Question

задание вычислите arctg
номера во вложении

Answer 1

Ответ:

1.

$6arctg( \sqrt{3} ) - 4arcsin( - \frac{1}{ \sqrt{2} } ) = \\ = 6 \times \frac{\pi}{3} + 4arcsin( \frac{ \sqrt{2} }{2}) = \\ = 2\pi + 4 \times \frac{\pi}{4} = 3\pi$

2.

$2arctg1 + 3arcsin( - \frac{1}{2} ) = \\ = 3 \times \frac{\pi}{4} - 3 \times \frac{\pi}{6} = \frac{3\pi}{4} - \frac{\pi}{2} = \\ = \frac{3\pi - 2\pi}{4} = \frac{\pi}{4}$

3.

$3arctg( - \frac{1}{ \sqrt{3} } ) + 2arccos( - \frac{ \sqrt{3} }{2} ) = \\ = - 3arctg( \frac{ \sqrt{3} }{ 3} ) - 2 \times \frac{5\pi}{6} = \\ = - 3 \times \frac{\pi}{6} - \frac{5\pi}{3} = \frac{\pi}{2} - \frac{5\pi}{3} = - \frac{7\pi}{6}$

4.

$5arctg( - \sqrt{3} ) - 3arccos( - \frac{ \sqrt{2} }{2} ) = \\ = - 5 \times \frac{\pi}{3} + 3 \times \frac{3\pi}{4} = - \frac{5\pi}{3} + \frac{9\pi}{4} = \\ = \frac{ - 20\pi + 27\pi}{12} = \frac{7\pi}{12}$

Answer 2

Ответ:

$1)\ \ 6\, arctg\sqrt3-4\, arcsin(-\dfrac{1}{\sqrt2})=6\cdot \dfrac{\pi}{3}+4\cdot \dfrac{\pi}{4}=2\pi +\pi =3\pi \\\\\\2)\ \ 2\, arctg\, 1+3\, arcsin(-\dfrac{1}{2})=2\cdot \dfrac{\pi}{4}-3\cdot \dfrac{\pi}{6}=\dfrac{\pi}{2}-\dfrac{\pi}{2}=0\\\\\\3)\ \ 3\, arctg(-\dfrac{1}{\sqrt3})+2\, arccos(-\dfrac{\sqrt3}{2})=-3\cdot \dfrac{\pi}{6}+2\cdot (\pi -\dfrac{\pi}{6})=-\dfrac{3\pi}{6}+\dfrac{10\pi }{6}=-\dfrac{7\pi}{6}$

$4)\ \ 5\, arctg(-\sqrt3)-3\, arccos(-\dfrac{\sqrt2}{2})=-5\cdot \dfrac{\pi}{3}-3\cdot (\pi-\dfrac{\pi}{4})=-\dfrac{5\pi }{3}-\dfrac{9\pi}{4}=-\dfrac{47\pi}{12}$