Question

Arccos(-√2/2)+arcsin(-0.5)​

Answer 1

Ответ:

решение смотри на фотографии

Пошаговое объяснение:

Answer 2

To calculate the values of $\arccos\left(-\frac{\sqrt{2}}{2}\right)$ and $\arcsin(-0.5)$, you can use trigonometric identities and properties.

1. $\arccos\left(-\frac{\sqrt{2}}{2}\right)$:

Recall that $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$. Therefore, $\arccos\left(-\frac{\sqrt{2}}{2}\right)$ is the same as finding the angle whose cosine is $-\frac{\sqrt{2}}{2}$.

Since the cosine function is negative in the second and third quadrants of the unit circle, we can say that $\arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$.

2. $\arcsin(-0.5)$:

Recall that $\sin\left(-\frac{\pi}{6}\right) = -0.5$. Therefore, $\arcsin(-0.5)$ is the same as finding the angle whose sine is $-0.5$.

Since the sine function is negative in the third and fourth quadrants of the unit circle, we can say that $\arcsin(-0.5) = -\frac{\pi}{6}$.

So, $\arccos\left(-\frac{\sqrt{2}}{2}\right) + \arcsin(-0.5) = \frac{3\pi}{4} - \frac{\pi}{6} = \frac{5\pi}{12}$.