计算2cosπ/2-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6+sin3π/2
问题描述：
计算2cosπ/2-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6+sin3π/2
同上 写出过程
答
cosπ/2=0
tanπ/4=1
(tanπ/6)^2=(√3/3)^2=1/3
sinπ/6=1/2
(cosπ/6)^2=(√3/2)^2=3/4
sin3π/2=-1
所以原式
=2*0-1+(3/4)*(1/3)-1/2+3/4-1
=-3/2