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How to prove the following trignometric identity? $$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$

Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$.

I would however like to learn the technique of dealing with surds such as these, especially in trignometric problems as I have a lot of similar problems and I don't have a clue as to how to deal with those.

Hints please!

EDIT:

What I have done using half angles is this: (and please note, for convenience, I am dropping the degree symbols. The angles here are in degrees however).

I know that $$ \cos 15 = \dfrac{\sqrt3+1}{2\sqrt2}$$

So,

$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$ $$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$

$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$

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    $\begingroup$ What number are you getting? If you tell us that, we'll be better able to help you see how to show they're equal. $\endgroup$ – Cameron Buie Aug 21 '13 at 6:56
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$$\text{As } \cot x =\frac{\cos x}{\sin x}$$ $$ =\frac{2\cos^2x}{2\sin x\cos x}(\text{ multiplying the numerator & the denominator by }2\cos7\frac12 ^\circ)$$

$$=\frac{1+\cos2x}{\sin2x}(\text{using }\sin2A=2\sin A\cos A,\cos2A=2\cos^2A-1$$

$$ \cot7\frac12 ^\circ =\frac{1+\cos15^\circ}{\sin15^\circ}$$

$\cos15^\circ=\cos(45-30)^\circ=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ=\frac{\sqrt3+1}{2\sqrt2}$

$\sin15^\circ=\sin(45-30)^\circ=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ=\frac{\sqrt3-1}{2\sqrt2}$

Method $1:$

$$\frac{1+\cos15^\circ}{\sin15^\circ}=\csc15^\circ+\cot15^\circ$$

$$\cot15^\circ=\frac{\cos15^\circ}{\sin15^\circ}=\frac{\sqrt3+1}{\sqrt3-1}=\frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)}=2+\sqrt3$$

$$\csc15^\circ=\frac{2\sqrt2}{\sqrt3-1}=\frac{2\sqrt2(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\sqrt2(\sqrt3+1)=\sqrt6+\sqrt2$$

Method $2:$

$$\implies \frac{1+\cos15^\circ}{\sin15^\circ}=\frac{1+\frac{\sqrt3+1}{2\sqrt2}}{\frac{\sqrt3-1}{2\sqrt2}}=\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}=\frac{(2\sqrt2+\sqrt3+1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}(\text{ rationalizing the denominator })$$

$$=\frac{2\sqrt6+4+2\sqrt3+2\sqrt2}2$$

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  • $\begingroup$ @ParthThakkar, how about this method? $\endgroup$ – lab bhattacharjee Aug 21 '13 at 8:32
  • $\begingroup$ Mast hai boss! ;) Really. Much lesser work than the other methods especially if you don't know the value of $\cos 15^{\circ}$ and $\sin 15^{\circ}$. +1 $\endgroup$ – Parth Thakkar Aug 21 '13 at 8:38
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    $\begingroup$ @ParthThakkar, also we don't need to solve any Equation and choose the value based on the Quadrant of $7.5^\circ$ $\endgroup$ – lab bhattacharjee Aug 21 '13 at 8:46
  • $\begingroup$ @ParthThakkar, added another method $\endgroup$ – lab bhattacharjee Aug 21 '13 at 19:02
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Here is an elementary (almost without words) proof that does not rely explicitly on half-angle or double-angle formulas. What is used is the fact that $\cot(30^\circ) = \sqrt{3}$, the exterior angle theorem, isoceles triangle theorems and Pythaogorean theorem of Euclidean geometry, and the fact that $8 + 4\sqrt{3} = \left(\sqrt{2}+\sqrt{6}\right)^2$ (cf. Robert Israel's answer). The crude, not-to-scale, diagram below is, I hope, self-explanatory especially if you start from the right side and work your way to the left.

The length of the base of the triangle is $$\cot(7.5^\circ) = \sqrt{8+4\sqrt{3}} +2+\sqrt{3} = \sqrt{2}+\sqrt{6} + \sqrt{4}+\sqrt{3}$$

enter image description here

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    $\begingroup$ +1. That's an amazing proof! Completely different from the others and elegant :D $\endgroup$ – Parth Thakkar Aug 24 '13 at 9:10
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Start from $\displaystyle\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{2\tan{15^\circ}}{1-\tan^2{15^\circ}}$.

If $x=\tan{15^\circ}$, then $\displaystyle\tan 15^\circ = x = \frac{2\tan{(\frac{15}{2})^\circ}}{1-\tan^2{(\frac{15}{2})^\circ}}$.

If $y=\tan{(\frac{15}{2})^\circ}$, then $x=\frac{2y}{1-y^2}$. Hence

$\displaystyle\frac{1}{\sqrt{3}} = \frac{2(\frac{2y}{1-y^2})}{1-(\frac{2y}{1-y^2})^2}$.

Simplify the above equation and solve for $y$, then find the reciprocal to find $\cot{(\frac{15}{2})^\circ}$.

EDIT: To simplify your surd, try to multiply both the numerator and denominator by $\sqrt{2\sqrt{2}−\sqrt{3}+1}$.

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  • $\begingroup$ Ok. I should've thought of multiplication by conjugate, really! Thanks :D $\endgroup$ – Parth Thakkar Aug 21 '13 at 7:14
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Note that if $\cot(x) = c$, $$\cot(4x) = \frac{1-6c^2+c^4}{4 c^3 - 4 c}$$ In this case $\cot(4x) = \sqrt{3}$. thus you want $$ c^4 - 6 c^2 + 1 - (4 c^3 - 4 c) \sqrt{3} = 0 $$ The quartic happens to factor as $$ (c^2 + (4-2 \sqrt{3}) c - 1)(c^2 + (-4 - 2 \sqrt{3}) c - 1)$$ Use numerical approximation to see which quadratic factor you want to be $0$, and solve. You may also find it useful to note that $4 + 2 \sqrt{3} = (1+\sqrt{3})^2 $.

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  • $\begingroup$ Factorizing that quadratic?! How did you get there? I understand that both the expressions are equal, but how did you factorize it? $\endgroup$ – Parth Thakkar Aug 21 '13 at 7:18
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    $\begingroup$ I admit that I used Maple to do the factoring in this case. But if you're looking for a factorization into two quadratics, it's natural to try the form $(c^2 + a c \pm 1)(c^2 + b c \pm 1)$, and then it's not hard to figure out what $a$ and $b$ and the $\pm$ should be. $\endgroup$ – Robert Israel Aug 21 '13 at 15:23
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using half angle identities, $$\cot(x)=\frac{1}{\tan(x)}=\frac{\sin(2x)}{1-\cos(2x)}=\frac{\left(\frac{1-\cos(4x)}{2}\right)^{\frac{1}{2}}}{1-\left(\frac{1+\cos(4x)}{2}\right)^\frac{1}{2}}$$ with $x=7.5^o=\frac{\pi}{24}$ and $\cos(\frac{4\pi}{24})=\frac{\sqrt{3}}{2}$ we can substitute and simplify by multiplying through by $\sqrt{2}$ twice, $$ \cot(\frac{\pi}{24}) = \frac { \frac { \left( 1-\frac{\sqrt{3}}{2} \right) ^{\frac{1}{2}} } { ( {2} )^{\frac{1}{2}} } } { 1- \frac { \left( 1+\frac{\sqrt{3}}{2} \right) ^{\frac{1}{2}} } { ( {2} )^{\frac{1}{2}} } } = \frac { \frac { ( 2-\sqrt{3} ) ^{\frac{1}{2}} } { ( {2} ) ^{\frac{1}{2}} } } { \sqrt{2} - \frac {(2+\sqrt{3}) ^{\frac{1}{2}}} {{(2)}^{\frac{1}{2}}} } = \frac { ( 2-\sqrt{3} ) ^{\frac{1}{2}} } { 2 - (2+\sqrt{3}) ^{\frac{1}{2}} } $$

to simplify from here, we could look to 'complete the square' under the radicals (leaving NO remainder) in order for the the powers $2$ and $1/2$ to multiply and cancel

look for $a, b$ such that $(a + b)^2 = 2+\sqrt{3}$ ideally (and similarly for the $2-\sqrt{3}$ radical), but note that any scaler multiple of this RHS would do as this multiple can be either factored out, i.e. completing for $k(2-\sqrt{3}) = (a-b)^2$ works if we replace $(2-\sqrt{3})$ with $\left(\frac{1}{\sqrt{k}}\right)^2(a-b)^2$ instead, or alternatively pulled in from outside of the radical, i.e. by multiplying the main equation through by the factor $\sqrt{k}$

this is the weakest part of the solution, as only trial and error led me to $k=2, a=1, b=\pm\sqrt{3}$

$$ \cot(\frac{\pi}{24}) = \frac {\left(\left(\frac{1}{\sqrt{2}}\right)^2(1-\sqrt{3})^2\right)^\frac{1}{2}} {2-\left(\left(\frac{1}{\sqrt{2}}\right)^2(1+\sqrt{3})^2\right)^\frac{1}{2}} = \frac {\frac{1}{\sqrt{2}}|1-\sqrt{3}|} {2-\frac{1}{\sqrt{2}}(1+\sqrt{3})} = \frac {|1-\sqrt{3}|} {2\sqrt{2}-(1+\sqrt{3})} $$ to simplify the (disgusting) modulus (which exists to maintain the exact radical solution), multiply through by its conjugate. being very careful with signs (positive $|1-\sqrt{3}|$ scaled by positive $(1+\sqrt{3})$ is positive) we have, $$ \cot(\frac{\pi}{24}) = \frac { |1-\sqrt{3}|(1+\sqrt{3}) } { 2\sqrt{2}(1+\sqrt{3})-(1+\sqrt{3})^2 } =\frac{2}{2\sqrt{2}+2\sqrt{6}-(4+2\sqrt{3})} $$ finally, multiplying through by $\frac{1}{2}(\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6})$, the denominator simplifies to $1$, and we're left with your required solution

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  • $\begingroup$ use the other solutions! $\endgroup$ – elbeardmorez Aug 22 '13 at 12:44

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