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Math Help - Trig Identity

  1. #1
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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. #2
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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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  3. #3
    Senior Member mollymcf2009's Avatar
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    Do you just need to simplify this using trig identities? Not sure what your question is.
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  4. #4
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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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  5. #5
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    Quote 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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  6. #6
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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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    is up to his old tricks again! Jhevon's Avatar
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    Quote 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?
    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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  8. #8
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    Talking

    yes that makes sense! thanks both of you for your help!
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