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Math Help - Please find exact value

  1. #1
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    Please find exact value

    Please find tan(B+C) given that sin C= \frac {1}{4}, with C in quadrant 2, and sin B= \frac {-1}{2}, with B in quadrant 4.
    The correct answer is \frac {4 \sqrt{3} + \sqrt{15}}{-11}

    I have fount that tan C= \frac{1}{- \sqrt{15}} and tan B= \frac {-1}{\sqrt{3}}
    \frac {-1}{\sqrt{3}}+ \frac{1}{-\sqrt{15}} over 1- \frac{-1}{\sqrt{3}} (\frac{1}{-\sqrt{15}})
    Is this correct? What do I do next to get the correct answer?

    Thank you
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oceanmd View Post
    Please find tan(B+C) given that sin C= \frac {1}{4}, with C in quadrant 2, and sin B= \frac {-1}{2}, with B in quadrant 4.
    The correct answer is \frac {4 \sqrt{3} + \sqrt{15}}{-11}

    I have fount that tan C= \frac{1}{- \sqrt{15}} and tan B= \frac {-1}{\sqrt{3}}
    \frac {-1}{\sqrt{3}}+ \frac{1}{-\sqrt{15}} over 1- \frac{-1}{\sqrt{3}} (\frac{1}{-\sqrt{15}})
    Is this correct? What do I do next to get the correct answer?

    Thank you
    umm...you don't have a lot of options here. combine the fractions in the numerator and denominator and simplify (note: you may need to conjugate). for the record, i used a different formula and got the answer, it's still a pain to simplify though. i always have trouble remembering the addition formula for tangent, so i used \tan (B + C) = \frac {\sin (B + C)}{\cos (B + C)}. but i'm sure you'll get the answer doing it your way. your values for tanB and tanC are correct
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  3. #3
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    Hello, oceanmd!

    You're doing great !!


    Find \tan(B+C), given that: \sin C = \frac {1}{4}\text{, with C in quadrant 2}
    and \sin B = -\frac {1}{2}\text{, with B in quadrant 4.}

    The correct answer is: . \frac {4 \sqrt{3} + \sqrt{15}}{-11}
    UGH! . . . Who writes fractions that way? . {\color{blue}-\,\frac{4\sqrt{3}+\sqrt{15}}{11}}

    I have found that: . \tan C\:=\:-\frac{1}{\sqrt{15}}\:\text{ and }\:\tan B\:=\:-\frac {1}{\sqrt{3}}

    Then: . \frac{-\frac {1}{\sqrt{3}} -\frac{1}{\sqrt{15}}}{1 - \left(-\frac{1}{\sqrt{3}}\right)\left(-\frac{1}{\sqrt{15}}\right)}

    Is this correct? . . . . Yes!
    What do I do next to get the correct answer? . . . . Simplify!

    We have: . -\,\frac{\dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{15}}}{1 - \dfrac{1}{\sqrt{3}\sqrt{15}}}


    Multiply top and bottom by \sqrt{3}\sqrt{15}

    . . -\,\frac{\dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{15}}}{1 - \dfrac{1}{\sqrt{3}\sqrt{15}}}\,\cdot\,\frac{\sqrt{  3}\sqrt{15}}{\sqrt{3}\sqrt{15}} \;=\;-\,\frac{\sqrt{15} + \sqrt{3}} {\sqrt{3}\sqrt{15} - 1} \;=\;-\,\frac{\sqrt{15} + \sqrt{3}}{\sqrt{45} - 1} \;=\;-\,\frac{\sqrt{15}+\sqrt{3}}{3\sqrt{5}-1}


    Rationalize: .multiply top and botttom by (3\sqrt{5} + 1)

    -\frac{\sqrt{15}+\sqrt{3}}{3\sqrt{5} - 1}\,\cdot\,\frac{3\sqrt{5} + 1}{3\sqrt{5}+1} \;=\;-\,\frac{(\sqrt{15})(3\sqrt{5}) + (\sqrt{15})(1) + (\sqrt{3})(3\sqrt{5}) + (\sqrt{3})(1)}{(3\sqrt{5})^2 - 1^2}

    . . = \;-\,\frac{{\color{red}3\sqrt{75}} +\sqrt{15} +3\sqrt{15} + \sqrt{3}}{45 - 1} \;=\;-\,\frac{{\color{red}15\sqrt{3}} + \sqrt{15} + 3\sqrt{15} + \sqrt{3}}{44}\;\;{\color{red}**}

    . . = \;-\,\frac{16\sqrt{3} + 4\sqrt{15}}{44} \;=\;-\,\frac{4(4\sqrt{3} + \sqrt{15})}{44} \;=\;\boxed{-\,\frac{4\sqrt{3} + \sqrt{15}}{11}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    ** . 3\!\cdot\!\sqrt{75} \:=\:3\!\cdot\!\sqrt{25\cdot3} \;=\;3\!\cdot\!\sqrt{25}\!\cdot\!\sqrt{3} \;=\;3\!\cdot\!5\!\cdot\!\sqrt{3} \;=\;15\sqrt{3}
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  4. #4
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    Thank you very much.
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