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 March 18th, 2015, 09:06 PM #1 Newbie   Joined: Mar 2015 From: England Posts: 18 Thanks: 1 tan pi/12 Find tan pi/12= t through sin pi/6. Sin pi/6= 2t/ (1+t^2). Solving this equation yields t= 2+/-sq root 3. But tan pi/12 is only 2- sq root 3. My question is why do we exclude 2+ sq root 3? Thank you March 18th, 2015, 09:26 PM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs We know that on $0\le x\le\frac{\pi}{4}$, we have $0\le\tan(x)\le1$. Thanks from axerexa March 18th, 2015, 09:55 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 $\sin\frac{5\pi}{6} = \sin\frac{\pi}{6}\!$. Hence $\tan\frac{5\pi}{12}$ (which is 2 + √3) needs to be excluded. Thanks from MarkFL March 18th, 2015, 10:30 PM   #4
Senior Member

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
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Math Focus: Calculus/ODEs
Quote:
 Originally Posted by axerexa Find tan pi/12= t through sin pi/6...
Another way to approach this:

$\displaystyle \sin\left(\frac{\pi}{12}\right)=\sin\left(\frac{1} {2}\cdot\frac{\pi}{6}\right)=\sqrt{\frac{1-\cos\left(\frac{\pi}{6}\right)}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{2-\sqrt{3}}}{2}$

$\displaystyle \cos\left(\frac{\pi}{12}\right)=\sqrt{1-\sin^2\left(\frac{\pi}{12}\right)}=\sqrt{1-\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{2+\sqrt{3}}}{2}$

Hence:

$\displaystyle \tan\left(\frac{\pi}{12}\right)= \frac{\sin\left(\frac{\pi}{12}\right)}{ \cos\left(\frac{\pi}{12}\right)}= \frac{\frac{\sqrt{2-\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}}= \frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\cdot\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}= \frac{2-\sqrt{3}}{\sqrt{4-3}}= 2-\sqrt{3}$ March 18th, 2015, 11:40 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 $\displaystyle \text{If }\sin(2x) \ne 0, \,\tan(x) = \frac{2\sin^2(x)}{2\sin(x)\cos(x)} = \frac{1 - \cos(2x)}{\sin(2x)}.$ $\displaystyle \therefore \tan\left(\frac{\pi}{12}\right) = \frac{1 - \cos\left(\frac{\pi}{6}\right)} {\sin\left(\frac{\pi}{6}\right)} = \frac{1 - \sqrt{3}/2}{1/2} = 2 - \sqrt3.$ Thanks from greg1313 Tags pi or 12, tan ,

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how find tan phyi/12

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