Where does this trig identiy come from?

[vincent_vega]

Our teacher used the following on a problem solution:

cos(x-a)cos(b-x) = \frac{1}{2}[cos(a+b-2x) + cos(a-b)]

Where does this come from? I can't find it in anywhere (except for wolframalpha). Thanks.

 

[Borek]

Images don't show, please try to enter the identity using Latex (see https://www.physicsforums.com/showthread.php?t=546968).

 

[vincent_vega]

thanks for the heads up

 

[DonAntonio]

Our teacher used the following on a problem solution:

cos(x-a)cos(b-x) = \frac{1}{2}[cos(a+b-2x) + cos(a-b)]

Where does this come from? I can't find it in anywhere (except for wolframalpha). Thanks.

It comes from the known identity for sum of cosines: \cos \alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}

DonAntonio