I've some argument like:

$$ \arctan(\sqrt3 + 2) $$ and as explained here How to calculate $\arctan(\sqrt3-2)$ to be $-15°$ by hand? i made an assumption and found that it should be $ \frac{5\pi}{12} $. I found the exact value of $ \sqrt3 + 2 $ then by table i found it's equal to $ \tan(75^{\circ}) $ and finally i found $ \arctan(\tan(75^{\circ})) $ But what if i have something more complicated to do assumption on like:

$$ \arccos({\frac{\sqrt{\sqrt3+2}}{2}}) $$

So i'm interesting if there is any sequence of operations on argument itself to convert it into fraction of $ \pi $ without guessing about how many it should be?


The best bet is to "unravel" the expression, try to simplify it by getting rid of roots and recognizing patterns:

$$\phi=\arccos\frac{\sqrt{\sqrt{3}+2}}{2}$$ $$\cos\phi=\frac{\sqrt{\sqrt{3}+2}}{2}$$ $$4\cos^2\phi=\sqrt{3}+2$$ $$2(2\cos^2\phi-1)=\sqrt{3}$$ $$2\cos2\phi=\sqrt{3}$$ $$\cos2\phi=\frac{\sqrt{3}}{2}$$ $$2\phi=\arccos\frac{\sqrt3}{2}=\frac{\pi}{6}$$ $$\phi=\frac{\pi}{12}$$

You'll always get some sort of polynomial in trigonometric functions. The goal is to reduce the degree of polynomial by using multiple angle formulas. If you don't recognize them yourself, you can take a look at Chebyshev polynomials of the first kind which tell you the multiple angle formulas: $\cos n\phi= T_n(\cos\phi)$. In this case, we recognized $T_2(x)=2x^2-1$.

Depending on how well twisted the expression is, you may have to get creative. There may be more than just multiple angle tricks: possibly, you have to use addition theorems as well, recognizing $\cos(x+y)$ where $x$ and $y$ are not equal, and so on. But the main trick is still just "undoing" the operations that you don't want to see.

  • $\begingroup$ Yes, that's exactly what i was looking for. Thanks a lot! $\endgroup$
    – Dmitrii
    Dec 6 '17 at 9:06
  • 1
    $\begingroup$ very nice derivation ! $\endgroup$
    – user
    Dec 6 '17 at 9:10

As $2+\sqrt3=\dfrac{(\sqrt3+1)^2}2$



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