How do you prove $\cos (\pi - x) = - \cos x$?

Hint: Here in the above proof we can apply the formula in the left hand side which is $\cos (\pi - x)$ of $\cos (A - B) = \cos A\cos B + \sin A\sin B$ and then we can substitute the values accordingly and get the simplified form of the answer. This way can lead us easily to its proof.

Complete step-by-step answer:
Here we are given to prove $\cos (\pi - x) = - \cos x$
Therefore we need to analyse the left hand side of the given proof which is $\cos (\pi - x)$
As it is of the form $\cos (A - B)$
Hence we can apply the formula here of the form:
$\cos (A - B) = \cos A\cos B + \sin A\sin B$$ - - - - (1)$
Hence we can compare $(A - B){\text{ with (}}\pi - x)$
We will get:
$
  A = \pi \\
  B = x \\
 $
So after substituting these values in the equation (1) we will get the equation as:
$\cos (A - B) = \cos A\cos B + \sin A\sin B$
$\cos (\pi - x) = \cos \pi \cos x + \sin \pi \sin x$$ - - - - - (2)$-
Now we know that $\sin (n\pi ) = 0,n \in Z$
So when we put in it $n = 1$ we will get $\sin \pi = 0 - - - (3)$
Also we know that value of $\cos \pi = - 1$$ - - - - (4)$ according to its general values.
So we can put these values from equation (3) and (4) in the equation (2) and we will get:
$\cos (\pi - x) = ( - 1)\cos x + (0)\sin x$
Now we have got the equation after the substitution of the value of the trigonometric functions and now we just need to solve it and simplify it as much as we can. So we will get:
$\cos (\pi - x) = ( - 1)\cos x + (0)\sin x$
$
  ( - 1)\cos x + 0 \\
   - \cos x \\
 $
Hence we get the value after substitution as $ - \cos x$
Hence we have proved that $\cos (\pi - x) = - \cos x$.

Note: Here in these types of problems the student must know how to compare the formula with the problem given so that we can understand the formula that is to be applied. As here also we have compared $(A - B){\text{ with (}}\pi - x)$ and we applied the formula accordingly. The student must know the formula of all the trigonometric functions.