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Math Help - I need help evaluating an expression under specific conditions

  1. #1
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    I need help evaluating an expression under specific conditions



    Evaluate the expression under the given conditions.tan(θ + ); cos θ = −1/3, θ in Quadrant III, sin = 1/4, in Quadrant 4



    Copied from my math homework, I'm not getting the right answer.

    Last edited by AKPunk49; April 15th 2013 at 07:30 PM.
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    Re: I need help evaluating an expression under specific conditions

    Hey AKPunk49.

    Can you show us what you tried: (Hint: did you try using the angle addition formulae?)?
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    Re: I need help evaluating an expression under specific conditions

    \displaystyle \sin{(\phi)} can not possibly be \displaystyle \frac{1}{4} in the fourth quadrant. Why?
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    Re: I need help evaluating an expression under specific conditions

    Fixed it, sorry.
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    Re: I need help evaluating an expression under specific conditions

    It doesn't change the fact that \displaystyle \sin{(\phi)} can not possibly be positive in the fourth quadrant. In fact, I don't see anything different at all...
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    Re: I need help evaluating an expression under specific conditions

    Oh my-lanta...its in Quad. 2. I'm having an odd week, sorry for that.

    The book tells me to make two right triangles in the corresponding quads and then find the identities, which I did. Then I plugged it all into the addition formula for tangent. I got -sr8/3+1/4 all divided by 1-(-sr8/3)(1/4)

    Sorry if that's confusing...
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    Re: I need help evaluating an expression under specific conditions

    Quote Originally Posted by chiro View Post
    Hey AKPunk49.

    Can you show us what you tried: (Hint: did you try using the angle addition formulae?)?
    I did, I don't know if I'm not doing it correctly or if I just don't know how to properly reduce.
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    Re: I need help evaluating an expression under specific conditions

    Hello, AKPunk49!

    \text{Given: }\:\cos\theta = -\tfrac{1}{3}\,\text{ in Quadrant 3},\,\sin\phi = \tfrac{1}{4}\,\text{ in Quadrant 2}

    \text{Find: }\:\tan(\theta + \phi)

    We are given: . \cos\theta \,=\, \frac{\text{-}1}{3}\,=\,\frac{adj}{hyp} in Quad 3.
    . . Then: . opp = \text{-}\sqrt{8} \quad\Rightarrow\quad \tan\theta \,=\, \frac{opp}{adj} \,=\,\frac{\text{-}\sqrt{8}}{\text{-}1}
    . . \tan\theta \,=\,\sqrt{8}

    We are given: . \sin\phi \,=\,\frac{1}{4} \,=\,\frac{opp}{hyp}
    . . Then: . adj \,=\,\text{-}\sqrt{15} \quad\Rightarrow\quad \tan\phi \,=\,\frac{opp}{adj} \,=\,\frac{1}{\text{-}\sqrt{15}}
    . . \tan\phi \,=\,-\frac{1}{\sqrt{15}}

    We have: . \tan(\theta + \phi) \:=\:\frac{\tan\theta + \tan\phi}{1-\tan\theta\tan\phi}

    . . . . . . . . . . . . . . . . =\;\frac{\sqrt{8} + \left(\text{-}\frac{1}{\sqrt{15}}\right)}{1 - \sqrt{8}\left(-\tfrac{1}{\sqrt{15}}\right)} \;=\;\frac{\sqrt{8} - \frac{1}{\sqrt{15}}}{1 + \frac{\sqrt{8}}{\sqrt{15}}}

    Multiply by \frac{\sqrt{15}}{\sqrt{15}}\!:\;\;\frac{\sqrt{120} - 1}{\sqrt{15} + \sqrt{8}}

    Therefore: . \tan(\theta + \phi) \;=\;\frac{2\sqrt{30} - 1}{\sqrt{15} - 2\sqrt{2}}
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    Re: I need help evaluating an expression under specific conditions

    Quote Originally Posted by Soroban View Post
    Hello, AKPunk49!


    We are given: . \cos\theta \,=\, \frac{\text{-}1}{3}\,=\,\frac{adj}{hyp} in Quad 3.
    . . Then: . opp = \text{-}\sqrt{8} \quad\Rightarrow\quad \tan\theta \,=\, \frac{opp}{adj} \,=\,\frac{\text{-}\sqrt{8}}{\text{-}1}
    . . \tan\theta \,=\,\sqrt{8}

    We are given: . \sin\phi \,=\,\frac{1}{4} \,=\,\frac{opp}{hyp}
    . . Then: . adj \,=\,\text{-}\sqrt{15} \quad\Rightarrow\quad \tan\phi \,=\,\frac{opp}{adj} \,=\,\frac{1}{\text{-}\sqrt{15}}
    . . \tan\phi \,=\,-\frac{1}{\sqrt{15}}

    We have: . \tan(\theta + \phi) \:=\:\frac{\tan\theta + \tan\phi}{1-\tan\theta\tan\phi}

    . . . . . . . . . . . . . . . . =\;\frac{\sqrt{8} + \left(\text{-}\frac{1}{\sqrt{15}}\right)}{1 - \sqrt{8}\left(-\tfrac{1}{\sqrt{15}}\right)} \;=\;\frac{\sqrt{8} - \frac{1}{\sqrt{15}}}{1 + \frac{\sqrt{8}}{\sqrt{15}}}

    Multiply by \frac{\sqrt{15}}{\sqrt{15}}\!:\;\;\frac{\sqrt{120} - 1}{\sqrt{15} + \sqrt{8}}

    Therefore: . \tan(\theta + \phi) \;=\;\frac{2\sqrt{30} - 1}{\sqrt{15} - 2\sqrt{2}}
    which of course you should then simplify...

    \displaystyle \begin{align*} \frac{2\sqrt{30} - 1}{\sqrt{15} - 2\sqrt{2}} &= \frac{\left( 2\sqrt{30} - 1 \right) \left( \sqrt{15} + 2\sqrt{2} \right) }{ \left( \sqrt{15} - 2\sqrt{2} \right) \left( \sqrt{15} + 2\sqrt{2} \right) } \\ &= \frac{2\sqrt{15 \cdot 30} + 4\sqrt{ 2 \cdot 30 } - \sqrt{15} - 2\sqrt{2} }{15 - 8} \\ &= \frac{2\sqrt{225 \cdot 2} + 4\sqrt{4 \cdot 15} - \sqrt{15} - 2\sqrt{2}}{7} \\ &= \frac{2 \cdot 15\sqrt{2} + 4\cdot 2 \sqrt{15} - \sqrt{15} - 2\sqrt{2}}{7} \\ &= \frac{28\sqrt{2} + 7\sqrt{15} }{7} \\ &= 4\sqrt{2} + \sqrt{15} \end{align*}
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