高一数学题
函数y=sinx[1+tanx·tan(x/2)]的最小正周期是______ 请写下步骤,谢谢!
(以tan(x/2)=t) sinx=2t/(1+t^2), tanx=2t/(1-t^2) y=sinx[1+tanx·tan(x/2)] =2t/(1 + t^2) · [1 + 2t^2/(1-t^2)] =2t/(1 + t^2) · [(1-t^2)/(1-t^2) + 2t^2/(1-t^2)] =2t/(1 + t^2) · (1 + t^2)/(1-t^2) =2t/(1-t^2) =tanx 所以最小正周期为T=派
是pai tanx*tan(x/2)+1=[2tan(x/2)/1-tan(x/2)^2]*tan(x/2)+1 =[2tan(x/2)^2/1-tan(x/2)^2]+1 =[1+tan(x/2)^2]/[1-tan(x/2)^2] =1/cosx 所以所求为sinx/cosx=tanx 最小正周期为pai
问:高一数学题函数f(x)在(-∞,0)和(0,+∞)上单调减,且f(-2)=f(2)=0,则f(x-1) 〉0的解集( ) 请教解题步骤,谢谢!
答:f(-2)=0,(-∞,0)上单调减,故f(x)在(-∞,-2)上大于0 f(2)=0,(0,+∞)上单调减,故f(x)在(0,2)上大于0 f(x)>0的范围...详情>>
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