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Trigonometric Identities

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(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
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Associated Topics:
High School Trigonometry

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