Question

F(x)=2cosx-5sinx у точці x0=π/2

Answer 1

Ответ:

$x_0=\dfrac{\pi}{2}\ \ ,\ \ \ f(x)=2cosx-5sinx\\\\ f(x_0)=2cosx_0-5sinx_0=2cos\dfrac{\pi}{2}-5sin\dfrac{\pi}{2}=2\cdot 0-5\cdot 1=-5$

Answer 2

Ответ:

Решение смотри на фотографии

Answer 3

To find the value of $F(x)$ at the point $x_0 = \frac{\pi}{2}$, we need to substitute $x_0$ into the given function $F(x) = 2\cos(x) - 5\sin(x)$.

$F\left(\frac{\pi}{2}\right) = 2\cos\left(\frac{\pi}{2}\right) - 5\sin\left(\frac{\pi}{2}\right)$

Recall that $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$, so we can substitute these values:

$F\left(\frac{\pi}{2}\right) = 2 \times 0 - 5 \times 1 = -5$

So, $F\left(\frac{\pi}{2}\right) = -5$.