Question

Буду благодарна за помощь

Answer 1

Ответ:

1.

$arccos( - 1) + arccos0 = \pi + \frac{\pi}{2} = \frac{3\pi}{2}$

2.

$arccos \frac{1}{2} - arccos \frac{ \sqrt{3} }{2} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$

3.

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

4.

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

5.

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

6.

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

7.

$\cos(2arccos \frac{1}{2} - 3arccos0 - arccos( - \frac{1}{2} ) ) = \\ = \cos(2 \times \frac{\pi}{3} - \frac{3\pi}{2} - \frac{2\pi}{3} ) = \cos( - \frac{3\pi}{2} ) = 0$

Answer 2

Ответ:

1. π + π/2 = 3π/2

2. π/3 - π/6 = π/6

3. 3π/4 + π/4 = π

4. 2π/3 - π/3 = π/3

5. tg(π/6) = $\frac{\sqrt{3} }{3}$

6. sin(π/4) = $\frac{\sqrt{2} }{2}$

7. cos(2π/3 - 3π/2 - 2π/3) = cos(- 3π/2) =cos(3π/2)= 0