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Date: Friday, January 24, 1997 7:45 PM
From: Alex Bogomolny
Dear Ronald:
Regardless of where you live, I am quite uncomfortable with the manner of your question. I'd be much happier if you explained the nature of your difficulty. Say, you tried to solve the problem in such-and-such way but got stuck at such-and-such point. Following is a solution. It's up to you to learn something from it.
Two formulas are needed for the reference sake:
- cos(a-b)=cos(a)*cos(b)+sin(a)*sin(b),
- cos(2a)=cos(a)2-sin(a)2
The latter is of course a particular case of the former. Now, multiply both sides of the identity by cos(a)*sin(a). After simplifications, you'll get
cos(3a)*cos(a)+sin(3a)*sin(a)=
(cos(a)/sin(a)-sin(a)/cos(a))*cos(a)*sin(a)
or, from 1.,
cos(3a-a)=cos(a)*cos(a)-sin(a)*sin(a)
which is true from 2.
Regards.
Copyright © 1996-2008 Alexander Bogomolny
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