Deriving Trigonometric IdentitiesDate: 07/20/2008 at 22:50:58 From: Andy Subject: Trigonometry | Identities How do you derive the sum-to-product and product-to-sum identities in trigonometry? I just want to know how you derive the formulas, because I don't want to memorize them--I hate rote memorization >:( Date: 07/21/2008 at 09:42:49 From: Doctor Ali Subject: Re: Trigonometry | Identities Hi Andy! Thanks for writing to Dr. Math. Since the method is exactly the same in all of the proofs, let me guide you through proving one of them. Then you can prove the others yourself! They actually come from the sum formulas. We know that: Sin(a + b) = Sin(a) Cos(b) + Cos(a) Sin(b) Cos(a + b) = Cos(a) Cos(b) - Sin(a) Sin(b) Let's use the first one. Sin(a + b) = Sin(a) Cos(b) + Cos(a) Sin(b) We can replace b with -b. This gives: Sin(a - b) = Sin(a) Cos(b) - Cos(a) Sin(b) Let's add these two equalities together: Sin(a + b) + Sin(a - b) = 2 Sin(a) Cos(b) Or, 1 / \ Sin(a) Cos(b) = --- | Sin(a + b) + Sin(a - b) | 2 \ / And that is the product-to-sum formula for Sin(a) Cos(b). We know that this equality holds for all a and b. We can assume x + y a = ------- 2 x - y b = ------- 2 You might ask why I made these change of variables. This is mainly because I was looking for two expressions whose sum is x and whose difference is y. Does that make sense? Now, if you put these expressions in for a and b and do a little simplification, you'll have, x + y x - y Sin(x) + Sin(y) = 2 Sin(-------) Cos(-------) 2 2 Here is the sum-to-product formula. Does that make sense? Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ |
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