\tan{2\alpha}=24/7

**dwsmith** · January 12th 2011, 12:48 PM

How can I show that $\displaystyle\tan{2\alpha}=\frac{24}{7}\equiv\tan{ \alpha}=\frac{3}{4}\mbox{?}$

**mr fantastic** · January 12th 2011, 12:52 PM

Originally Posted by dwsmith

How can I show that $\displaystyle\tan{2\alpha}=\frac{24}{7}\equiv\tan{ \alpha}=\frac{3}{4}\mbox{?}$

Apply the usual double angle formula, substitute $\tan{\alpha}=\frac{3}{4}$ and simplify.

**dwsmith** · January 12th 2011, 01:06 PM

Originally Posted by mr fantastic

Apply the usual double angle formula, substitute $\tan{\alpha}=\frac{3}{4}$ and simplify.

$\displaystyle\tan{2\alpha}=\frac{2\tan{\alpha}}{1-\tan^2{\alpha}}=\frac{24}{7}\Rightarrow 7\tan{\alpha}=12-12\tan^2{\alpha}$

$\displaystyle\Rightarrow 12\tan^2{\alpha}+7\tan{\alpha}-12=(3\tan{\alpha}+4)(4\tan{\alpha}-3)=0$

$\displaystyle\tan{\alpha}=\frac{3}{4}, \ \frac{-4}{3}$

These are both 3,4,5 right triangles.

This comes from PDE so it doesn't matter which one I would pick for the overall solution of the PDE.

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