Question

Алгебра.............

Answer 1

Ответ:

$-1; \quad \frac{\sqrt{2}}{2}; \quad \frac{-9-\sqrt{2}}{2};$

Объяснение:

$sin \pi+sin(-\frac{\pi}{2})=sin \pi-sin(\frac{\pi}{2})=0-1=-1;$

$cos\frac{\pi}{4}+ctg(\frac{3\pi}{2})=\frac{\sqrt{2}}{2}+ctg(\pi+\frac{\pi}{2})=\frac{\sqrt{2}}{2}+ctg(\frac{\pi}{2})=\frac{\sqrt{2}}{2}+0=\frac{\sqrt{2}}{2};$

$5sin\frac{\pi}{6}+3tg\frac{\pi}{4}-cos\frac{\pi}{4}-10tg\frac{\pi}{4}=5 \cdot \frac{1}{2}-7 \cdot 1-\frac{\sqrt{2}}{2}=\frac{5-\sqrt{2}}{2}-7=\frac{-9-\sqrt{2}}{2};$