已知α为第三象限角,且f(α)=sin(π-α)cos(2π-α)tan(-α+π)/sin(π+α)tan(2π-α)1）化简f_数学解答

已知α为第三象限角,且f(α)=sin(π-α)cos(2π-α)tan(-α+π)/sin(π+α)tan(2π-α)
1）化简f(α) （2）cos（α-3π/2）=1/5,求f(α)的值（3）若α=1920°,求f(α)的值

人气：433 ℃　时间：2020-04-03 18:44:10

解答

f(α)=sin(π-α)cos(2π-α)tan(-α+π)/sin(π+α)tan(2π-α)
=sinacosa(-tana)/(-sina)(-tana)
=-(sinacosatana/sinatana)
=cosa
cos（α-3π/2）=1/5
=cosacos3π/2+sinasin3π/2
=-sina=1/5
sina=-1/5
a在第三象限
cosa=负的五分之二倍根六
a=1920°
cos1920°=cos(11π-60°)=cos(π-60°)=-cos60°=-1/2