yinyinx
幼苗
共回答了27个问题采纳率:81.5% 举报
sin(π/6-α)=2cos(α-4π)
sin(π/6-α)=2cosα
sinπ/6cosα-cosπ/6sinα-=2cosα
1/2cosα-√3/2sinα-=2cosα
3/2cosα=-√3/2sinα
3cosα=-√3sinα
sinα=-√3cosα
[2cos(π/2+α)-sin(-α)]/[cos(π/2-α)+5sin(π/2+α)]
=(-2sinα+sinα)/(sinα+5cosα)
=(-sinα)/(sinα+5cosα)
=(-√3cosα)/(-√3cosα+5cosα)
=-√3/(-√3+5)
=-√3/(5-√3)
=-√3(5+√3)/(5-√3)(5+√3)
=-(5√3+3)/22
1年前
追问
6
嘿嘿哦
举报
我前面那块打错了,应该是(3Π+α)=2cos(α-4Π)应该怎么做呀?
举报
yinyinx
sin(3π+α)=2cos(α-4π) sin(π+α)=2cosα -sinα=2cosα sinα=-2cosα [2cos(π/2+α)-sin(-α)]/[cos(π/2-α)+5sin(π/2+α)] =(-2sinα+sinα)/(sinα+5cosα) =(-sinα)/(sinα+5cosα) =[-(-2cosα)]/(-2cosα+5cosα) =2cosα/(3cosα) =-2/3