Упр.19.11 ГДЗ Мордкович 10-11 класс (Алгебра)
а) $$\cos\frac{5\pi}{8}\cdot\cos\frac{3\pi}{8}+\sin\frac{5\pi}{8}\cdot\sin\frac{3\pi}{8}$$;
б) $$\sin\frac{2\pi}{15}\cdot\cos\frac{\pi}{5}+\cos\frac{2\pi}{15}\cdot\sin\frac{\pi}{5}$$;
в) $$\cos\frac{\pi}{12}\cdot\cos\frac{\pi}{4}-\sin\frac{\pi}{12}\cdot\sin\frac{\pi}{4}$$;
г) $$\sin\frac{\pi}{12}\cdot\cos\frac{\pi}{4}-\cos\frac{\pi}{12}\cdot\sin\frac{\pi}{4}$$.
а) $$\cos \frac{5\pi}{8}\cos \frac{3\pi}{8}+\sin \frac{5\pi}{8}\sin \frac{3\pi}{8}=\cos\left(\frac{5\pi}{8}-\frac{3\pi}{8}\right)=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}.$$
б) $$\sin \frac{2\pi}{15}\cos \frac{\pi}{5}+\cos \frac{2\pi}{15}\sin \frac{\pi}{5}=\sin\left(\frac{2\pi}{15}+\frac{\pi}{5}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}.$$
в) $$\cos \frac{\pi}{12}\cos \frac{\pi}{4}-\sin \frac{\pi}{12}\sin \frac{\pi}{4}=\cos\left(\frac{\pi}{12}+\frac{\pi}{4}\right)=\cos \frac{\pi}{3}=\frac{1}{2}.$$
г) $$\sin \frac{\pi}{12}\cos \frac{\pi}{4}-\cos \frac{\pi}{12}\sin \frac{\pi}{4}=\sin\left(\frac{\pi}{12}-\frac{\pi}{4}\right)=\sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2}.$$
Ответ: а) $$\frac{\sqrt{2}}{2}$$; б) $$\frac{\sqrt{3}}{2}$$; в) $$\frac{1}{2}$$; г) $$-\frac{1}{2}$$.
