【1】化简:sin(a-5π)/cos(3π-a)×cos(π/2-a)/sin(a-3π)×cos(8π-a)/sin(-a-4π)
问题描述:
【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)
答
(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...