Question

Решить математику с фото.
Задание нужно расписать.
С 1 по 4 нужно решить.
В пятом нужно определить знак произведения функций.

Answer 1

$1) ~ \sin 420^{\circ} = \sin (60^{\circ} + 360^{\circ}) = \sin 60^{\circ} = \dfrac{\sqrt{3}}{2}$

$2) ~ \text{tg} \, 240^{\circ} = \text{tg} \,(60^{\circ} + 180^{\circ}) = \text{tg} \, 60^{\circ} = \sqrt{3}$

$3) ~ \text{ctg}\, \dfrac{7\pi}{3} = \text{ctg}\left(\dfrac{\pi + 6\pi}{3} \right) = \text{ctg}\left(\dfrac{\pi}{3} + \dfrac{6\pi}{3} \right) = \text{ctg} \left(\dfrac{\pi}{3} + 2\pi \right) = \text{ctg}\, \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{3}$

$4) ~ \cos \dfrac{4\pi}{3} = \cos \dfrac{\pi + 3\pi}{3} = \cos \left(\dfrac{\pi}{3} + \dfrac{3\pi}{3} \right) = \cos \left(\dfrac{\pi}{3} + \pi \right) = \cos \dfrac{\pi}{3} = \dfrac{1}{2}$

$5) ~ \text{ctg} \, 390^{\circ} = \text{ctg}\, (30^{\circ} + 360^{\circ}) = \text{ctg}\, 30^{\circ}$

$\sin \dfrac{9\pi}{4} = \sin \dfrac{\pi + 8\pi}{4} = \sin \left(\dfrac{\pi}{4} + \dfrac{8\pi}{4} \right)=\sin \left(\dfrac{\pi}{4} + 2\pi \right) = \sin \dfrac{\pi}{4}$

$30^{\circ} \in (0^{\circ}; ~ 90^{\circ}) \Rightarrow \text{ctg} \, 30^{\circ} > 0$

$\dfrac{\pi}{4} = \dfrac{180^{\circ}}{4} = 45^{\circ} \in (0^{\circ}; ~ 90^{\circ}) \Rightarrow \sin \dfrac{\pi}{4} > 0$

$\text{ctg} \, 390^{\circ} \cdot \sin \dfrac{9 \pi}{4} > 0$