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What is the simplest way to solve:

$\tan{a}=1/2$

I know from Mathematica that the answer is around 0.463648, but how can I achieve this result using only pencil and paper?

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    $\begingroup$ What is the range for $a$? $\endgroup$ – Element118 Jan 10 '16 at 0:22
  • $\begingroup$ Locate an old-fashioned table. Sines are the most likely to be available. $\endgroup$ – André Nicolas Jan 10 '16 at 0:24
  • $\begingroup$ You can draw the triangle and look at a protractor. Or you can use the power series of arctan to estimate $arctan(1/2)$. $\endgroup$ – Ahmed S. Attaalla Jan 10 '16 at 1:54
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There's not exactly a nice form, other than $a = \arctan \frac 1 2$. If you prefer some alternate expressions, you can write this as an inverse sine as well; since $2 \sin a = \cos a$, the Pythagorean identity implies that

$$1 = \sin^2 a + \cos^2 a = 5 \sin^2 a$$

so that $\sin a = \pm \frac 1 {\sqrt 5}$.

You can interpret $a$ as being the small angle in a right triangle whose legs have length $1$ and $2$.

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  • $\begingroup$ But without tables ... are we supposed to know what the angle of such a triangle is supposed to be? I'm not sure that I do. Nice answer though. $\endgroup$ – fleablood Jan 10 '16 at 1:12
  • $\begingroup$ @fleablood Indeed, and I must admit I'm a little disappointed that such a simply represented angle doesn't have some nice representation in terms of $\pi$. Thanks. $\endgroup$ – user296602 Jan 10 '16 at 1:17
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hint: $\tan a = \dfrac{1}{2} \Rightarrow \cos a = 2\sin a\Rightarrow \cos^2 a = 4\sin^2 a = 4(1-\cos^2a) \Rightarrow 5\cos^2 a = 4\Rightarrow \cos a = \pm \dfrac{2}{\sqrt{5}}$. From this you can look up $\arccos \dfrac{2}{\sqrt{5}}$ and take it from there.

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Just as Ahmed S. Attaalla commented, the problem is simple if you write $$\tan(a)=\frac 12 \implies a=\tan^{-1}(\frac 12)$$ Now, remebering Taylor series $$\tan^{-1}(x)=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+O\left(x^9\right)$$ you can get a very accurate solution (using the four terms given above, you have six significant figures since $\frac{6229}{13440}\approx 0.46346726$, the exact value being $\approx0.46364761$).

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