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Just a small thought popped up in my mind; and now I'm stuck on it. Any idea on how to find the value of $\tan 20^\circ$? I tried doing it by using the multiple angle formulas, but I didn't get an answer... How do I proceed?

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  • $\begingroup$ Is your $20$ in degrees or radians? $\endgroup$ – Henning Makholm Jul 8 '15 at 14:15
  • $\begingroup$ It is well-known that this cannot be written in terms of square roots, this being equivalent to the fact that the regular nonagon is not constructible with ruler and compass. $\endgroup$ – Keith Jul 8 '15 at 14:15
  • $\begingroup$ Aditya, I assume you mean $20\circ$ and I have edited your question. Please correct me if I am wrong. $\endgroup$ – Thomas Jul 8 '15 at 14:19
  • $\begingroup$ Thanks Thomas,.It is degrees $\endgroup$ – Aditya Pai Jul 8 '15 at 14:28
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$20°$ is a third of $60°$, for which the value of the tangent is well known to be $\sqrt3$. Let us denote $x:=\tan(20°)$.

Then, by the addition formula,

$$\tan(40°)=\frac{2x}{1-x^2},$$ and $$\tan(60°)=\frac{x+\dfrac{2x}{1-x^2}}{1-x\dfrac{2x}{1-x^2}}=\frac{3x-x^3}{1-3x^2}=\sqrt3,$$ or $$x^3-3\sqrt3x^2-3x+\sqrt3=0.$$

It turns out that this cubic equation cannot be solved by real radicals, so you need to use numerical methods such as Newton's.

Starting from $x=\dfrac\pi9$ ($20°$ in radians), the iterates are

$0.349065850399\cdots\\ 0.364116885850\cdots\\ 0.363970248087\cdots\\ 0.363970234266\cdots\\ 0.363970234266\cdots\\ 0.363970234266\cdots\\ \cdots$

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  • $\begingroup$ You misunderstood the meaning of the Wikipedia article. Specifically, it does not mean that it cannot be solved by radicals but that it cannot be solved without roots of complex numbers. Radicals include taking roots of complex numbers. But yes in any case to numerically find such values we end up having to use something like Newton's approximation. $\endgroup$ – user21820 Jul 8 '15 at 15:51
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$\sin(20^\circ)$ and $\cos(20^\circ)$ can both be found using the triple-angle formulas. (I'm assuming your "20" was in degrees, not radians.)

Unfortunately, both of those will end up with you having to solve a cubic; you can use Cardano's formula to do that.

You can't solve it with just quadratics and algebra, for if you could, it'd be possible to trisect a 60-degree angle; but that is in fact exactly the example used generally to show that trisection is impossible, because $\cos(60^{\circ})$ is not a surd.

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  • $\begingroup$ What you say about $\cos(60^\circ)$ is correct, but a little weird since people who don't know Galois theory will not understand why it not being a surd means that trisection is impossible! $\endgroup$ – user21820 Jul 8 '15 at 14:17
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    $\begingroup$ Are you sure Cardano's formula will work? You might need to take the cube root of a complex number, which is something of at least equivalent difficulty to trisecting an angle. $\endgroup$ – Keith Jul 8 '15 at 14:18
  • $\begingroup$ @Keith: It's generally allowed to take roots of complex numbers, otherwise there is no closed form for the roots of those cubics. $\endgroup$ – user21820 Jul 8 '15 at 14:19
  • $\begingroup$ You can also write down the cubic directly; $\tan 20^\circ$ is the positive $x$ such that the real part of $(1+ix)^3$ is $\frac12(1+x^2)^{3/2}$, which simplifies to a cubic equation in $x^2$. $\endgroup$ – Henning Makholm Jul 8 '15 at 14:21
  • $\begingroup$ @Keith: And by the way, taking cube roots of a complex number is not at all equivalent in power to trisection. Trisection is equivalent to finding roots of irreducible cubics in general, including $\sqrt[3]{2}$. $\endgroup$ – user21820 Jul 8 '15 at 14:23
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We have that $\tan 20^{\circ} = \dfrac{\sin 20^{\circ}}{\cos 20^{\circ}} = \dfrac{\sin 20^{\circ}}{\sin 70^{\circ}}$.

This list here, gives you the exact value of the sine of every integer angle between $1$ and $90$. This allows you to compute $\tan 20^{\circ}$.


Whilst this section doesn't answer your question the way you want it, it does prove an alternative from brutally disgusting surds or horrible calculators.

You can construct a $20$-$70$-$90$ triangle yourself using a protractor and a ruler. Then take the ratio of the sides to get $\tan 20^{\circ}$. This won't be accurate, but you may find it amusing to do. :-)

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    $\begingroup$ Haha your linked pdf is amusing. I think he just churned it out using some computer algebra system. $\endgroup$ – user21820 Jul 8 '15 at 14:26
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    $\begingroup$ @user21820 - He most certainly did, that would explain the horrid formatting. ;-) $\endgroup$ – Zain Patel Jul 8 '15 at 14:27
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    $\begingroup$ Though the list provides closed forms for all angles multiple of $1°$, when the angle isn't a multiple of $3°$, cubic roots of complex numbers appear. To effectively compute these, you need... trigonometric functions, radicals aren't enough, and the solution is somewhat circular. $\endgroup$ – Yves Daoust Jul 8 '15 at 14:55
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    $\begingroup$ @YvesDaoust the $I$ that appears is the complex $i$ so. $\endgroup$ – snulty Jul 8 '15 at 15:09
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    $\begingroup$ @YvesDaoust nothing, just it took me a minute to realise it when I was looking at the linked pdf! Just weird formatting :P $\endgroup$ – snulty Jul 11 '15 at 19:29
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$\cos(3t) = 4 \cos(t)^3 - 3 \cos(t)$

$\sin(3t) = 3 \sin(t) - 4 \sin(t)^3$.

Substituting $t = 20^\circ$ gives cubics that can be solved in closed form if you allow the operation of taking cube roots.

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I'm assuming you mean via multiple angle formulas, and I'm assuming you want an exact answer. The simple answer is, it would require a ridiculous amount of work. You can write:

$$ \tan \left(45-30+5\right)$$

You could perform the double angle formula twice, and reduce the problem to finding tan(5), which still isn't easily solvable. Heres a fun site to give you an idea of how bad these things can get :)

http://www.jdawiseman.com/papers/easymath/surds_tan.html

Note that if you are happy with an approximate solution, you should look into the CORDIC algorithm, which is how many hand-held calculators still do trigonometry :P

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  • $\begingroup$ Yeah, I get it.... Complex equations. Any generalisation for finding roots of a cubic equation? (If there is any) $\endgroup$ – Aditya Pai Jul 8 '15 at 14:33
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Since, obviously, numerical calculations would be required, I cannot resist the pleasure of reusing a 1400 years old approximation proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}$$ A similar one $$\cos(x)\simeq \frac{\pi^2-4x^2}{x^2+\pi^2}$$ Applied to $x=\frac \pi 9$, this leads to $$\tan(\frac \pi 9)\simeq \frac{10496}{28721} \approx 0.365447$$ Using this as a starting point $x_0$ for Newton method, as Yves Daoust proposed, the iterative scheme will converge in a couple of iterations $$x_1 \approx 0.363971632203$$ $$x_2\approx0.363970234267$$ $$x_3\approx0.363970234266$$.

Another solution would be to use Taylor expansion around $x=a$ $$\tan(x)=\tan (a)+(x-a) \left(\tan ^2(a)+1\right)+(x-a)^2 \left(\tan ^3(a)+\tan (a)\right)+$$ $$(x-a)^3 \left(\tan ^4(a)+\frac{4 \tan ^2(a)}{3}+\frac{1}{3}\right)+$$ $$(x-a)^4 \left(\tan ^5(a)+\frac{5 \tan ^3(a)}{3}+\frac{2 \tan (a)}{3}\right)+O\left((x-a)^5\right)$$ and to use $a=\frac \pi 8$ for which the tangent is $\sqrt 2 -1$ ( easily obtained using the double angle formula). Using one term, the result is $\approx 0.363094$; with two terms $\approx 0.364018$; with three terms $\approx 0.363969$; with four terms $\approx 0.363970$.

Another solution is based on Pade approximants. The simplest would be $$\tan(x)\simeq \frac{(x-a)+\tan (a)}{1-(x-a) \tan (a)}$$ Using $a=\frac \pi 8$ would give $\approx 0.364002$.

The next Pade approximant would be $$\tan(x)\simeq \frac{(x-a)+\tan(a)-\frac{1}{3} (x-a)^2 \tan (a)}{1-(x-a) \tan (a)-\frac{1}{3} (x-a)^2}$$ which would give $\approx 0.363970238$.

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