468(1) 469(1,3) )))))))

Решите задачу:

$1)\qquad cos2a=?\\\\ \frac{cosa-2sina}{sina-2cosa} =-0,5\; \; \; \Rightarrow \\\\ \frac{cosa-2sina}{sina-2cosa} =\Big [ \frac{:cosa}{:cosa} \Big ]= \frac{1-2tga}{tga-2}=-\frac{1}{2}\; \; \; \Rightarrow \\\\2\cdot ( 1-2tga)=-(tga-2)\\\\2-2tga=-tga+2\\\\tga=0\\\\Formyla:\quad cosx=\frac{1-tg^2\frac{x}{2}}{1+tg^2\frac{x}{2}}\; \; \; \Rightarrow \\\\cos2a=\frac{1-tg^2a}{1+tg^2a}= \frac{1-0}{1+0} =1$

$2)\quad 8\cdot sin^2\frac{15\pi }{6}\cdot cos^2\frac{17\pi }{6}=8\sin^2(2\pi +\frac{3\pi }{6})\cdot cos^2(3\pi -\frac{\pi }{6})=\\\\=8\cdot sin^2\frac{\pi}{2}\cdot cos^2\frac{\pi}{6}=8\cdot 1^2\cdot (\frac{\sqrt3}{2})^2=8\cdot \frac{3}{4}=6\\\\3)\quad tg\frac{7\pi}{8}+ctg\frac{7\pi}{8}=tg(\pi -\frac{\pi}8})+ctg(\pi -\frac{\pi}{8})=$

$=-tg\frac{\pi}{8}-ctg\frac{\pi}{8}=-\frac{sin\frac{\pi}{8}}{cos\frac{\pi}{8}}-\frac{cos\frac{\pi}{8}}{sin\frac{\pi}{8}}=- \frac{sin^2\frac{\pi}{8}+cos^2\frac{\pi}{8}}{sin\frac{\pi}{8}\cdot cos\frac{\pi }{8}}=$

$=-\frac{1}{\frac{1}{2}sin\frac{\pi}{4}}=-\frac{2}{\frac{\sqrt2}{2}}=-\frac{2\cdot 2}{\sqrt2}=-2\sqrt2$