Упр.659 ГДЗ Никольский Потапов 9 класс (Алгебра)
а) $$\sin\frac{2\pi}{8}-\cos^2\frac{\pi}{8}$$
б) $$2\sin 50^\circ\cdot\sin 40^\circ$$
в) $$\cos^2 15^\circ-\cos^2 75^\circ$$
г) $$(\sin 80+\sin 10)(\cos 80^\circ-\cos 10^\circ)$$
$$\sin^2\frac{\pi}{8}-\cos^2\frac{\pi}{8}=-(\cos^2\frac{\pi}{8}-\sin^2\frac{\pi}{8})=-\cos\frac{\pi}{4}=-\frac{\sqrt2}{2}.$$
$$2\sin50^\circ\cdot\sin40^\circ=2\sin(90^\circ-40^\circ)\sin40^\circ=2\sin40^\circ\cos40^\circ=\sin80^\circ.$$
$$\cos^215^\circ-\cos^275^\circ=\cos^215^\circ-\cos^2(90^\circ-15^\circ)=\cos^215^\circ-\sin^215^\circ=\cos30^\circ=\frac{\sqrt3}{2}.$$
$$(\sin80^\circ+\sin10^\circ)(\cos80^\circ-\cos10^\circ)=$$
$$=\sin80^\circ\cos80^\circ-\sin80^\circ\cos10^\circ+\sin10^\circ\cos80^\circ-\sin10^\circ\cos10^\circ$$
$$=\frac12\sin160^\circ-\bigl(\sin80^\circ\cos10^\circ-\cos80^\circ\sin10^\circ\bigr)-\frac12\sin20^\circ$$
$$=\frac12\sin160^\circ-\frac12\sin20^\circ-\sin(80^\circ-10^\circ)$$
$$=\frac12(\sin160^\circ-\sin20^\circ)-\sin70^\circ$$
$$=\frac12\cdot 2\sin\frac{160^\circ+20^\circ}{2}\cos\frac{160^\circ-20^\circ}{2}-\sin70^\circ =\sin90^\circ\cos70^\circ-\sin70^\circ$$
$$=\cos70^\circ-\sin70^\circ.$$
Ответ
а) $$-\frac{\sqrt2}{2}$$; б) $$\sin80^\circ$$; в) $$\frac{\sqrt3}{2}$$; г) $$\cos70^\circ-\sin70^\circ$$.
