How do I find the value of cot pi/12?

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How do I find the value of cot pi/12?

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Answer 1 · 2015-10-26T00:13:40

Find exact value of $cot (\frac{π}{12})$ $cot ((pi)/12)$

Ans: $(2 - \sqrt{3})$ $(2 - sqrt3)$

Explanation:

$cot (\frac{π}{12}) = \frac{1}{tan (\frac{π}{12})}$ $cot ((pi)/12) = 1/(tan ((pi)/12)$ . First find $tan (\frac{π}{12})$ $tan ((pi)/12)$
Call $tan (\frac{π}{12}) = t$ $tan ((pi)/12) = t$
$tan (2 t) = tan (\frac{π}{6}) = \frac{1}{\sqrt{3}}$ $tan (2t) = tan ((pi)/6) = 1/sqrt3$
Apply the trig identity: $tan 2 a = \frac{2 tan a}{1 - {tan}^{2} a}$ $tan 2a = (2tan a)/(1 - tan^2 a)$
We get:
$\frac{1}{\sqrt{3}} = \frac{2 t}{1 - t^{2}}$ $1/sqrt3 = (2t)/(1 - t^2)$
$1 - t^{2} = 2 \sqrt{3} t$ $1 - t^2 = 2sqrt3t$ . Solve the quadratic equation in t.

$t^{2} + 2 \sqrt{3} t - 1 = 0$ $t^2 + 2sqrt3t - 1 = 0$
$D = d^{2} = b^{2} - 4 a c = 12 + 4 = 16$ $D = d^2 = b^2 - 4ac = 12 + 4 = 16$ --> $d = \pm 4$ $d = +- 4$
There are 2 real roots:
$tan (\frac{π}{12}) = t = - 2 \frac{\sqrt{3}}{2} \pm \frac{4}{2} = - \sqrt{3} \pm 2$ $tan ((pi)/12) = t = - 2sqrt3/2 +- 4/2 = -sqrt3 +- 2$
Since the arc (pi/12) is in Quadrant I, its tan is positive, then
$tan (\frac{π}{12}) = (- \sqrt{3} + 2) .$ $tan ((pi)/12) = (-sqrt3 + 2).$
Check by calculator:
$tan (\frac{π}{12}) = tan 15 = 0.27 .$ $tan ((pi)/12) = tan 15 = 0.27.$
$(2 - \sqrt{3}) = 0.27$ $(2 - sqrt3) = 0.27$ . OK

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How do I find the value of cot pi/12?

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