Упр.659 ГДЗ Никольский Потапов 9 класс (Алгебра)

а)sin2пи/8- cos^2пи/8
б) 2sin50°•sin40°
в) cos^215°-cos^275°
г) (sin80+sin10)(cos80°-cos10°)

1. $$\sin^2\frac{\pi}{8}-\cos^2\frac{\pi}{8}=-(\cos^2\frac{\pi}{8}-\sin^2\frac{\pi}{8})=-\cos\frac{\pi}{4}=-\frac{\sqrt2}{2}.$$

2. $$2\sin50^\circ\cdot\sin40^\circ=2\sin(90^\circ-40^\circ)\sin40^\circ=2\sin40^\circ\cos40^\circ=\sin80^\circ.$$

3. $$\cos^215^\circ-\cos^275^\circ=\cos^215^\circ-\cos^2(90^\circ-15^\circ)=\cos^215^\circ-\sin^215^\circ=\cos30^\circ=\frac{\sqrt3}{2}.$$

4. $$ (\sin80^\circ+\sin10^\circ)(\cos80^\circ-\cos10^\circ)= $$

   $$ =\sin80^\circ\cos80^\circ-\sin80^\circ\cos10^\circ+\sin10^\circ\cos80^\circ-\sin10^\circ\cos10^\circ $$

   $$ =\frac12\sin160^\circ-\bigl(\sin80^\circ\cos10^\circ-\cos80^\circ\sin10^\circ\bigr)-\frac12\sin20^\circ $$

   $$ =\frac12\sin160^\circ-\frac12\sin20^\circ-\sin(80^\circ-10^\circ) $$

   $$ =\frac12(\sin160^\circ-\sin20^\circ)-\sin70^\circ $$

   $$ =\frac12\cdot 2\sin\frac{160^\circ+20^\circ}{2}\cos\frac{160^\circ-20^\circ}{2}-\sin70^\circ =\sin90^\circ\cos70^\circ-\sin70^\circ $$

   $$ =\cos70^\circ-\sin70^\circ. $$

Ответ

а) $$-\frac{\sqrt2}{2}$$; б) $$\sin80^\circ$$; в) $$\frac{\sqrt3}{2}$$; г) $$\cos70^\circ-\sin70^\circ$$.