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

Рассмотрим вариант решения задания из учебника Мордкович, Семенов 10 класс, Мнемозина: a) (cos пи/8 + sin пи/8) * (cos3 пи/8 — sin3 пи/8); б) sin 7пи/8 (cos4 7пи/16 — sin4 7пи/16); в) (cos пи/12 — sin пи/12) * (cos3 пи/12 + sin3 пи/12); г) sin пи/12 (cos6 7пи/24 — sin6 7пи/24).

1. а)

   $$\left(\cos \frac{\pi}{8}+\sin \frac{\pi}{8}\right)\left(\cos^3 \frac{\pi}{8}-\sin^3 \frac{\pi}{8}\right)$$
   $$=\left(\cos \frac{\pi}{8}+\sin \frac{\pi}{8}\right)\left(\cos \frac{\pi}{8}-\sin \frac{\pi}{8}\right)\left(\cos^2 \frac{\pi}{8}+\cos \frac{\pi}{8}\sin \frac{\pi}{8}+\sin^2 \frac{\pi}{8}\right)$$
   $$=\left(\cos^2 \frac{\pi}{8}-\sin^2 \frac{\pi}{8}\right)\left(1+\sin \frac{\pi}{8}\cos \frac{\pi}{8}\right)$$
   $$=\cos \frac{\pi}{4}\left(1+\frac12\sin \frac{\pi}{4}\right)$$
   $$=\frac{\sqrt2}{2}\left(1+\frac12\cdot\frac{\sqrt2}{2}\right)=\frac{\sqrt2}{2}+\frac14=\frac{2\sqrt2+1}{4}.$$

2. б)

   $$\sin \frac{7\pi}{8}\left(\cos^4 \frac{7\pi}{16}-\sin^4 \frac{7\pi}{16}\right)$$
   $$=\sin \frac{7\pi}{8}\left(\cos^2 \frac{7\pi}{16}-\sin^2 \frac{7\pi}{16}\right)\left(\cos^2 \frac{7\pi}{16}+\sin^2 \frac{7\pi}{16}\right)$$
   $$=\sin \frac{7\pi}{8}\cos \frac{7\pi}{8}$$
   $$=\frac12\sin \frac{7\pi}{4}=-\frac{\sqrt2}{4}.$$

3. в)

   $$\left(\cos \frac{\pi}{12}-\sin \frac{\pi}{12}\right)\left(\cos^3 \frac{\pi}{12}+\sin^3 \frac{\pi}{12}\right)$$
   $$=\left(\cos \frac{\pi}{12}-\sin \frac{\pi}{12}\right)\left(\cos \frac{\pi}{12}+\sin \frac{\pi}{12}\right)\left(\cos^2 \frac{\pi}{12}-\cos \frac{\pi}{12}\sin \frac{\pi}{12}+\sin^2 \frac{\pi}{12}\right)$$
   $$=\left(\cos^2 \frac{\pi}{12}-\sin^2 \frac{\pi}{12}\right)\left(1-\sin \frac{\pi}{12}\cos \frac{\pi}{12}\right)$$
   $$=\cos \frac{\pi}{6}\left(1-\frac12\sin \frac{\pi}{6}\right)$$
   $$=\frac{\sqrt3}{2}\left(1-\frac14\right)=\frac{3\sqrt3}{8}.$$

4. г)

   $$\sin \frac{\pi}{12}\left(\cos^6 \frac{\pi}{24}-\sin^6 \frac{\pi}{24}\right)$$
   $$=\sin \frac{\pi}{12}\left(\cos^2 \frac{\pi}{24}-\sin^2 \frac{\pi}{24}\right)\left(\cos^4 \frac{\pi}{24}+\cos^2 \frac{\pi}{24}\sin^2 \frac{\pi}{24}+\sin^4 \frac{\pi}{24}\right)$$
   $$=\sin \frac{\pi}{12}\cos \frac{\pi}{12}\left(1-\cos^2 \frac{\pi}{24}\sin^2 \frac{\pi}{24}\right)$$
   $$=\frac12\sin \frac{\pi}{6}\left(1-\frac14\sin^2 \frac{\pi}{12}\right)$$
   $$=\frac14\left(1-\frac14\cdot\frac{1-\cos \frac{\pi}{6}}{2}\right)$$
   $$=\frac14\left(1-\frac{1-\frac{\sqrt3}{2}}{8}\right)=\frac{14+\sqrt3}{64}.$$

Ответ

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