题目

已知sina-2cosa=0求sin(π-a)cos(2π-a)sin(-a+2分之3π)除以tan(-a-π)sin(-π-a） 求答案阿

提问时间：2021-01-02

答案

sin(π-a)cos(2π-a)sin(-a+1.5π)/(tan(-a-π)sin(-π-a）)
=sin(a)cos(a)sin(-a-0.5π)/(tan(-a)sin(π-a-2π）)
=(-1)^2*sin(a)cos(a)cos(a)/(tan(a)sin(π-a）)
=sin(a)cos(a)cos(a)/(tan(a)sin(a）)
=cos(a)cos(a)/tan(a)
sina-2cosa=0==>tan(a)=2
sina-2cosa=0==>sin(a)^2=4cos(a)^2==>1=5cos(a)^2==>cos(a)^2=1/5
所以
cos(a)cos(a)/tan(a)=0.1