题目

tan2a=-2根号2,2a属于(90`,180`),求 [2cos^2(a/2)-sina-1]/根号2sin(a+45`)
急用啊

提问时间：2020-09-07

答案

tan2a=-2√2 (cos2a) ^2=1/(1+(tan2a)^2)=1/9
cos2a=-1/3
(sina)^2=(1-cosa)/2=2/3 sina=√2/√3=√6/3
(cosa)^2=(1+cosa)/2=1/3 cosa=1/√3=√3/3
2(cos(a/2))^2-1=cosa
sin(a+45)=(√2/2)(√6+√3)/3
[2(cos(a/2))^2-sina-1]/[√2sin(a+45)]= [(√3-√6)/3]/(√6+√3)/3=-(√6-√3)^2/3=-3+2√2