verify cos(x+y)+cos(x-y)=2cosxcosy?

.. cos(x+y) + cos(x-y)
= cos(x) cos(y) + sin(x) sin(y) + cos(x) cos(-y) + sin(x) sin(-y)
= cos(x) cos(y) + sin(x) sin(y) + cos(x) cos(y) - sin(x) sin(y)
= cos(x) cos(y) + cos(x) cos(y)
= 2 cos(x) cos(y)

Expand the left hand side using your "sum of cosine" formula.

This should be memorized:
cos (x+y) = cos x * cos y - sin x * sin y

This should NOT be memorized:
cos (x-y) = cos x * cos y + sin x * sin y

It's too easy to get it wrong. To derive cos (x-y), simply take the first equation and replace 'y' with '-y':
cos (x + -y) = cos x * cos -y - sin x * sin -y

Then from your unit circle, replace cos -y with cos y, and sin -y with -sin y:
cos (x + -y) = cos x cos y - sin x * - sin y

Cancel out the negatives, and you get your identity:
cos (x - y) = cos x cos y + sin x sin y

You can also quickly derive cos 2A from this, by simply setting x = y = A. Try to limit the identites you memorize...all those +/- signs will confuse you at test time!

To answer your question, just add those two identities, and cancel out the stuff, and you'll be able to get 2 cos x cos y.

Another form of this equation is:
cos x cos y = 1/2 * [cos (x + y) + cos (x - y)]

It converts a product of cosines into a sum of cosines.

One other tip in trig in general...remember that identities go BOTH ways! This may seem trivial, but it's important. It's easy to see that cos (45 + 30) = cos 45 * cos 30 - sin 45 sin 30. But you can also put things back together. If you ever happen to see a problem that asks you to evaluate cos 10 cos 20 - sin 10 sin 20, don't try to evaluate sin 10! Stuff them back together to get cos (10 + 20).

cos(x+y)+cos(x-y) = cos(x) cos(y) - sin(x)sin(y) + cos(x) cos(y) + sin(x)sin(y)
= 2 cos(x) cos(y)

Taking x=0 & y=pi/2, the LHS=
cos(0+pi/2)+cos(0-pi/2)=
0+0=0
the RHS=
2cos(0)cos(pi/2)=
2*1*0=0
So, the given equation is true.
for x=0 & y=pi/2. It is likely that
the equation is true for other
values of x,y. For this, it needs
a proof.

I'm assuming you can use the rules:
we know that---- cos(x+y) = cos(x)cos(y) - sin(x)sin(y)
and------------------ cos(x-y) = cos(x)cos(y) + sin(x)sin(y)
so: cos(x+y) + cos(x-y) = cos(x)cos(y) - sin(x)sin(y) + cos(x)cos(y) + sin(x)sin(y)
---- cos(x+y) + cos(x-y) = 2cos(x)cos(y)

LHS:
cos(x + y) + cos(x - y)
= [cos(x)cos(y) - sin(x)sin(y)] + [cos(x)cos(y) + sin(x)sin(y)]
= cos(x)cos(y) + cos(x)cos(y) - sin(x)sin(y) + sin(x)sin(y)
= 2cos(x)cos(y)

RHS:
2cos(x)cos(y)

Therefore, cos(x + y) + cos(x - y) ≡ 2cos(x)cos(y)

there is a direct formula for cosxcosy:
cosxcosy = (cox(x+y) + cox(x-y)) / 2

there's nothing much to write

cos(x + y) = cos(x)cos(y) - sin(x)sin(y)
cos(x - y) = cos(x)cos(y) + sin(x)sin(y)

cos(x + y) + cos(x - y) =>
cos(x)cos(y) - sin(x)sin(y) + cos(x)cos(y) + sin(x)sin(y) =>
2cos(x)cos(y)