proszę o pomoc - zadanie z matematyki z tematu trygonometrii. do rozwiązania zadanie 8.62 podpunkt b help :)

[latex]cos{frac{pi}{6}}=frac{sqrt{3}}{2}\cos{frac{pi}{6}}=2cos^2{frac{pi}{12}}-1\2cos^2{frac{pi}{12}}-1=frac{sqrt{3}}{2}\2cos^2{frac{pi}{12}}=frac{sqrt{3}}{2}+1=frac{sqrt{3}+2}{2}\cos^2{frac{pi}{12}}=frac{sqrt{3}+2}{4}\cos{frac{pi}{12}}=frac{sqrt{sqrt{3}+2}}{2}[/latex] [latex]cos{frac{11}{12}pi}=cos(pi-frac{pi}{12})=-cos{frac{pi}{12}}=-frac{sqrt{sqrt{3}+2}}{2}[/latex]

cos11/12π=cos(π-π/12)=- cosπ/12= -cos(π/4-π/6)=-(cosπ/4cosπ/6+sinπ/4sinπ/6)=-(√2/2*√3/2+√2/2*1/2)=-(√6/4+√2/4)=-[latex] frac{ sqrt{6}+ sqrt{2} }{4} [/latex]