MHB How can you prove that cot 7.5 degrees equals the sum of four square roots?

[anemone]

Prove that $\cot 7\dfrac{1}{2}^{\circ}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$.

 

[kaliprasad]

Spoiler

we have $\cos\ 15^\circ = \cos(60- 45)^\circ = \cos\ 60^\circ \cos\ 45^\circ + \sin \ 60^\circ \sin\ 45^\circ$

= $\dfrac{1}{2}\dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2}\dfrac{1}{\sqrt{2}}$

= $\dfrac{1}{4}(\sqrt{2} + \sqrt{6})$

also

$\sin \ 15^\circ = \sin (60- 45)^\circ = \sin \ 60^\circ \cos\ 45^\circ - \cos \ 60^\circ \sin\ 45^\circ$

= $\dfrac{\sqrt{3}}{2}\dfrac{1}{\sqrt{2}} - \dfrac{1}{2}\dfrac{1}{\sqrt{2}}$

= $\dfrac{1}{4}(\sqrt{6} - \sqrt{2})$now $\cot\ x= \dfrac{\cos\ x }{\sin \ x} = \dfrac{2\cos^2\ x }{2\sin \ x\cos \ x}$

= $\dfrac{1+\cos 2x}{\sin 2x}$ Hence $\cot \ 7\frac{1}{2}^\circ = \dfrac{1+\cos\ 15^\circ}{\sin\ 15^\circ}$

= $\dfrac{1+\dfrac{1}{4}(\sqrt{2} + \sqrt{6})}{\dfrac{1}{4}(\sqrt{6} - \sqrt{2})}$

= $\dfrac{4+(\sqrt{2} + \sqrt{6})}{(\sqrt{6} - \sqrt{2})}$

= $\dfrac{4+(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})}$

= $\dfrac{4(\sqrt{6} + \sqrt{2})+(\sqrt{6} + \sqrt{2})^2}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})}$

= $\dfrac{4(\sqrt{6} + \sqrt{2})+(\sqrt{6} + \sqrt{2})^2}{4}$

= $(\sqrt{6} + \sqrt{2})+\dfrac{(\sqrt{6} + \sqrt{2})^2}{4}$

= $(\sqrt{6} + \sqrt{2})+\dfrac{6 + 2 + 4 \sqrt{3}}{4}$

= $(\sqrt{6} + \sqrt{2})+\dfrac{8 + 4 \sqrt{3}}{4}$

= $(\sqrt{6} + \sqrt{2})+ 2 + \sqrt{3}$

= $\sqrt{2} + \sqrt{3}+ 2 + \sqrt{6}$

= $\sqrt{2} + \sqrt{3}+\sqrt{4} + \sqrt{6}$

 

[anemone]

Nice one, kaliprasad!(Yes)

I can't wait to show how I saw someone approached the problem by only drawing up a triangle and the rest is so self-explanatory! :)

Spoiler

View attachment 3075
Start to construct the rightmost triangle and work to the left, with creating two isosceles triangles and note that $\sqrt{8+4\sqrt{3}}=\sqrt{(2+2\sqrt{2}\sqrt{6}+6)}=\sqrt{(\sqrt{2}+\sqrt{6})^2}=\sqrt{2}+\sqrt{6}$

Therefore $\cot7\dfrac{1}{2}^{\circ}=\dfrac{1}{\dfrac{1}{\sqrt{3}+\sqrt{4}+\sqrt{8+4\sqrt{3}}}}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$.