Question

Тригонометрия
Задание 16)

Answer 1

$\frac{cos\frac{3\pi}{8}cos\frac{\pi}{8}-sin\frac{3\pi}{8}sin\frac{\pi}{8}}{tg(\frac{\pi}{4}+\beta)} = \frac{cos(\frac{3\pi}{8}+\frac{\pi}{8})}{tg(\frac{\pi}{4}+\beta)} = \frac{cos\frac{\pi}{2}}{tg(\frac{\pi}{4}+\beta)}=\frac{0}{tg(\frac{\pi}{4}+\beta)}=0$

Answer 2

$\frac{cos\frac{3\pi }{8}\cdot cos\frac{\pi }{8}-sin\frac{3\pi }{8}\cdot sin\frac{\pi }{8}}{tg(\frac{\pi}{4}+\beta )}=\frac{cos(\frac{3\pi}{8}+\frac{\pi}{8})}{tg(\frac{\pi}{4}+\beta )}=\frac{cos\frac{\pi}{2}}{tg(\frac{\pi}{4}+\beta )}=0\; ,\\\\cos\frac{\pi }{2}=0$