已知f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)(n属于Z)化简f(x)

问题描述:

已知f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)(n属于Z)化简f(x)

当n为偶数时f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)=-[sinxcosx/cos(π-x)]tanx=
sinxtanx
当n为奇数时,f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)=-[sinxcosx/cosx]tanx
=-sinxtanx