Упр.27.24 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

а) $$\sin^2\frac{3\pi}{8}+\cos^2\frac{3\pi}{8}+\sin^4\frac{3\pi}{8}+\cos^4\frac{3\pi}{8}+\sin^6\frac{3\pi}{8}+\cos^6\frac{3\pi}{8}$$;

б) $$\cos^2\frac{5\pi}{8}-\sin^2\frac{5\pi}{8}+\cos^4\frac{5\pi}{8}-\sin^4\frac{5\pi}{8}+\cos^4\frac{5\pi}{8}-\sin^6\frac{5\pi}{8}$$.

а) Обозначим

$$S=\left(\sin^2\frac{3\pi}{8}+\cos^2\frac{3\pi}{8}\right)+\left(\sin^4\frac{3\pi}{8}+\cos^4\frac{3\pi}{8}\right)+\left(\sin^6\frac{3\pi}{8}+\cos^6\frac{3\pi}{8}\right).$$

Тогда

$$\sin^2\frac{3\pi}{8}+\cos^2\frac{3\pi}{8}=1.$$

Далее

$$\sin^4\frac{3\pi}{8}+\cos^4\frac{3\pi}{8}=\left(\sin^2\frac{3\pi}{8}+\cos^2\frac{3\pi}{8}\right)^2-2\sin^2\frac{3\pi}{8}\cos^2\frac{3\pi}{8}$$

$$=1-2\sin^2\frac{3\pi}{8}\cos^2\frac{3\pi}{8}.$$

Так как

$$\sin\frac{3\pi}{8}\cos\frac{3\pi}{8}=\frac12\sin\frac{3\pi}{4}=\frac{\sqrt2}{4},$$

то

$$\sin^4\frac{3\pi}{8}+\cos^4\frac{3\pi}{8}=1-2\cdot\frac18=\frac34.$$

Теперь

$$\sin^6\frac{3\pi}{8}+\cos^6\frac{3\pi}{8}=\left(\sin^2\frac{3\pi}{8}+\cos^2\frac{3\pi}{8}\right)\left(\sin^4\frac{3\pi}{8}-\sin^2\frac{3\pi}{8}\cos^2\frac{3\pi}{8}+\cos^4\frac{3\pi}{8}\right)$$

$$=1\left(\frac34-\frac18\right)=\frac58.$$

Следовательно,

$$S=1+\frac34+\frac58=\frac{19}{8}=2\frac38.$$

б) Обозначим

$$T=\left(\cos^2\frac{5\pi}{8}-\sin^2\frac{5\pi}{8}\right)+\left(\cos^4\frac{5\pi}{8}-\sin^4\frac{5\pi}{8}\right)+\left(\cos^6\frac{5\pi}{8}-\sin^6\frac{5\pi}{8}\right).$$

Первое слагаемое:

$$\cos^2\frac{5\pi}{8}-\sin^2\frac{5\pi}{8}=\cos\frac{5\pi}{4}=-\frac{\sqrt2}{2}.$$

Второе слагаемое:

$$\cos^4\frac{5\pi}{8}-\sin^4\frac{5\pi}{8}=\left(\cos^2\frac{5\pi}{8}-\sin^2\frac{5\pi}{8}\right)\left(\cos^2\frac{5\pi}{8}+\sin^2\frac{5\pi}{8}\right)$$

$$=-\frac{\sqrt2}{2}.$$

Третье слагаемое:

$$\cos^6\frac{5\pi}{8}-\sin^6\frac{5\pi}{8}=\left(\cos^2\frac{5\pi}{8}-\sin^2\frac{5\pi}{8}\right)\left(\cos^4\frac{5\pi}{8}+\cos^2\frac{5\pi}{8}\sin^2\frac{5\pi}{8}+\sin^4\frac{5\pi}{8}\right).$$

Так как

$$\cos^4\frac{5\pi}{8}+\cos^2\frac{5\pi}{8}\sin^2\frac{5\pi}{8}+\sin^4\frac{5\pi}{8}=1-\sin^2\frac{5\pi}{8}\cos^2\frac{5\pi}{8},$$

а

$$\sin\frac{5\pi}{8}\cos\frac{5\pi}{8}=\frac12\sin\frac{5\pi}{4}=-\frac{\sqrt2}{4},$$

то

$$\cos^6\frac{5\pi}{8}-\sin^6\frac{5\pi}{8}=-\frac{\sqrt2}{2}\left(1-\frac18\right)=-\frac{7\sqrt2}{16}.$$

Тогда

$$T=-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}-\frac{7\sqrt2}{16}=-\frac{23\sqrt2}{16}.$$

Ответ: а) $$2\frac38$$; б) $$-\frac{23\sqrt2}{16}$$.