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December 21st, 2007, 04:34 PM   #1
n30
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trignometric identities.

((4sin x)(sin y))/((4cos x)(cos y)) = (tan x)(tan y)

How is that trig function equal to those tan functions?
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December 21st, 2007, 04:54 PM   #2
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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.
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December 21st, 2007, 09:28 PM   #3
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You mean the circle with equation x² + y² = r². That's not the same (position or radius) as "a unit circle".
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December 22nd, 2007, 01:09 PM   #4
n30
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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!
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December 22nd, 2007, 04:12 PM   #5
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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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