Using the double angle formula of $\tan(x)$ and knowing that $\cot(\dfrac{\pi}{6} ) = \sqrt3$ we can make progress towards finding the value of $\cot(\dfrac{\pi}{12})$.
$\cot(2x) = \dfrac{1-\tan^2(x)}{2\tan(x)}$
$\cot(\dfrac{\pi}{6}) = \dfrac{1-\dfrac{1}{\cot^2(\pi/12)}}{2(\dfrac{1}{\cot(\pi/12)})}$
$2\sqrt3=\cot(\dfrac{\pi}{12})-\dfrac{1}{\cot(\pi/12)}$
$\cot^2(\dfrac{\pi}{12})-2\sqrt3 \cot(\dfrac{\pi}{12})-1=0$
We can solve the last line with the quadratic formula and it tells us that:
$\cot(\dfrac{\pi}{12})=2+\sqrt3$ or $-2+\sqrt3$
But this second solution is wrong? Of course $\cot$ can't have two values for one input, but where does this second solution come from?
$\endgroup$ 33 Answers
$\begingroup$There are two angles $\theta$ in $[0,2\pi)$ with a cotangent of $\sqrt 3$. One is $\frac \pi 6$ and one is $\frac {7 \pi}6$. This is because adding $\pi$ to the angle changes both the cosine and sine to their negatives, so the signs divide out. Half of the first is $\frac \pi{12}$ and half of the second is $\frac {7 \pi}{12}$. They are both solutions to an angle that is half an angle with cotangent of $\sqrt 3$. You need to choose the right one for your problem.
$\endgroup$ $\begingroup$There are two angles in the fundamental domain $(0,\pi)$ of the cotangent so that $2x=\frac\pi6+k\pi$. These are $\frac\pi{12}$ and $\frac\pi{12}+\frac\pi2=\frac{7\pi}{12}$. Your quadratic equation has roots at the cotangent of both of these angles.
$\endgroup$ $\begingroup$It comes from the fact that you solved a quadratic with nonzero discriminant, in this case being $$(-2\sqrt 3)^2-4(1)(-1)=12+4=16.$$
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