• Jan 12th 2011, 12:48 PM
dwsmith
How can I show that $\displaystyle \displaystyle\tan{2\alpha}=\frac{24}{7}\equiv\tan{ \alpha}=\frac{3}{4}\mbox{?}$
• Jan 12th 2011, 12:52 PM
mr fantastic
Quote:

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

Apply the usual double angle formula, substitute $\displaystyle \tan{\alpha}=\frac{3}{4}$ and simplify.
• Jan 12th 2011, 01:06 PM
dwsmith
Quote:

Originally Posted by mr fantastic
Apply the usual double angle formula, substitute $\displaystyle \tan{\alpha}=\frac{3}{4}$ and simplify.

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

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

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

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.