Find the exact value of tan(`pi/12).

Does this need to be done in radians or degrees or am I starting on the wrong track??

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- Jul 8th 2008, 05:33 PM #1

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- Jul 8th 2008, 05:46 PM #2
here are the two ways i'd go about this:

(1) Note that $\displaystyle \tan \bigg( \frac {\pi}{12} \bigg) = \tan \bigg( \frac 12 \cdot \frac {\pi}6 \bigg)$

now use the half angle formula for tangent, that is, the expansion for $\displaystyle \tan \frac {\theta}2$. here, $\displaystyle \theta = \frac {\pi}6$

or

(2) Note that $\displaystyle \tan \bigg( \frac {\pi}{12} \bigg) = \tan \bigg( \frac {4 \pi}{12} - \frac {3 \pi}{12} \bigg) = \tan \bigg( \frac {\pi}3 - \frac {\pi}4 \bigg)$.

now use the addition formula for tangent

can you continue?

- Jul 8th 2008, 05:51 PM #3

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Use radians as the angle is in terms of $\displaystyle \pi$.

$\displaystyle \tan \left( \frac{ \pi}{12} \right) = \tan \left( \frac{1}{2} \cdot \frac{ \pi}{6} \right)$

You will need the half angle identity for tan for this problem.

$\displaystyle \tan \left( \frac{\theta}{2} \right) = \frac{\sin \left( \frac{\theta}{2} \right)}{\cos \left( \frac{\theta}{2} \right)} $

Multiply by $\displaystyle \frac{\cos \left( \frac{\theta}{2} \right)}{\cos \left( \frac{\theta}{2} \right)} $

$\displaystyle \frac{\sin \left( \frac{\theta}{2} \right)}{\cos \left( \frac{\theta}{2} \right)} \cdot \frac{\cos \left( \frac{\theta}{2} \right)}{\cos \left( \frac{\theta}{2} \right)} \ \ \Rightarrow \ \ \frac{\sin \left( \theta \right)}{2 \cos^2 \left( \frac{\theta}{2} \right)} $

$\displaystyle \therefore \tan \left( \frac{\theta}{2} \right) = \frac{\sin \theta }{1 + \cos \theta}$

You should know $\displaystyle \sin \left( \frac{ \pi}{6} \right)$ and $\displaystyle \cos \left( \frac{ \pi}{6} \right)$.

Bobak

- Jul 8th 2008, 09:23 PM #4

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- Jul 9th 2008, 01:38 AM #5

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