Wednesday, May 18, 2011

Note about the sum of cosines formula

I was showing someone the derivation of the formulas for sums and differences of sines and cosines (my post here). Unfortunately, I have some trouble remembering these. The trick I used was to try to recall that the derivation started by analyzing the difference cos(s-t) and it's a particularly easy form:

cos(s-t) = cos s cos t + sin s sin t

Then it occurred to me that there is a fairly obvious point about this that should make it even clearer. Just remember that pattern is sine sine, cosine cosine, both terms positive.
Then suppose s = t, we have

cos2(s) + sin2(s) = 1

So, which function and for what combination of s with itself would we always get 1? Well, it's obviously the difference, which always equals zero (the sum, 2s, could be any angle). And which function always gives 1 with an argument of 0? The cosine, of course.

cos(s-t) = cos s cos t + sin s sin t

Getting to the formula for cos(s+t) just involves realizing that if we plug in u = -t we have

cos(s+u) = cos s cos(-u) + sin s sin(-u)

but

cos(-u) = cos(u)
sin(-u) = - sin(u)

So it's the sine term in the formula that changes sign when we add.

cos(s+u) = cos s cos u - sin s sin u

As for the other one, perhaps the easiest is Euler:

eis = cos s + i sin s
ei(s+t) = cos(s+t) + i sin(s+t)

ei(s+t) = eis eit
= (cos s + i sin s) (cos t + i sin t)
= cos s cos t - sin s sin t + i (sin s cos t) + i (cos s sin t)

The real part gives us what we had before,

cos(s+t) = cos s cos t - sin s sin t

and the imaginary part is equal to the imaginary part of the sum from the previous line:

i sin(s+t) = i (sin s cos t) + i (cos s sin t)
sin(s+t) = sin s cos t + cos s sin t

In fact, maybe this is enough by itself. :)

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