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Imagining a scalene triangle with sides $\sin(x), \cos(x)$ and $\tan(x)$, how would you find angle $x$ if it was between $\cos(x)$ and $\sin(x)$ when $0<x<\frac{\pi}{2}$?

I tried using the law of cosines but it lead nowhere and honestly haven't gone very far. $$\cos(x)=\frac{\sin^2(x)+\cos^2(x)-\tan^2(x)}{2\sin(x)\cos(x)}\\ \cos(x)=\frac{1-\tan^2(x)}{\sin(2x)}$$

Math
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Tiago André
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    Please show the work you did using the Law of Cosines. (That certainly seems the appropriate way to approach the problem.) Someone may be able to easily identify a simple algebraic or conceptual error, or suggest a next step, without having to duplicate your effort. ... Since comments are easily overlooked, edit your question to add details. – Blue Dec 15 '19 at 16:35
  • We must have $x\le 54.717469^{\circ}$ – The Demonix _ Hermit Dec 15 '19 at 16:41
  • Your work is a good start! Rather than combining $2\sin x\cos x$ into $\sin 2x$ (a valid operation), I cleared the fractions and wrote everything in terms of $\sin x$. The result isn't pretty, however. – Blue Dec 15 '19 at 17:06
  • I guess simplifying $tan^2(x)$ into $\frac{sin^2(x)}{cos^2(x)}$ could help, but I think it becomes too messy and doesn't add anything to the process so I omitted it. – Tiago André Dec 15 '19 at 17:08
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    At some point, you want to solving for something, so you'll need to write the equation in terms of a single something. Upon clearing fractions (and re-writing $\tan^2x$), you find yourself with even powers of cosines and odd powers of sine; this suggests writing everything in terms of $\sin x$. I agree that it's messy, though. I got a quintic equation that had to be solved numerically ... but perhaps I made an error somewhere. – Blue Dec 15 '19 at 17:19
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    @Blue The same here – Reinhard Meier Dec 15 '19 at 17:21
  • Numerically it's $x \approx 0.362803$ which is about $20.7871^\circ$. – Michael Hoppe Dec 15 '19 at 21:17
  • Where you write "how would you find angle $x$ if it was between $\cos(x)$ and $\sin(x)$", are you saying that either $\cos(x) < x < \sin(x)$ or $\sin(x) < x < \cos(x)$ must be true? – Varun Vejalla Dec 16 '19 at 01:08
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    @automaticallyGenerated No, I'm referring to the physical position of the angle as it's between the sides $\sin(x)$ and $\cos(x)$. – Tiago André Dec 16 '19 at 01:15

3 Answers3

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From the law of cosines, we have that $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ In this case, we have that $c = \tan(x), a = \cos(x), b = \sin(x)$, and $C = x$.

Plugging those in, we get $$\tan^2(x) = \cos^2(x) + \sin^2(x) - 2\cos(x)\sin(x)\cos(x)$$

After making the simplification $\cos^2(x) + \sin^2(x) = 1$ and multiplying by $\cos^2(x)$, the result is $$\sin^2(x) = \cos^2(x)-2\cos^4(x)\sin(x)$$

Making the substitution $u = \sin(x)$, we get $$2u^{5}-4u^{3}+2u^{2}+2u-1=0$$

This is a quintic equation with no closed form for the roots. However, WolframAlpha says the relevant root is approximately $0.463$. $x$ is then the $\arcsin$ of this, which means that $$x \approx 0.481$$

Varun Vejalla
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    WA also gives a solution $u \approx -0.785$, which corresponds to a triangle having sides $\sin x', \cos x', \tan x'$ where $x'$ is the exterior angle formed by the first two of those sides. – Oscar Lanzi Dec 16 '19 at 12:45
  • There is no closed form solution for the roots - and by extension, $x$ - in radicals, but there is a solution in special functions. – Jam Dec 19 '19 at 18:12
  • @Jam What would the solution be with special functions? – Varun Vejalla Dec 20 '19 at 19:48
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Elaborating on the answers by Automatically Generated and Claude Leibovici, despite quintics not being generally solvable in radicals, we can obtain a closed form solution by permitting special functions, albeit an unreadably long solution that goes beyond the scope of the original question. Following the excellent answer on Question 542108 by Tito Piezas III, we can reduce the quintic $p_1(x)=2x^5−4x^3+2x^2+2x−1$, first to principal quintic form, then to Bring quintic form. We achieve this by applying a quadratic Tschirnhausen transformation to $p_1$, attaining $p_2(y)=y^{5}-0.455y^{2}-0.531y-0.116$, where $y=x^{2}+0.148x-0.8$ *. We then follow this with a quartic Tschirnhausen transformation on $p_2$, for $p_3(z)=z^{5}-0.108z+0.0131$, where $z=y^{4}-0.477y^{3}+0.317y^{2}-1.219y-0.295$. This is finally converted to Bring quintic form with the substitution $z=0.108^{{1}/{4}}t$, giving us $p_4(t)=t^5-t+0.211$. We can then return to the desired root through the correctly chosen chain of roots, $t=0.939\to z=0.539\to y=-0.510\to x=0.463$.

Then, we may solve $p_4$ using the technique from Piezas III's answer in Question 540964. Using the notation from that answer, we would have $k = (0.268-0.625 i)$ and $\frac{(k^2)^{1/8}}{2\cdot 5^{3/4}\sqrt{k(1-k^2)}}=(0.139 + 0.0234 i)$. The root we desire corresponds to $n=4$, so $p=(-0.784 + 1.177 i)+16\cdot 4$, so then for $j=0,1,2,3,4,5$, we have $\small S_j=(-0.418-1.074i),\ (3.533-0.341i),\ (0.495+0.848i), (-0.000+1.000i),\ (-0.706+0.704i),\ (0.005-0.140i)$. Using this technique, the root $t=0.939$ is expressible in a closed form, using radicals, the complete elliptic integral of the first kind, $K(x)$, and the Dedekind eta function, $\eta(x)$. Therefore, so too is the root $x=0.463$ of $p_1$, as is the size of the angle in question.

* Each coefficient from this point onwards has a closed form in radicals but as they are unreasonably large, I've given rounded decimals only. See Tito Piezas III's answer for a full account of the substitutions, transformations and derivations of the coefficients. I'd advise using a computer algebra solver, like Maple or Mathematica when using this technique due to the volume of operations and terms. NB: Some values I've listed may have rounding errors.

Jam
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Starting from @automaticallyGenerated's answer, facing a quintic polynomial, tou will need a numerical method.

Let us consider that we look for the zero of function $$f(x)=\sin^2(x)- \cos^2(x)+2\cos^4(x)\sin(x)$$ and use Newton method starting with $x_0=0$; this will provide the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.000000 \\ 1 & 0.500000 \\ 2 & 0.480810 \\ 3 & 0.481195 \end{array} \right)$$

Edit

Graphing or using inspection, we can notice that the solution is close to $\frac \pi 6$. So, making a Taylor expansion around this point, we have $$f(x)=\frac{1}{16}+t-\frac{392 }{507}t^2+\frac{7552 }{19773}t^3+\frac{504320 }{771147}t^4+O\left(t^5\right)$$ where $t=\frac{13\sqrt{3}}{16} \left(x-\frac{\pi }{6}\right)$.

Now, using series reversion (using $f(x)=y$), we have $$x=\frac{\pi }{6}+u+\frac{49 }{26 \sqrt{3}}u^2+\frac{817}{507} u^3+\frac{22975 }{26364 \sqrt{3}}u^4+O\left(u^5\right)$$ where $u=\frac{16 y-1}{13 \sqrt{3}}$.

Make $y=0$ to get the approximation $$x \sim \frac{\pi }{6}-\frac{497738471}{6776839836 \sqrt{3}}=0.481194$$

Claude Leibovici
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