Упр.21.4 ГДЗ Мордкович 10-11 класс (Алгебра)

• а) $$2\sin\frac{\pi}{8}\cdot\cos\frac{\pi}{8}$$;
  б) $$\sin\frac{\pi}{8}\cdot\cos\frac{\pi}{8}+\frac{1}{4}$$;
  в) $$\cos^2\frac{\pi}{8}-\sin^2\frac{\pi}{8}$$;
  г) $$\frac{\sqrt{2}}{2}-\left(\cos\frac{\pi}{8}+\sin\frac{\pi}{8}\right)^2$$.

Используем формулы двойного аргумента:

$$\sin 2\alpha = 2\sin \alpha \cos \alpha,$$
$$\cos 2\alpha = \cos^2 \alpha — \sin^2 \alpha.$$

1. $$2\sin \frac{\pi}{8}\cos \frac{\pi}{8}=\sin \left(2\cdot \frac{\pi}{8}\right)=\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}.$$

2. $$\sin \frac{\pi}{8}\cos \frac{\pi}{8}+\frac14=\frac12\cdot 2\sin \frac{\pi}{8}\cos \frac{\pi}{8}+\frac14=\frac12\sin \frac{\pi}{4}+\frac14$$
   $$=\frac12\cdot \frac{\sqrt{2}}{2}+\frac14=\frac{\sqrt{2}+1}{4}.$$

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

4. $$\frac{\sqrt{2}}{2}-\left(\cos \frac{\pi}{8}+\sin \frac{\pi}{8}\right)^2$$
   $$=\frac{\sqrt{2}}{2}-\left(\cos^2 \frac{\pi}{8}+2\sin \frac{\pi}{8}\cos \frac{\pi}{8}+\sin^2 \frac{\pi}{8}\right)$$
   $$=\frac{\sqrt{2}}{2}-\left(1+\sin \frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}-1-\frac{\sqrt{2}}{2}=-1.$$

Ответ: а) $$\frac{\sqrt{2}}{2}$$; б) $$\frac{\sqrt{2}+1}{4}$$; в) $$\frac{\sqrt{2}}{2}$$; г) $$-1$$.