C)/2]cos[(B-C)/2] 2sin[(B+C)/2]=4sin(A/2)cos(A/2)cos[(B+C)/2] sin[(B+C)/2]=cos(A/2) sin(A/2)=cos[(B+C)/2] 2sin[(B+C)/2]=4sin(A/2)cos(A/2)cos[(B+C)/2] [sin(A/2)]^2=1/2,A是三角形内角 sin(A/2)=√2/2 A/2=π/4 A=π/2
sinB+sinC=sinA(cosB+cosC)
sinB+sinC=sin(B+c)*(cosB+cosC)
=[sinBcosC+cosBsinC]*(cosB+cosC)
=sinBcosBcosC+siBcosC^2+cosB^2sinC+cosBsinCcosC
sinB(1-cosC^2)+sinC(1-cosB^2)=sinBcosBcosC+cosBsinCcosC
sinb*sinc^2+sinc*sinb^2=sinBcosBcosC+cosBsinCcosC
由正弦定理和余弦定理有： 原式===>b+c=a*[(a^2+c^2-b^2)/(2ac)+(a^2+b^2-c^2)/(2ab)] ===>b+c=(a^2+c^2-b^2)/(2c)+(a^2+b^2-c^2)/(2b) ===>(a^2-c^2-b^2)/(2c)+(a^2-b^2-c^2)/(2b)=0 ===>(a^2-c^2-b^2)*[1/(2c)+1/(2b)]=0 则a^2-c^2-b^2＝0，即a^2＝c^2＋b^2 所以，角A＝90度
答：sina =2*tan(a/2)]/{1 +[tan(a/2)]^2} cosa ={1 -[tan(a/2)]^2}/{1 +[tan(a/2)]^2} 代入...详情>>
