Question

срочно!!!!!!!!!!!
$arccos( - \frac{1}{2} ) - arc \tan( - \frac{ \sqrt{3} }{2})$
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Answer 1

Ответ:

Ответ:

\begin{gathered}\displaystyle 2arcsin(-\frac{\sqrt3}{2})+arctg(-1)+arccos\frac{\sqrt2}{2}=-2\cdot \frac{\pi}{3}-\frac{\pi}{4}+\frac{\pi}{4}=-\frac{2\pi}{3}\\\\\\\\3arcsin\frac{1}{2}+4\, arccos(-\frac{1}{\sqrt2})-arcctg(-\sqrt3)=3\cdot \frac{\pi}{6} +4\cdot (\pi -\frac{\pi}{4})-(\pi -\frac{\pi}{6})=\\\\\\=\frac{\pi}{2}+3\pi -\frac{5\pi}{6}=\frac{16\pi}{6}=\frac{8\pi}{3}\end{gathered}

2arcsin(−

2

3

)+arctg(−1)+arccos

2

2

=−2⋅

3

π

−

4

π

+

4

π

=−

3

2π

3arcsin

2

1

+4arccos(−

2

1

)−arcctg(−

3

)=3⋅

6

π

+4⋅(π−

4

π

)−(π−

6

π

)=

=

2

π

+3π−

6

5π

=

6

16π

=

3

8π

Пошаговое объяснение:

пон