Question

已知cos(15+α)=3/5,α为锐角

已知：cos（15°+α）=3/5 , α为锐角
求【tan(435°-α)+sin(α-165°)】/【cos(195°+α)×sin(105°+α)】

Answer 1

因为tana的周期是180，cosa . sina的是360
故原式=[tan(90-(15+a))+sin((15+a)-180)]/.[cos(180+15+a)><sin(90+15+a)]
=[cos(15+a)/sin(15+a)-sin(15+a)]/[-cos(15+a)><cos(15+a)]
因为a是锐角，故sin(15+a)=4/5 与 cos(15+a)=3/5代回原式 得5/36

Answer 2

α为锐角，cos（15°+α）=3/5>0，
则15°+α是锐角，
sin(15°+α)=4/5，
tan(15°+α)=4/3
[tan(435°-α)+sin(α-165°)]/[cos(195°+α)×sin(105°+α)]
={tan[360°-(15°+α)]+sin[-180°+(15°+α)]}/{cos[180°+(15°+α)]×sin[90°+(15°+α)]}
=[-tan(15°+α)-sin(15°+α)]/[-cos(15°+α)×cos(15°+α)]
=(-4/3-4/5)/[(-3/5)*(3/5)]
=-(32/15)/(-9/25)
=160/27

追问

写错了，最后答案是5/36