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I need your help.

I don't know how to simplify: $\frac{-1+\sqrt{3}+\sqrt{4+2\sqrt{3}}}{2\sqrt{3}} $ and $\frac{-1+\sqrt{3}-\sqrt{4+2\sqrt{3}}}{2\sqrt{3}}$

Thank you in advance.

I found $1$ and $\frac{-1}{\sqrt{3}}$. Then I have to show that these are solutions of the following equation: $arctan(x)+arctan(x\sqrt{3})= \frac{7\pi}{12}$. How can I do that ?

Gerry Myerson
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vev78
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    To begin with, multiply the numerator and the denominator by $\sqrt{3}$. That way, your denominator will become rational. – Manolito Pérez Oct 14 '14 at 08:50
  • Thank you ! I found 1 and -1/(sqrt{3}). Then I have to show that these are solutions of the following equation: arctan(x)+arctan(x*sqrt{3})= (7pi/12). How can I do that ? – vev78 Oct 14 '14 at 09:02
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    Just replace $x$ in the equation $\arctan(x)+\arctan(\sqrt{3}x) = \frac{7\pi}{2}$ by the numbers you found, and verify that equality holds. – Manolito Pérez Oct 14 '14 at 09:34
  • @vev78, Be careful for $\arctan$ addition. See http://math.stackexchange.com/questions/138310/show-that-2-tan-12-pi-cos-1-frac35/583359#583359 and http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values – lab bhattacharjee Oct 14 '14 at 17:43

1 Answers1

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HINT:

$$4+2\sqrt3=(\sqrt3)^2+1^2+2\cdot1\cdot\sqrt3=(\sqrt3+1)^2$$

  • Thank you !! But I never see this kind of simplification: do you have any trick to help me to see it easier ? – vev78 Oct 14 '14 at 08:56
  • Here is a way that you can cheat... Suppose that you suspect that either $$\frac{-1+\sqrt{3}+\sqrt{4+2\sqrt{3}}}{2\sqrt{3}}\text{ or }\frac{-1+\sqrt{3}-\sqrt{4+2\sqrt{3}}}{2\sqrt{3}}=1$$ Equating the latter fraction with $1$ will give us $-\sqrt{4+2\sqrt{3}}=\sqrt{3}+1$, which is clearly impossible. So let's equate the former with $1$ instead. $$\begin{array}{lll} \frac{-1+\sqrt{3}+\sqrt{4+2\sqrt{3}}}{2\sqrt{3}}&=&1\ -1+\sqrt{3}+\sqrt{4+2\sqrt{3}}&=&2\sqrt{3}\ \sqrt{4+2\sqrt{3}}&=&\sqrt{3}+1\ 4+2\sqrt{3}&=&(\sqrt{3}+1)^2\ \end{array}$$ Now you can "see" the simplification. – John Joy Oct 14 '14 at 15:21