【1】化简:sin(a-5π)/cos(3π-a)×cos(π/2-a)/sin(a-3π)×cos(8π-a)/sin
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/04/28 02:02:01
【1】化简:sin(a-5π)/cos(3π-a)×cos(π/2-a)/sin(a-3π)×cos(8π-a)/sin(-a-4π)
【2】已知f(x)=sin(nπ-x)cos(nx+x)/cos[(n+1)π-x]×tan(x-nπ)cot(nπ/2+x)(n∈Z),求f(7/6π)
【2】已知f(x)=sin(nπ-x)cos(nx+x)/cos[(n+1)π-x]×tan(x-nπ)cot(nπ/2+x)(n∈Z),求f(7/6π)
(1)=sin(a-π)/cos(π-a)*sin(a)/sin(a-π)*cos(a)/sin(-a)
=sin(a)/cos(a)*sin(a)/[-sin(a)]*cos(a)/[-sin(a)]
=1
(2) f(x)=sin(nπ-x)cos(nx+x)/cos[(n+1)π-x]×tan(x-nπ)cot(nπ/2+x)
当n为偶数时:f(x)=sin(-x)cos(x)/cos(π-x)*tan(x)*cot(x)
=sinx
当n为奇数时:f(s)=sin(π-x)cos(π+x)/cos(x)*tan(x)*cot(π/2+x)
=sin(x)[-cos(x)]/cos(x)*tan(x)*[-tan(x)]
=sin(x)[tan(x)]^2
所以f(7/6π)=-1/2(n为偶数)
=(-1/2)*(1/3)
=-1/6
=sin(a)/cos(a)*sin(a)/[-sin(a)]*cos(a)/[-sin(a)]
=1
(2) f(x)=sin(nπ-x)cos(nx+x)/cos[(n+1)π-x]×tan(x-nπ)cot(nπ/2+x)
当n为偶数时:f(x)=sin(-x)cos(x)/cos(π-x)*tan(x)*cot(x)
=sinx
当n为奇数时:f(s)=sin(π-x)cos(π+x)/cos(x)*tan(x)*cot(π/2+x)
=sin(x)[-cos(x)]/cos(x)*tan(x)*[-tan(x)]
=sin(x)[tan(x)]^2
所以f(7/6π)=-1/2(n为偶数)
=(-1/2)*(1/3)
=-1/6
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