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

 

[CellCoree]

This trigonometric identity is known as the cosine addition formula and it can be derived using basic trigonometric identities. One way to obtain this identity is by using the double angle formula for cosine, which states that cos(2x) = cos^2(x) - sin^2(x). By substituting (x-a) for x in the double angle formula and expanding, we can obtain the first term of the given identity, cos(a+b-2x). To obtain the second term, cos(a-b), we can use the identity cos(-x) = cos(x) and substitute (-x) for (b-x) in the original identity. The final step is to use the identity cos(x)cos(y) = (1/2)[cos(x+y) + cos(x-y)] and substitute (a-b) for y to obtain the second term. Therefore, the given identity is a combination of multiple trigonometric identities and can be derived using basic algebraic manipulations.