2 π\nB) cos-sin,\n8\n2\n8\nπ\nr) (cos-+sin).

B) $$cos^2\frac{\pi}{8}-sin^2\frac{\pi}{8}=cos(2\cdot\frac{\pi}{8})=cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$$.

Ответ:$$\frac{\sqrt{2}}{2}$$

r) $$\frac{\sqrt{2}}{2}(cos\frac{\pi}{8}+sin\frac{\pi}{8})^2=\frac{\sqrt{2}}{2}(cos^2\frac{\pi}{8}+2cos\frac{\pi}{8}sin\frac{\pi}{8}+sin^2\frac{\pi}{8})=\frac{\sqrt{2}}{2}(1+sin(2\cdot\frac{\pi}{8}))=\frac{\sqrt{2}}{2}(1+sin\frac{\pi}{4})=\frac{\sqrt{2}}{2}(1+\frac{\sqrt{2}}{2})=\frac{\sqrt{2}}{2}+\frac{1}{2}=\frac{\sqrt{2}+1}{2}$$.

Ответ:$$\frac{\sqrt{2}+1}{2}$$