Question

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
写过程

Answer 1

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

Answer 2

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
cos(π/2+a-2π）=-sina=1/5
a第三象限，cosa=-2√6/5
1920-5*360=120
f(120)=-cos120=cos60=1/2