Date: 12/10/97 at 00:31:03 From: chris henderson Subject: Trigonometric identities How do you prove sin(t+s)sin(t-s) = sin^2t-sin^2s ?
Date: 12/10/97 at 01:51:36 From: Doctor Luis Subject: Re: Trigonometric identities Prove the identity: sin(t+s)sin(t-s) = sin^2t-sin^2s One way to approach this problem is by factoring the righthand side like this: (sin(t))^2-(sin(s))^2 = (sin(t)+sin(s))(sin(t)-sin(s)) In order to simplify this, use the trigonometric addition identities: sin(a+b) = sin(a)cos(b) + sin(b)cos(a) sin(a-b) = sin(a)cos(b) - sin(b)cos(a) (the last equation is easily derived by substituting -b for b in the previous equation). This means that sin(a+b) + sin(a-b) = 2sin(a)cos(b) sin(a+b) - sin(a-b) = 2sin(b)cos(a) which is the same thing we have on the righthand side of the identity we are trying to prove, except that we have t = a+b s = a-b Now (through a temporary change of variables, to help us visualize better what's going on) the righthand side becomes (sin(t)+sin(s))(sin(t)-sin(s)) = (sin(a+b)+sin(a-b))(sin(a+b)-sin(a-b)) renaming variables = (2sin(a)cos(b))(2sin(b)cos(a)) by the identity derived above = (2sin(a)cos(a))(2sin(b)cos(b)) rearranging factors = sin(2a)sin(2b) using the fact that sin(2M)=2sin(M)cos(M) = sin(t+s)sin(t-s) by definition, t+s = 2a and t-s = 2b and this establishes the identity! If you have any question regarding the method of proving identities feel free to ask again. -Doctor Luis, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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