Question

Cos(3п/2 + а) cos(n-a) sin(180*+a)

Answer 1

Ответ:

Ответ смотри во вложении

Answer 2

It seems like you've provided a mathematical expression involving trigonometric functions. Let's break it down step by step:

1. $\cos\left(\frac{3\pi}{2} + a\right)$:

   • Recall that $\cos$ is a periodic function with a period of $2\pi$, meaning $\cos(x) = \cos(x + 2\pi k)$ for any integer $k$.
   • $\frac{3\pi}{2}$ is in the third quadrant where $\cos$ is negative, so $\cos\left(\frac{3\pi}{2}\right) = 0$ and $\sin\left(\frac{3\pi}{2}\right) = -1$.
   • Therefore, $\cos\left(\frac{3\pi}{2} + a\right) = \cos(a)$ and $\sin\left(\frac{3\pi}{2} + a\right) = -\sin(a)$.

2. $\cos(n - a)$:

   • This is just $\cos(n - a)$, and it cannot be simplified further without knowing the values of $n$ and $a$.

3. $\sin(180^\circ + a)$:

   • Remember that $\sin$ is periodic with a period of $2\pi$ (or $360^\circ$). So $\sin(x) = \sin(x + 2\pi k)$ for any integer $k$.
   • $180^\circ$ is equivalent to $\pi$ radians.
   • In the second quadrant where $\sin$ is positive, $\sin(180^\circ) = 1$ and $\cos(180^\circ) = 0$.
   • Therefore, $\sin(180^\circ + a) = \sin(a)$ and $\cos(180^\circ + a) = -\cos(a)$.

Putting it all together:

$\cos\left(\frac{3\pi}{2} + a\right) \cos(n - a) \sin(180^\circ + a) = \cos(a) \cos(n - a) \sin(a)$

Keep in mind that this expression is simplified based on the identities mentioned above. If you have specific values for $n$ and $a$, you can substitute them in to get a numerical result.