I discovered some trig identities this week; as they don't look even a little bit familiar, I assume they didn't come up when I was paying attention in highschool.
In short: for any product consisting of sines and cosines: if there are an odd number of sines, the expression can be rewritten as a sum of sines. If there are an even number of sines, the expression can be rewritten as a sum of cosines.
You can get a quick sense of how this might kind of work out by considering the exponential expression for sign, which has a factor i in the denominator. An odd number of sines in the product produces an odd number of i factors, producing an imaginary number. To get real numbers from this, you'll need to pop one of the i's off again by creating sine terms.
It's all pretty straight forward to do with exponentials and a little bit of induction; or you can demonstrate it using the more common identities....
cos(A)cos(B)
= (1/2) * 2 cos(A)cos(B)
= (1/2) * ( 2 cos(A)cos(B) - sin(A)sin(B) + sin(A)sin(B) )
= (1/2) * ( cos(A)cos(B) - sin(A)sin(B) + cos(A)cos(B) + sin(A)sin(B) )
= (1/2) * ( cos(A)cos(B) - sin(A)sin(B) + cos(A)cos(-B) - sin(A)sin(-B))
= (1/2) * ( cos(A+B) + cos (A-B) )
sin(A)sin(B)
= (1/2) * 2 sin(A)sin(B)
= (1/2) * ( 2 sin(A)sin(B) - cos(A)cos(B) + cos(A)cos(B) )
= (1/2) * ( sin(A)sin(B) - cos(A)cos(B) + cos(A)cos(B) + sin(A)sin(B))
= (1/2) * ( sin(A)sin(B) - cos(A)cos(B) + cos(A)cos(-B) - sin(A)sin(-B))
= (1/2) * ( - (cos(A)cos(B) - sin(A)sin(B)) + (cos(A)cos(-B) - sin(A)sin(-B)))
= (1/2) * ( cos(A-B) - cos(A+B) )
sin(A)cos(B)
= (1/2) * 2 sin(A)cos(B)
= (1/2) * ( 2 * sin(A)cos(B) + cos(A)sin(B) - cos(A)sin(B) )
= (1/2) * ( sin(A)cos(B) + cos(A)sin(B) + sin(A)cos(B) - cos(A)sin(B))
= (1/2) * ( sin(A)cos(B) + cos(A)sin(B) + sin(A)cos(-B) + cos(A)sin(-B))
= (1/2) * (sin(A+B) + sin(A-B))
September 10, 2005 2:18 PM
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