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To prove the providedexpression, we'll use the identities:
cos (a+b) = cos a*cos b - sin a*sin b
cos (a-b) = cos a*cos b + sin a*sin b
Let a = `Pi` and b = x
cos (`Pi`-x) = cos `Pi`*cos x + sin `Pi`*sin x
But cos `Pi` = -1 and sin `Pi` = 0
cos (` `-x) =-1*cos x + 0 ` `*sin x
cos (` `-x) =-cos x
cos (` `+x) = cos ` `*cos x - sin ` `*sin x
cos (` `+x) = -cos x
We can think in this way: if we'll subtract from 180 degrees an angle smaller than 180 degrees and bigger that 90 degrees, the angle will be located in the 2nd quadrant, where the values of cosine function are negative.
If we'll add to 180 degrees an angle bigger than 180 degrees and smaller that 270 degrees, the angle will be located in the 3rd quadrant, where the values of cosine function are negative, too.
Therefore, cos (`pi`-x) = cos (`pi` +x) = - cos x.
From the picture, it's not difficult to see that cos(pi-x)=-cos(x)
If you drew a similar diagram, it would also be pretty easy to see that cos(pi-x)=cos(pi+x)
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