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Thread: [SOLVED] tangent sum identity issue

  1. #1
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    [SOLVED] tangent sum identity issue

    I've been working through a tanget sum identity problem and I've mostly got it, except I've got the signs reversed. I was curious if someone could point out the mistake I made. I apologize for how long this is. All assistance is appreciated.

    Problem is
    $\displaystyle tan(A+B)$
    where
    $\displaystyle sinA = \frac{-3}{5}$
    is in QIII
    and
    $\displaystyle cosB = \frac{-1}{4}$
    is in QII

    sinA= $\displaystyle \frac{-3}{5}$ so the hypotenuse is 5, the oppose side is -3, making the adacent side 4. Thus tan(A) = $\displaystyle \frac{-3}{4}$.

    for B, cosB=$\displaystyle \frac{-1}{4}$ so the hypotenuse is 4, the adjacent side -1, making the opposite side $\displaystyle \sqrt15$. Thus tan(b) = $\displaystyle \frac{\sqrt15}{-1}$ or just $\displaystyle -\sqrt15$.

    The tangent sum identity is $\displaystyle \frac{tanA+tanB}{1-tanAtanB}$

    Substituting in, I get
    $\displaystyle \frac{\frac{-3}{4}+ -\sqrt15}{1-\frac{-3}{4}(-\sqrt15)}$

    which is
    $\displaystyle \frac{\frac{-3-4\sqrt15}{4}}{1-\frac{3\sqrt15}{4}}$
    and then
    $\displaystyle \frac{\frac{-3-4\sqrt15}{4}}{\frac{4-3\sqrt15}{4}}$
    followed by
    $\displaystyle \frac{-3-4\sqrt15}{4}*\frac{4}{4-3\sqrt15}$
    and finally
    $\displaystyle \frac{-3-4\sqrt15}{4-3\sqrt15}$

    unfortunately, the answer would appear to be
    $\displaystyle \frac{3-4\sqrt15}{4+3\sqrt15}$

    Can someone please point out my mistake?
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  2. #2
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    Quote Originally Posted by satis View Post
    I've been working through a tanget sum identity problem and I've mostly got it, except I've got the signs reversed. I was curious if someone could point out the mistake I made. I apologize for how long this is. All assistance is appreciated.

    Problem is
    $\displaystyle tan(A+B)$
    where
    $\displaystyle sinA = \frac{-3}{5}$
    is in QIII
    and
    $\displaystyle cosB = \frac{-1}{4}$
    is in QII

    sinA= $\displaystyle \frac{-3}{5}$ so the hypotenuse is 5, the oppose side is -3, making the adacent side (negative) 4. Thus tan(A) = $\displaystyle \textcolor{red}{\frac{-3}{-4}}$.

    here is your mistake ... in quad III, tangent is positive



    Can someone please point out my mistake?
    ...
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  3. #3
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    awesome, thank you very much. I seem to be making that mistake a lot. I'm just happy my methodology is sound (other than screwing up signs).
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