作业帮f(x)=sin(nπ-x)cos(nπ+x)/cos((n+1)π-x)*tan(x-nπ)*cot(nπ/2+x),求f(π/6)的值
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f(x)=sin(nπ-x)cos(nπ+x)/cos((n+1)π-x)*tan(x-nπ)*cot(nπ/2+x),求f(π/6)的值
f(x)=sin(nπ-x)cos(nπ+x)/cos((n+1)π-x)*tan(x-nπ)*cot(nπ/2+x),求f(π/6)的值
f(x)=sin(nπ-x)cos(nπ+x)/cos((n+1)π-x)*tan(x-nπ)*cot(nπ/2+x),求f(π/6)的值
当n为偶数
f(x)
=sin(nπ-x)cos(nπ+x)/cos((n+1)π-x)*tan(x-nπ)*cot(nπ/2+x)
=sin(-x)cosx/[-cos(-x)]tanxcotx
=sinx
=sinπ/6
=1/2
当n为奇数
f(x)
=sin(nπ-x)cos(nπ+x)/cos((n+1)π-x)*tan(x-nπ)*cot(nπ/2+x)
=[-sin(-x)](-cosx)/cos(-x)tanx(-tanx)
=sinxtanx^2
=sin(π/6)tan(π/6)^2
=1/6