1. ## Tan 2 theta

I know from my conic that $Tan2\Theta=\frac{1}{4}$, can you show me how I arrive at:

$Cos2\Theta=\frac{4}{\sqrt17}$ from the above please so that I can work out my transformed conic.

Thank you.

2. Hello, bigroo!

$\text{I know from my conic that: }\:\tan2\theta\:=\:\dfrac{1}{4}$

$\text{Can you show me how I arrive at: }\:\cos2\theta\:=\:\dfrac{4}{\sqrt17}}$

We have: . $\tan2\theta \:=\:\dfrac{1}{4} \:=\:\dfrac{opp}{adj}$

$\,2\theta$ is in a right triangle with: . $opp = 1,\;adj = 4$

Code:
*
|   *        __
|       *   √17
1 |           *
|               *
|              2@   *
* - - - - - - - - - - - *
4

Pythagorus says: . $hyp^2 \:=\:1^2 + 4^2 \:=\:17 \quad\Rightarrow\quad hyp \:=\:\sqrt{17}$

Therefore: . $\cos2\theta \;=\;\dfrac{adj}{hyp} \;=\;\dfrac{4}{\sqrt{17}}$

3. Thank you soroban, it was basic Pythagorus after all. I guess I was looking too deep.