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January 21st, 2009, 04:29 PM
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| | 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! | 
January 21st, 2009, 04:34 PM
|  | Primero Espada | | Join Date: Mar 2008 Location: Canada
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| | This looks awfully familiar to this identity:
Let  and  and the conclusion follows. | 
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. | 
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. | 
January 21st, 2009, 10:37 PM
|  | Flow Master | | Join Date: Dec 2007 Location: Zeitgeist
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| | Quote:
Originally Posted by DaCoo911 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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January 22nd, 2009, 03:09 PM
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| | hint ...
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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January 22nd, 2009, 03:13 PM
|  | vs Jhevon | | Join Date: Feb 2007 Location: New York, USA
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Originally Posted by DaCoo911 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,
apply this to the left hand side, then multiply the resulting expression by  . 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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January 22nd, 2009, 03:16 PM
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| | yes that makes sense! thanks both of you for your help! | | Thread Tools | | | | Display Modes | Linear Mode |
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