Question

(arccos(-1))/(arcsin((1)/(3))+arccos((1)/(3)))

Answer 1

(arccos(-1))/(arcsin(1/3))+arccos(1/3)= 0/(arcsin(1/3)) + arccos(1/3) = cos(1/3).
Дальше не знаю как (от cos(1/3) нет значений в обычной таблице, может стоит заглянуть в таблицу Брадиса?)

Answer 2

180/(19.5+70.5)=180/90=2 градуса

Answer 3

To find the value of the expression $\frac{\arccos(-1)}{\arcsin(\frac{1}{3}) + \arccos(\frac{1}{3})}$, let's start by evaluating the individual components step by step:

1. $\arccos(-1)$: The arccosine function $\arccos(x)$ gives the angle whose cosine is $x$. Since $\cos(\pi) = -1$, we have $\arccos(-1) = \pi$.

2. $\arcsin(\frac{1}{3})$: The arcsine function $\arcsin(x)$ gives the angle whose sine is $x$. So, $\arcsin(\frac{1}{3})$ is the angle whose sine is $\frac{1}{3}$.

3. $\arccos(\frac{1}{3})$: Similar to the above, $\arccos(\frac{1}{3})$ is the angle whose cosine is $\frac{1}{3}$.

Now, we don't know the exact values of $\arcsin(\frac{1}{3})$ and $\arccos(\frac{1}{3})$, but we do know that they are both acute angles (between 0 and $\frac{\pi}{2}$). Therefore, their sum $\arcsin(\frac{1}{3}) + \arccos(\frac{1}{3})$ is also an acute angle.

So, the final expression becomes:

$\frac{\pi}{\text{(acute angle)}}$.

Since the denominator is an acute angle, its value is less than $\frac{\pi}{2}$. Thus, the whole expression is larger than $\frac{\pi}{\frac{\pi}{2}}$, which simplifies to 2.

Therefore, $\frac{\arccos(-1)}{\arcsin(\frac{1}{3}) + \arccos(\frac{1}{3})} > 2$.