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After some calculations I have $$\arctan\biggl(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\biggr)-\arcsin\biggl(\frac{3}{2\sqrt{4+\sqrt{3}}}\biggr).$$ Mathematica simplyfies this to $\pi/12$ and I wonder how it is done.

Edit: The results comes from calculating $x$ in the figure below. There might be simple ways to do it than mine. enter image description here

I add my solution:

Cosinus Theorem: $$ |AB|^2 =1^2+(\sqrt{3}\,)^2-2\cdot1\cdot\sqrt{3}\cdot\cos(120°) =1+3-2\sqrt{3}(-\tfrac{1}{2}) =4+\sqrt{3}. $$

Pythagoras: $$ |AC|^2+|AB|^2=|BC|^2 \quad\Leftrightarrow\quad |AC|^2 =|BC|^2-|AB|^2 =4+\sqrt{3}-(\sqrt{2}\,)^2 =4+\sqrt{3}-2 =2+\sqrt{3} $$ i.e. $$\textstyle|AC|=\sqrt{2+\sqrt{3}}.$$

Sinus Theorem $$ \frac{\sin(y)}{\sqrt{3}}=\frac{\sin(120°)}{\sqrt{4+\sqrt{3}}} \quad\Leftrightarrow\quad \frac{\sin(y)}{\sqrt{3}}=\frac{\frac{\sqrt{3}}{2}}{\sqrt{4+\sqrt{3}}} \quad\Leftrightarrow\quad \sin(y) =\frac{3}{2\sqrt{4+\sqrt{3}}} $$ i.e. $$ y=\arcsin\biggl(\frac{3}{2\sqrt{4+\sqrt{3}}}\biggr) $$

Further, $$ \tan(x+y) =\frac{|AC|}{|AB|} =\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}} $$ i.e. $$ x+y=\arctan\biggl(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\biggr). $$ and $$ x=\arctan\biggl(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\biggr)-\arcsin\biggl(\frac{3}{2\sqrt{4+\sqrt{3}}}\biggr) $$

mf67
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    Can you provide some sort of context for how you got this (e.g. is it from a question? If so, provide the question with the question source and how you got to this stage)? Otherwise, it is likely that your question will get closed. Also, add any attempts that you have done prior to getting help. – Yajat Shamji Jan 02 '24 at 13:29
  • I would like to see the geometry that led to this expression. I can concoct figures that correspond to the individual terms, but I don't see a "natural" way to visualize the difference. For instance, in a triangle with sides $1$ and $\sqrt{3}$ bounding an angle of $120^\circ$, the second term gives the angle opposite the side $\sqrt{3}$. Noting that $\sqrt{2+\sqrt{3}}/\sqrt{2}=(1+\sqrt{3})/2$, the first angle lies in a right triangle formed from the same elements. Knowing how these were supposed to fit together, might make the target value straightforward to determine geometrically. – Blue Jan 02 '24 at 15:03
  • @YajatShamji - I updated the question. My solution is in Swedish, I can post a condensed version in English if you want. – mf67 Jan 02 '24 at 15:06
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    @Blue - I updated the question with the geometry problem. – mf67 Jan 02 '24 at 15:07
  • @mf67, English is the preferred language of MSE (if I am not mistaken...). – Yajat Shamji Jan 02 '24 at 15:08
  • @YajatShamji - now added, in 'some kind of English' – mf67 Jan 02 '24 at 15:34

1 Answers1

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Try to take some trig function of all expression. For example, $\tan$: $$\tan\left(\underbrace{\arctan \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2}}}_{=:\alpha} - \underbrace{\arcsin\frac{3}{2\sqrt{4+\sqrt{3}}}}_{=:\beta}\right)=$$ $$=\tan(\alpha-\beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\cdot\tan\beta}$$ There is no problem is computing $\tan\alpha$ because it is $\tan(\arctan(\cdot))$. We need to find $\tan(\arcsin(\cdot))$: $$\tan(\arcsin x) = \{x = \sin\gamma\}=\tan\gamma = \frac{\sin\gamma}{\cos\gamma} = \frac{x}{\pm\sqrt{1-x^2}}$$ Plus or minus choice depends on $\gamma$. After that you will know $\tan$ of the angle which should be equal to $$\tan\frac{\pi}{12} = \frac{\sin\frac{\pi}{12}}{\cos\frac{\pi}{12}} = \frac{2\sin\frac{\pi}{12}\cos\frac{\pi}{12}}{2\cos^2\frac{\pi}{12}}= \frac{\sin\frac{\pi}{6}}{1+\cos\frac{\pi}{6}} = \frac{1}{2 + \sqrt{3}}=2 - \sqrt{3}$$

Egor Ivanov
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    It's not a bad idea, but note 1/ $\arcsin(x)\in[-\pi/2,\pi/2]$ and has the same sign as $x$, therefore $\tan(\arcsin(x))=\dfrac{x}{\sqrt{1-x^2}}$, no $\pm$. If you prefer, $\cos(\arcsin x)$ is always $\ge0$. And 2/, even if it's almost obvious in this case, it should be emphasized that when you have $\tan \varphi$, you only know $\varphi$ modulo $\pi$. Here of course both terms are in $(0,\pi/2)$, so the difference is in $(-\pi/2,\pi/2)$, and you can find the exact value of $\varphi$. – Jean-Claude Arbaut Jan 02 '24 at 13:52