cos(x + y) + cos(x − y) = 2 cos(x) cos(y)
\(\texttt{We use formula: }~~~ \boxed{\cos(x\pm y) = \cos (x) \cos (y) ~\mp ~\sin (x) \sin (y)}\\\\ \cos(x + y) + \cos(x - y) =\\\\ = \underbrace{\cos (x) \cos (y) -\sin (x) \sin (y)}_{\cos(x + y) } +\underbrace{\cos (x) \cos (y) +\sin (x) \sin (y)}_{\cos(x - y)} =\\\\ =\cos (x) \cos (y)+\cos (x) \cos (y)-\sin (x) \sin (y)+\sin (x) \sin (y)=\\\\ =\cos (x) \cos (y)(1+1) + \sin (x) \sin (y)(-1+1) = \\\\ =2\cos (x) \cos (y) + 0\sin (x) \sin (y) = \boxed{\boxed{2\cos (x) \cos (y)}}\)
Answer:
\(2cos(\frac{x+y+x-y}{2}) cos(\frac{x+y-x+y}{2})\)
\(2cos(\frac{2x}{2})cos(\frac{x+y-x+y}{2})\)
\(2cos(x)cos(\frac{2y}{2})\)
\(2cos(x)cos(y)\)
