tan（13π/3）+sin(-5π/2)+cos(-23π/6)
问题描述：
tan（13π/3）+sin(-5π/2)+cos(-23π/6)
答
tan（13π/3）+sin(-5π/2)+cos(-23π/6)
=tan（π/3）+sin(-π/2)+cos(π/6)
=√3-1+√3/2
=3√3/2-1
答
原式=tan(13π/3) - sin(5π/2) + cos(23π/6)
=tan(4π + π/3) - sin(2π + π/2) + cos(4π - π/6)
=tan(π/3) - sin(π/2) + cos(π/6)
=√3 - 1 + √3/2
=(3√3)/2 - 1