已知cos(75+α)=1/3,若a是第三象限角求cos(105-a)+sin(255+a)-tan(a-105)

题目

已知cos(75+α)=1/3,若a是第三象限角求cos(105-a)+sin(255+a)-tan(a-105)

答案

若α是第三象限角,有：
180°+k*360°<α<270°+k*360°
则255°+k*360°<75°+α<345°+k*360°
又cos(75°+α)=1/3>0
可知75°+α是第四象限角
则sin(75°+α)<0,tan(75°+α)<0
所以由sin²(75°+α)+cos²(75°+α)=1易得：
sin(75°+α)=-2√2/3,tan(75°+α)=sin(75°+α)/cos(75°+α)=-2√2
所以：cos(105°-α)+sin(255°+α)-tan(α-105°)
＝cos[180°-(75°+α)]+sin[180°+(75°+α)]-tan[(α+75°)-180°]
＝-cos(75°+α)-sin(75°+α)-tan(α+75°)
＝－(1/3 -2√2/3 -2√2)
＝(8√2 －1)/3