若sinb=2sin(2a-b),(a≠丌/2+k丌,a-b≠丌/2+k丌,k属于Z),则tan(a-b)/tan a的值、 急-

求证：sin(a+b)cosa-(1/2)[sin(2a+b)-sinb]=sinb 急 sin^A=sin^B=sinC(sinB+sinC)三角形ABC中,若sin^A=sin^B=sinC(sinB+sinC),则角A=呃……是sin^2A=sin^2B+sinC(sinB+sinC) 证明sina sin(a+2b) - sinb sin(b+2a) =sin(a+b)sin(a-b) 证明:sin(2a+b)/sinb-2cos(a+b)=sinb/sina急 sinb/sina=cos(a+b),证明3sinb=sin(2a+b) 若tan(A+B)=2tanA,求证3sinB=sin(2A+B) 若tan(A+B)=2tanA.求证3sinB=sin(2A+B), 若tan(a+b)=2tana,求证：3sinb=sin(2a+b) 若cosA-2sin(A-B)=0,求证:tan(A-B)=cosB/(sinB+2) 若arcsin(sinA + sin B) + arcsin(sinA - sinB) = π /2,求 (sin²A + sin²B) 的值. sinA=1/2,sinB=1/3,sin(A+B)sin(A-B)=? 求证cosa*sinb=1/2[sin(a+b)-sin(a-b)] 已知A、B为锐角,且sinA*sinA+sinB*sinB=sin(A+B),求证A+B=∏/2 sin(2a-b)-2cos(a+b)sina/sinb (sinA+sinB)(sinA-sinB) = 1/2 sinC^2是怎么化简成为sin(A+B)sin(A-B) = 1/2sin(A+B)^2 在三角形ABC中,若(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),则三角形是?a/sinA=b/sinB=k则a=ksinA,b=ksinB代入(a^2+b^2)sin(A-B)=(a^2-b^2)sinC并把k约分(sin²A+sin²B)sin(A-B)=(sin²A-sin²B)sin(A+B) sin²A*[sin(A+B)-sin(A-B) 为什么sinA-sinB/sinA+sinB=cos[(A+B)/2]sin[(A-B)/2]/{sin[(A+B)/2]cos[(A-B)/2]? 已知tan(a+b)=2tan a 证明 3sinb=sin(2a+b)