三角函数的题目
已知siny=msin(2x+y),(m≠1),求证:tan(x+y)=tanx(1+m)/(1-m)
已知siny=msin(2x+y),(m≠1),求证:tan(x+y)=tanx(1+m)/(1-m) siny=msin(2x+y)--->m=siny/sin(2x+y) --->(1+m)/(1-m) =[sin(2x+y)+siny]/[sin(2x+y)-siny] =[2sin(x+y)cosy]/[2cos(x+y)siny] =tan(x+y)/tany --->tan(x+y)=tanx(1+m)/(1-m)
答:sin(x+y)=1 所以,x+y=k*pi+pi/2,x=k*pi+pi/2-y tan(2x+y)+tany =tan(2k*pi+pi-2y+y)+tan...详情>>
答:你可以自考啊.既学知识又拿文凭还省钱.详情>>