已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/24 07:22:31
已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ)(n∈Z),求f(7π/6)已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos

已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)
已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)

已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)
①当n=2k时,
f(x) =-sinxcosx/(-cosx)·(1/tanx)
=-cosx
f(7π/6)=√3/2
②当n=2k+1时,
f(x)=sinx·(-cosx)/cosx·(1/tanx)
=-cosx
f(7π/6)=√3/2
∴当n∈Z时,f(7π/6)=√3/2