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Thread: tan identity

  1. #1
    Member jacs's Avatar
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    tan identity

    I have this question
    Express tan (90 + x) in terms of tan x

    not really sure what to do with it, using the compound angles i have gotten this far

    $\displaystyle \frac{tan90+tanx}{1-tan90tanx}$

    = $\displaystyle \frac{\infty+tanx}{1-\infty tanx}$

    not really sure how to proceed from there, or if that is an acceptable answer, or i am going about this in completely the wrong way

    thanks
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  2. #2
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    The identity $\displaystyle \tan(a+b) = \frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$ is not valid if either $\displaystyle a$ or $\displaystyle b$ is equal to $\displaystyle \frac{\pi}{2}$.

    Why? Because if, for example, $\displaystyle a = \frac{\pi}{2}$, then $\displaystyle \tan{a} = \tan\left(\frac{\pi}{2}}\right) = \tan\left(\frac{\pi}{4}+\frac{\pi}{4}\right)$.

    Using the same identity we would get $\displaystyle \tan{a} = \frac{\tan{\frac{\pi}{4}}+\tan{\frac{\pi}{4}}}{1-\tan{\frac{\pi}{4}}\tan{\frac{\pi}{4}}}$.

    But since $\displaystyle \tan{\frac{\pi}{4}} = 1$, our denominator is zero. Therefore $\displaystyle \tan{\frac{\pi}{2}}$ is undefined.

    Don't worry though. By symmetry, or otherwise, you can get your $\displaystyle \tan\left(\frac{\pi}{2}+x\right)[/Math] to [Math]-\cot{x}.$
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  3. #3
    Member jacs's Avatar
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    I think i have figured out a way to do it without looking at the graphs, although i now see the transformation from tan x to tan (90 + x) = - cot x.

    Is this a sound approach?

    180 - (90 - x) = 90 + x
    hence
    tan (90 + x) = tan {180 - (90 - x)}
    = -tan (90 - x) (lying in the second quadrant)
    = - cot x (complementary angles)
    = $\displaystyle -\frac{1}{tanx}$

    thus expressing the original in terms of x
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  4. #4
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    Looks good to me. More neater perhaps would be to write:

    $\displaystyle \displaystyle \tan\left(\frac{\pi}{2}+x\right) = \frac{\sin(\frac{\pi}{2}+x)}{\cos(\frac{\pi}{2}+x) } [/Math] [Math] \displaystyle = \frac{\sin\frac{\pi}{2}\cos{x}+\cos\frac{\pi}{2}\s in{x}}{\cos\frac{\pi}{2}\cos{x}-\sin\frac{\pi}{2}\sin{x}} = \frac{(1)\cos{x}+(0)\sin{x}}{(0)\cos{x}-(1)\sin{x}} = -\frac{\cos{x}}{\sin{x}} = -\frac{1}{\tan{x}}.$

    Looks hot too.
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  5. #5
    Member jacs's Avatar
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    Hadn't even thought of rewriting it in terms of sin and cos, and now i see it, it is so glaringly obvious.

    Thanks so much for your help with this problem, i now have 3 distinct ways of looking at it,which is three more than i had yesterday, lol
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