Question

3arccos(1/2)+6arctg(-1/√3) решение ​

Answer 1

3*п/3 + 6*(-п/6) = 3п/3 - 6п/6 = п - п = 0

Answer 2

To solve this expression, let's first simplify the values of the trigonometric functions:

1. $\arccos\left(\frac{1}{2}\right)$: This is the angle whose cosine is $\frac{1}{2}$. The angle that meets this condition is $\frac{\pi}{3}$ (60 degrees), since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.

2. $\arctg\left(-\frac{1}{\sqrt{3}}\right)$: This is the angle whose tangent is $-\frac{1}{\sqrt{3}}$. The angle that meets this condition is $-\frac{\pi}{6}$ (-30 degrees), since $\tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$.

Now, we substitute these values into the given expression:

$\arccos\left(\frac{1}{2}\right) + 6\arctg\left(-\frac{1}{\sqrt{3}}\right)$

$= \frac{\pi}{3} + 6 \cdot \left(-\frac{\pi}{6}\right)$

$= \frac{\pi}{3} - \pi$

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

So, the solution to the expression $\arccos\left(\frac{1}{2}\right) + 6\arctg\left(-\frac{1}{\sqrt{3}}\right)$ is $-\frac{2\pi}{3}$.