Question

Вычислить: а) arcsin √2/2 ; б) arcsin (-1) ; в) arcсos 1/2 ; г) arcсos (- √3/2);
д) arcсos (sin 2π/3); е) arcсos (-1) - arcсos 1/2 - 3 arcсos (- √3/2 );
ж) cos (arcsin ( - √2/2 )) ; з) arcsin (- √2/2 ) + arcsin 1 – arcsin √3/2 ;
и) cos ( 1/2 arcsin 1 + 2arcсos √2/2 ).
С объяснением

Answer 1

а)

$arcsin( \frac{ \sqrt{2} }{2} ) = \frac{\pi}{4}$

б)

$arcsin( - 1) = - \frac{\pi}{2}$

в)

$arccos( \frac{1}{2} ) = \frac{\pi}{3}$

г)

$arccos( - \frac{ \sqrt{3} }{2} ) = \frac{5\pi}{6}$

д)

$arccos( \sin( \frac{2\pi}{3} ) ) = \\ = arccos( \frac{ \sqrt{3} }{2} ) = \frac{\pi}{6}$

е)

$arccos( - 1) - arccos( \frac{1}{2}) - 3arccos ( - \frac{ \sqrt{3} }{2} ) = \\ = \pi - \frac{\pi}{3} - 3 \times \frac{5\pi}{6} = \\ = \frac{2\pi}{3} - \frac{5\pi}{2} = \frac{4\pi - 15\pi}{6} = \\ = - \frac{11\pi}{6}$

ж)

$\cos(arcsin ( - \frac{ \sqrt{2} }{2} )) = \cos( - \frac{\pi}{4} ) = \frac{ \sqrt{2} }{2}$

з)

$arcsin( - \frac{ \sqrt{2} }{2} ) + arcsin1 - arcsin( \frac{ \sqrt{3} }{2} ) = \\ = - \frac{\pi}{4} + \frac{\pi}{2} - \frac{\pi}{3} = - \frac{\pi}{4} + \frac{\pi}{6} = \\ = \frac{ - 3\pi + 2\pi}{12} = - \frac{\pi}{12}$

и)

$\cos( \frac{1}{2} arcsin1 + 2arccos( \frac{ \sqrt{2} }{2} ) ) = \cos( \frac{1}{2} \times \frac{\pi}{2} + 2 \times \frac{\pi}{4} ) = \\ = \cos( \frac{\pi}{4} + \frac{\pi}{2} ) = \cos( \frac{3\pi}{4} ) = - \frac{ \sqrt{2} }{2}$