My Math Forum trignometric identities.

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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: 20,933 Thanks: 2207 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
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Quote:
 Originally Posted by johnny 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.
Thank YOU!

 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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