(Sin (п-x)+cos (0.5п+x)cosx)/sin^2 0.5п при x=п/6

Question

(Sin (п-x)+cos (0.5п+x)cosx)/sin^2 0.5п при x=п/6

Answer 1

(sin(π-x)+cos(π/2+x)cosx)/sin²π/2=(sinx-sinx*cosx)/1=sinx(1-cosx)=sinπ/6(1-cosπ/6)=1/2*(1-√3/2)=1/2-√3/4.