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Thread: Trig problem

  1. #1
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    Trig problem

    Alright, so I've got sin(5/13) and I've tan(-sqrt3)
    Find tan (alpha-beta), please help!!! for exact value
    I'm instructed to use tan(a+b)=tanx-tany/1+tanx(tany), and utilising properties of triangles I can resolve that sin(5/13) has a tan of (5/12)
    /I plug it in to my equation to get [(5/13)+(-sqrt3)]/[1+(5/12)(-sqrt3)].. man I'm lost
    Last edited by EngineeringGuy; Nov 19th 2017 at 04:18 PM.
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  2. #2
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    Re: Trig problem

    it would be easier to answer this if it made a lick of sense...

    wth are alpha and beta?
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  3. #3
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    Re: Trig problem

    sin is alpha, tan is beta
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  4. #4
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    Re: Trig problem

    Quote Originally Posted by EngineeringGuy View Post
    sin is alpha, tan is beta
    Do you mean $\sin{\alpha} = \dfrac{5}{13}$ and $\tan{\beta} =-\sqrt{3}$. ?

    If so, is $\alpha$ in quadrant I or II ? Is $\beta$ in quadrant II or IV ?
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  5. #5
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    Re: Trig problem

    bingo, and sine alpha is such that -3pi/2 <a<-pi, and tan beta is pi/2<B<pi
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  6. #6
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    Re: Trig problem

    Quote Originally Posted by EngineeringGuy View Post
    bingo, and sine alpha is such that -3pi/2 <a<-pi, and tan beta is pi/2<B<pi
    That means both $\alpha$ and $\beta$ are quadrant II angles, which also indicates $\tan{\alpha} < 0$

    use the difference identity for tangent ...

    $\tan(\alpha - \beta) = \dfrac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}$


    In future, please provide the problem statement with all given information in your initial post. Thank you for your cooperation.
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    Re: Trig problem

    yes thanks and will do but that isn't answering my question really, I'm getting stuck with the operations
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  8. #8
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    Re: Trig problem

    Quote Originally Posted by EngineeringGuy View Post
    yes thanks and will do but that isn't answering my question really, I'm getting stuck with the operations
    in quadrant II, $\sin{\alpha} = \dfrac{5}{13} \implies \tan{\alpha} = -\dfrac{5}{12}$

    you now have values for both $\tan{\alpha}$ and $\tan{\beta}$ ... use the given identity and calculate the value of $\tan(\alpha-\beta)$
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  9. #9
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    Re: Trig problem

    yeah... the answer in the back of the book is -240+169sqrt3/69, I get to -5+12sqrt3/12+5sqrt3 and don't know what to do
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  10. #10
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    Re: Trig problem

    Quote Originally Posted by EngineeringGuy View Post
    yeah... the answer in the back of the book is -240+169sqrt3/69, I get to -5+12sqrt3/12+5sqrt3 and don't know what to do
    rationalize the denominator ...

    $\dfrac{12\sqrt{3}-5}{5\sqrt{3}+12} \cdot \dfrac{5\sqrt{3}-12}{5\sqrt{3}-12}$



    also ...
    (-5+12sqrt3)/(12+5sqrt3)
    ... parentheses around the terms in the numerator and denominator in your rational expressions
    Thanks from EngineeringGuy
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  11. #11
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    Re: Trig problem

    that damn conjugate! thank you so much
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