已知α是第二象限的角,且cos(α-π/2)=1/5,求(sin(π+α)cos(π-α)tan(-3π/2-α))/(tan(π/2+α)cos(3π/2+α))的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/08/07 11:03:58
已知α是第二象限的角,且cos(α-π/2)=1/5,求(sin(π+α)cos(π-α)tan(-3π/2-α))/(tan(π/2+α)cos(3π/2+α))的值
x){}Km|6c5kyq˙jy|ΓS8Q|=:663stD1HDB 06cmڰ&HPOf D:v8D8a$=eUDd';\ U6V0 ]@= Z$ U0a/.H̳ ZF

已知α是第二象限的角,且cos(α-π/2)=1/5,求(sin(π+α)cos(π-α)tan(-3π/2-α))/(tan(π/2+α)cos(3π/2+α))的值
已知α是第二象限的角,且cos(α-π/2)=1/5,求(sin(π+α)cos(π-α)tan(-3π/2-α))/(tan(π/2+α)cos(3π/2+α))的值

已知α是第二象限的角,且cos(α-π/2)=1/5,求(sin(π+α)cos(π-α)tan(-3π/2-α))/(tan(π/2+α)cos(3π/2+α))的值
cos(a-π/2)=1/5
sina=1/5
α是第二象限的角
cosa=-2√6/5
(sin(π+a)cos(π-a)tan(-3π/2-a))/(tan(π/2+a)cos(3π/2+a))
=-sina(-cosa)tana/((-tana)sina)
=-cosa
=2√6/5