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  #1  
Old January 21st, 2009, 04:29 PM
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Exclamation Trig Identity

HELLLLLP! lol i've tried for soo long now. I'm still getting used to these things...they're just like a PUZZLE! thanks!


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  #2  
Old January 21st, 2009, 04:34 PM
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This looks awfully familiar to this identity: \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}

Let a = x - y and b = y and the conclusion follows.
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Old January 21st, 2009, 04:38 PM
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Do you just need to simplify this using trig identities? Not sure what your question is.
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Old January 21st, 2009, 04:54 PM
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ya its just a matter of expanding the left side and simplyfiying down to tanx, i just can't find a way to do it.
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Old January 21st, 2009, 10:37 PM
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Originally Posted by DaCoo911 View Post
ya its just a matter of expanding the left side and simplyfiying down to tanx, i just can't find a way to do it.
Post #2 tells you exactly how to do it. If you're still stuck please say where.
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Old January 22nd, 2009, 03:09 PM
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hint ...

\frac{\tan{a} + \tan{b}}{1 - \tan{a}\tan{b}} = \tan(a+b)

Last edited by mr fantastic; January 22nd, 2009 at 07:07 PM. Reason: No edit - just flagging the reply as having been moved from a duplicate post.
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Old January 22nd, 2009, 03:13 PM
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Originally Posted by DaCoo911 View Post
I have this identity in which i have to prove. Its supposed to be done by means of expanding the left side using sum and difference formulas and then simplifying down to tan x. Just can;t seem to figure it out. Thanks

recall, \tan (x - y) = \frac {\tan x - \tan y}{1 + \tan x \tan y}

apply this to the left hand side, then multiply the resulting expression by \frac {1 + \tan x \tan y}{1 + \tan x \tan y}. can you take it from there?
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Last edited by mr fantastic; January 22nd, 2009 at 07:07 PM. Reason: No edit - just flagging the reply as having been moved from a duplicate post.
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Old January 22nd, 2009, 03:16 PM
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yes that makes sense! thanks both of you for your help!
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trig identities prove, trigo identities, trigonometry

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