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December 21st, 2007, 04:34 PM  #1 
Newbie Joined: Dec 2007 Posts: 5 Thanks: 0  trignometric identities.
((4sin x)(sin y))/((4cos x)(cos y)) = (tan x)(tan y) How is that trig function equal to those tan functions? 
December 21st, 2007, 04:54 PM  #2 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  In general, tan u = sin u / cos u. Proof: In a unit circle, let sin u = y/r, and cos u = x/r. Also, we will say that tan u = y/x. We can algebraically do some division, where (y/r)/(x/r) = y/x, hence sin u / cos u = tan u. ((4sin x)(sin y))/((4cos x)(cos y)) = ((sin x)(sin y))/((cos x)(cos y)) = ((sin x)/(cos x))(sin y)/(cos y)) = (tan x)(tan y) Q. E. D. 
December 21st, 2007, 09:28 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
You mean the circle with equation x² + y² = r². That's not the same (position or radius) as "a unit circle".

December 22nd, 2007, 01:09 PM  #4  
Newbie Joined: Dec 2007 Posts: 5 Thanks: 0  Quote:
 
December 22nd, 2007, 04:12 PM  #5 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
You're welcome, and the x^2 + y^2 = r^2 should've been replaced over unit circle, so skipjack is right on that part. On the algebraic part, I think I showed you correctly, where the obvious general equation tan u = sin u / cos u comes along. 

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