Question

1.5 с 2 по 5 срочно , не по делу не писать буду жалобы кидать

Answer 1

$2)y=tg(x+\frac{\pi }{6})Sin(\frac{\pi }{3}-x)=tg(\frac{7\pi }{6}+\frac{\pi }{6})Sin(\frac{\pi }{3}-\frac{\pi }{6})=tg\frac{8\pi }{6}Sin\frac{\pi }{6}=\\\\=tg\frac{4\pi }{3}*Sin\frac{\pi }{6}=tg(\pi+\frac{\pi }{3})*\frac{1}{2}=tg\frac{\pi }{3}*\frac{1}{2}=\sqrt{3}*\frac{1}{2} =\boxed{\frac{\sqrt{3} }{2}} \\\\\\3)y=Sin^{2}x+Cos^{2}2x =Sin^{2}(-\frac{\pi }{12})+Cos^{2}(-\frac{\pi }{6})=$

$=\frac{1-Cos\frac{\pi }{6} }{2} +Cos^{2}\frac{\pi }{6}=\frac{1-\frac{\sqrt{3} }{2} }{2}+(\frac{\sqrt{3} }{2})^{2}=\frac{2-\sqrt{3} }{4}+\frac{3}{4} =\boxed{\frac{5-\sqrt{3} }{4}}$

$4)y=Sin(x-\frac{\pi }{4})Cos(\frac{\pi }{4} +x)=Sin(\frac{\pi }{12}-\frac{\pi }{4})Cos(\frac{\pi }{4}+\frac{\pi }{12})=\\\\=Sin(-\frac{\pi }{6} )*Cos\frac{\pi }{3}=-\frac{1}{2} *\frac{1}{2}=-\frac{1}{4}=\boxed{-0,25}\\\\\\5)y=Sin5xCos3x=Sin(5*\frac{\pi }{8})*Cos(3*\frac{\pi }{8})=Sin\frac{5\pi }{8}*Cos\frac{3\pi }{8}=\\\\=Sin(\pi -\frac{3\pi }{8})*Cos\frac{3\pi }{8}=-Sin\frac{3\pi }{8}*Cos\frac{3\pi }{8}=-\frac{1}{2}*2 Sin\frac{3\pi }{8}*Cos\frac{3\pi }{8}=$

$=-\frac{1}{2}Sin\frac{3\pi }{4}=-\frac{1}{2}Sin(\pi-\frac{\pi }{4})=-\frac{1}{2}*(-Sin\frac{\pi }{4})=\frac{1}{2} *\frac{\sqrt{2} }{2} =\boxed{\frac{\sqrt{2} }{4}}$