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Math Help - [SOLVED] Need help with Trig Sum Formulas Question

  1. #1
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    Exclamation [SOLVED] Need help with Trig Sum Formulas Question

    Can someone help me solve this:

    tan(a-b)

    cosa = -4/5 Lies in Quadrant 3

    sinb = 1/3 Lies in Quadrant 1
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  2. #2
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    Quote Originally Posted by gmorrow2 View Post
    Can someone help me solve this:

    tan(a-b)

    cosa = -4/5 Lies in Quadrant 3

    sinb = 1/3 Lies in Quadrant 1

    for angle a , sketch a reference triangle in quad III

    adjacent side = -4 , hypotenuse = 5 , opposite side = ?

    \tan{a} = \frac{opp}{adj}


    for angle b , sketch a reference triangle in quad I

    opposite side = 1 , hypotenuse = 3 , adjacent side = ?

    \tan{b} = \frac{opp}{adj}


    lastly, calculate \tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}
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  3. #3
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    I did all the math already, I guess I should of posted the answer I got (which is why I'm asking for help) to see if I got it right


    Here is what I got:


    <br /> <br />
\tan(a-b)=\frac{3-8\sqrt{2}}{4-6\sqrt{2}}<br /> <br />


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  4. #4
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    Quote Originally Posted by gmorrow2 View Post
    I did all the math already, I guess I should of posted the answer I got (which is why I'm asking for help) to see if I got it right


    Here is what I got:


    <br /> <br />
\tan(a-b)=\frac{3-8\sqrt{2}}{4-6\sqrt{2}}<br /> <br />


    Show your calculations of tan(a), tan(b) and tan(a-b)
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  5. #5
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    <br />
\tan{a} = \frac{3}{4}<br />

    <br />
\tan{b} = \frac{\sqrt{8}}{8}<br />

    The answer I posted above is the overall calculations. The radicals are messing with me I just need to know if I did them right.
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  6. #6
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    note that \frac{\sqrt{8}}{8} = \frac{2\sqrt{2}}{8} =  \frac{\sqrt{2}}{4}

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

    \tan(a-b)=\frac{\frac{3}{4}-\frac{\sqrt{2}}{4}}{1+\frac{3\sqrt{2}}{16}}

    multiply by \frac{16}{16} to clear the fractions ...

    \tan(a-b) = \frac{4(3-\sqrt{2})}{16+3\sqrt{2}}
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  7. #7
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    AHH that's what I was missing the radicals always get me.


    Thanks, I'll redo it and see if I get the same thing, Thanks!
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