Question

Arccos (-1/2)-arcsin/2​

Answer 1

Arccos(-1/2)=2П/3
Arcsin(корень из 3/2)=П/3
2П/3-П/3=П/3

Answer 2

It looks like there's a typo in your expression. I assume you mean:

$\arccos\left(-\frac{1}{2}\right) - \arcsin\left(\frac{1}{2}\right)$

Let's break this down:

1. $\arccos\left(-\frac{1}{2}\right)$: This is asking for the angle whose cosine is $-\frac{1}{2}$. In the unit circle, this corresponds to an angle of $\frac{2\pi}{3}$ radians or $120^\circ$.

2. $\arcsin\left(\frac{1}{2}\right)$: This is asking for the angle whose sine is $\frac{1}{2}$. In the unit circle, this corresponds to an angle of $\frac{\pi}{6}$ radians or $30^\circ$.

Now, substitute these values into the expression:

$\frac{2\pi}{3} - \frac{\pi}{6}$

To combine these fractions, find a common denominator, which is 6 in this case:

$\frac{4\pi}{6} - \frac{\pi}{6}$

Now subtract the numerators:

$\frac{4\pi - \pi}{6}$

Simplify:

$\frac{3\pi}{6}$

Finally, simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 3:

$\frac{\pi}{2}$

So, $\arccos\left(-\frac{1}{2}\right) - \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{2}$.