Упр.27.23 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

а) $$\left(\cos\frac{\pi}{8}+\sin\frac{\pi}{8}\right)\left(\cos^3\frac{\pi}{8}-\sin^3\frac{\pi}{8}\right)$$
б) $$\sin\frac{7\pi}{8}\left(\cos^4\frac{7\pi}{16}-\sin^4\frac{7\pi}{16}\right)$$
в) $$\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)\left(\cos^3\frac{\pi}{12}+\sin^3\frac{\pi}{12}\right)$$
г) $$\sin\frac{\pi}{12}\left(\cos^6\frac{7\pi}{24}-\sin^6\frac{7\pi}{24}\right)$$

а) $$\left(\cos \frac{\pi}{8}+\sin \frac{\pi}{8}\right)\left(\cos^3 \frac{\pi}{8}-\sin^3 \frac{\pi}{8}\right)$$

Используем формулу разности кубов:

$$\cos^3 \frac{\pi}{8}-\sin^3 \frac{\pi}{8}=\left(\cos \frac{\pi}{8}-\sin \frac{\pi}{8}\right)\left(\cos^2 \frac{\pi}{8}+\cos \frac{\pi}{8}\sin \frac{\pi}{8}+\sin^2 \frac{\pi}{8}\right).$$

Тогда

$$\left(\cos \frac{\pi}{8}+\sin \frac{\pi}{8}\right)\left(\cos \frac{\pi}{8}-\sin \frac{\pi}{8}\right)\left(1+\sin \frac{\pi}{8}\cos \frac{\pi}{8}\right)$$

$$=\left(\cos^2 \frac{\pi}{8}-\sin^2 \frac{\pi}{8}\right)\left(1+\sin \frac{\pi}{8}\cos \frac{\pi}{8}\right)$$

$$=\cos \frac{\pi}{4}\left(1+\frac12 \sin \frac{\pi}{4}\right)=\frac{\sqrt2}{2}\left(1+\frac12\cdot \frac{\sqrt2}{2}\right)=\frac{2\sqrt2+1}{4}.$$

б) $$\sin \frac{7\pi}{8}\left(\cos^4 \frac{7\pi}{16}-\sin^4 \frac{7\pi}{16}\right)$$

Разность четвёртых степеней:

$$\cos^4 \frac{7\pi}{16}-\sin^4 \frac{7\pi}{16}=\left(\cos^2 \frac{7\pi}{16}-\sin^2 \frac{7\pi}{16}\right)\left(\cos^2 \frac{7\pi}{16}+\sin^2 \frac{7\pi}{16}\right).$$

Следовательно,

$$\sin \frac{7\pi}{8}\left(\cos^2 \frac{7\pi}{16}-\sin^2 \frac{7\pi}{16}\right)=\sin \frac{7\pi}{8}\cos \frac{7\pi}{8}.$$

Так как $$\sin \frac{7\pi}{8}=\sin \frac{\pi}{8},$$ а $$\cos \frac{7\pi}{8}=-\cos \frac{\pi}{8},$$ то

$$\sin \frac{7\pi}{8}\cos \frac{7\pi}{8}=-\frac12 \sin \frac{\pi}{4}=-\frac{\sqrt2}{4}.$$

в) $$\left(\cos \frac{\pi}{12}-\sin \frac{\pi}{12}\right)\left(\cos^3 \frac{\pi}{12}+\sin^3 \frac{\pi}{12}\right)$$

Используем формулу суммы кубов:

$$\cos^3 \frac{\pi}{12}+\sin^3 \frac{\pi}{12}=\left(\cos \frac{\pi}{12}+\sin \frac{\pi}{12}\right)\left(\cos^2 \frac{\pi}{12}-\cos \frac{\pi}{12}\sin \frac{\pi}{12}+\sin^2 \frac{\pi}{12}\right).$$

Тогда

$$\left(\cos^2 \frac{\pi}{12}-\sin^2 \frac{\pi}{12}\right)\left(1-\sin \frac{\pi}{12}\cos \frac{\pi}{12}\right)$$

$$=\cos \frac{\pi}{6}\left(1-\frac12 \sin \frac{\pi}{6}\right)=\frac{\sqrt3}{2}\left(1-\frac14\right)=\frac{3\sqrt3}{8}.$$

г) $$\sin \frac{\pi}{12}\left(\cos^6 \frac{\pi}{24}-\sin^6 \frac{\pi}{24}\right)$$

Разложим разность шестых степеней:

$$\cos^6 \frac{\pi}{24}-\sin^6 \frac{\pi}{24}=\left(\cos^2 \frac{\pi}{24}-\sin^2 \frac{\pi}{24}\right)\left(\cos^4 \frac{\pi}{24}+\cos^2 \frac{\pi}{24}\sin^2 \frac{\pi}{24}+\sin^4 \frac{\pi}{24}\right).$$

После преобразований получаем

$$\sin \frac{\pi}{12}\cos \frac{\pi}{12}\left(1-\frac14 \sin^2 \frac{\pi}{12}\right).$$

Так как $$\sin \frac{\pi}{12}\cos \frac{\pi}{12}=\frac12 \sin \frac{\pi}{6}=\frac14,$$ а $$\sin^2 \frac{\pi}{12}=\frac{1-\cos \frac{\pi}{6}}{2}=\frac{2-\sqrt3}{4},$$ то

$$\frac14\left(1-\frac14\cdot \frac{2-\sqrt3}{4}\right)=\frac{14+\sqrt3}{64}.$$

Ответ: а) $$\frac{2\sqrt2+1}{4}$$; б) $$-\frac{\sqrt2}{4}$$; в) $$\frac{3\sqrt3}{8}$$; г) $$\frac{14+\sqrt3}{64}$$.