Вычислите нужна помощь ctg 15° sin 7пи/12

[latex]ctgx= \frac{cosx}{sinx}; cos(x-y)=cosxcosy+sinxsiny; [/latex] [latex]sin(x-y)=sinxcosy-cosxsiny[/latex]; вычисляем [latex]ctg15= \frac{cos(45-30)}{sin(45-30)}; \frac{cos45cos30+sin45sin30}{sin45cos30-cos45sin30}= \frac{ \frac{ \sqrt{2} }{2}( \frac{ \sqrt{3} }{2}+ \frac{1}{2}) }{ \frac{ \sqrt{2} }{2}( \frac{ \sqrt{3} }{2}- \frac{1}{2}) }= \frac{ \frac{ \sqrt{6} + \sqrt{2} }{4} }{ \frac{ \sqrt{6} - \sqrt{2} }{4}}= \frac{ \sqrt{6}+ \sqrt{2} }{ \sqrt{6}- \sqrt{2} } [/latex]=[latex] \frac{( \sqrt{6}+ \sqrt{2})^2 }{ (\sqrt{6}+ \sqrt{2})(\sqrt{6}- \sqrt{2}) }= \frac{6+2 \sqrt{12}+2 }{6-2}= \frac{8+2 \sqrt{12} }{4}=2+ \frac{2*2 \sqrt{3} }{4}=2+ \sqrt{3} [/latex] [latex]sin( \frac{7 \pi }{12})=sin( \frac{ \pi }{4}+ \frac{ \pi }{3})=sin \frac{ \pi }{4}cos \frac{ \pi }{3}+cos \frac{ \pi }{4}sin \frac{ \pi }{3}= \frac{ \sqrt{2} }{2}( \frac{1}{2}+ \frac{ \sqrt{3} }{2})= [/latex][latex] \frac{ \sqrt{6}+ \sqrt{2} }{4} [/latex]