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In my textbook a problem is given:

For $x\in [1,\,2]$ find the range of $f(x) = \tan{x}$

I knew that lowest and highest bound is $-\infty$ and $\infty$ since $\frac{\pi}{2}$ lies in the interval $[1,\,2]$ but the interval exclude some value between $[0,\,\tan{1})$ $\rightarrow$ positive and $[0,\,\tan{2})$ $\rightarrow$ negative.

But I can't evauluate $\tan 1$ and $\tan 2$. I've tried many trigonometric substitutions, identities, etc..

At last when I resolved to the answers page the answer is ($-\infty,\,\tan2] \cup (-\infty,\,\tan2]$

And this I coerced me into slapping myself. I've wasted an hour, not expecting this kind of solution. So, can anybody can solve $\tan 1$ and $\tan 2$ analytically (not by using graphs, etc... cause they won't be provided in an exam no matter how many time books solutions reference them freely.) :(

You can't use any mathematical gizmos like Taylor series, etc... :(

^I've already googled but couldn't find any solutions.

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    $\begingroup$ There are no closed form analytic solutions for non-zero integer arguments for the trig functions, though you might want to use a series expansion. Only rational multiples of $\pi$ as arguments give algebraic results. $\endgroup$
    – nathan.j.mcdougall
    Commented Oct 22, 2016 at 19:00
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    $\begingroup$ It is not totally clear what you are asking for. Tan 1 is an irrational number and you cannot write all decimal places. The shortest form to refer to it is just to write $\tan 1$. It's similar to $\sqrt{2}$ or $\pi$; you mentioned $\pi$ in your text and somehow you don't feel a need to rewrite or rename it. It's the same with $\tan 1$. $\endgroup$
    – Peter Franek
    Commented Oct 22, 2016 at 19:09
  • $\begingroup$ @PeterFranek:A clsoe approximation will be sufficient..I don't knwo whether they will accept the solution given in book in exams or not :((in many cases they don't ) $\endgroup$
    – Xasel
    Commented Oct 22, 2016 at 19:20
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    $\begingroup$ I guess they will accept the answer $[\tan 1, \infty)\cup (-\infty, \tan 2]$. $\endgroup$
    – Peter Franek
    Commented Oct 22, 2016 at 19:22
  • $\begingroup$ Well lets hope so @PeterFranek $\endgroup$
    – Xasel
    Commented Oct 22, 2016 at 19:32

1 Answer 1

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Using Euler's formula we can get \begin{align*}\sin 1=\frac{e^i-e^{-i}}{2i} && \cos 1=\frac{e^i+e^{-i}}{2} && \tan 1=\frac{i(e^{-i}-e^i)}{e^i+e^{-i}}\end{align*}

There are no simpler closed forms for these.

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  • $\begingroup$ This is true but you're just expressing the numbers in terms of a different function. It doesn't give a neater closed form solution. $\endgroup$
    – Jam
    Commented Oct 22, 2016 at 20:17
  • $\begingroup$ I think this would atleast this would suffice $\endgroup$
    – Xasel
    Commented Oct 23, 2016 at 7:17
  • $\begingroup$ for exmaination purpose $\endgroup$
    – Xasel
    Commented Oct 23, 2016 at 7:19

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