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March 18th, 2015, 09:06 PM   #1
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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?

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March 18th, 2015, 09:26 PM   #2
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We know that on $0\le x\le\frac{\pi}{4}$, we have $0\le\tan(x)\le1$.
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March 18th, 2015, 09:55 PM   #3
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$\sin\frac{5\pi}{6} = \sin\frac{\pi}{6}\!$. Hence $\tan\frac{5\pi}{12}$ (which is 2 + √3) needs to be excluded.
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March 18th, 2015, 10:30 PM   #4
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Quote:
Originally Posted by axerexa View Post
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:

$\displaystyle \tan\left(\frac{\pi}{12}\right)= \frac{\sin\left(\frac{\pi}{12}\right)}{ \cos\left(\frac{\pi}{12}\right)}= \frac{\frac{\sqrt{2-\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}}= \frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\cdot\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}= \frac{2-\sqrt{3}}{\sqrt{4-3}}= 2-\sqrt{3}$
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March 18th, 2015, 11:40 PM   #5
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$\displaystyle \text{If }\sin(2x) \ne 0, \,\tan(x) = \frac{2\sin^2(x)}{2\sin(x)\cos(x)} = \frac{1 - \cos(2x)}{\sin(2x)}.$

$\displaystyle \therefore \tan\left(\frac{\pi}{12}\right) = \frac{1 - \cos\left(\frac{\pi}{6}\right)} {\sin\left(\frac{\pi}{6}\right)} = \frac{1 - \sqrt{3}/2}{1/2} = 2 - \sqrt3.$
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