Oprava chyb v příkladech z FYI

KFY FYI1/Priklad01.md

@@ -35,7 +35,7 @@ pro $t = T$
 
 **Dráha**
 
-$\displaystyle s = \frac{1}{2}\cancel{a_{t}} \cdot \frac{(v_{1} - v_{2})^T}{a_{t}^{\cancel{2}}} + v_{0} \cdot \frac{v_{1} - v_{0}}{a_{t}} = \frac{v_{1}^2 - 2v_{1}v_{0} + v_{0}^2}{2a_{t}} + \frac{v_{0}v_{1} - v_{0}^2}{a_{t}} = \frac{v_{1}^2 - \cancel{2v_{1}v_{0}} + \cancel{v_{0}^2} + \cancel{2v_{1}v_{0}} - \cancel{2}v_{0}^2}{2a_{t}} = \frac{v_{1}^2 - v_{0}^2}{2a_{t}}$
+$\displaystyle s = \frac{1}{2}\cancel{a_{t}} \cdot \frac{(v_{1} - v_{2})^2}{a_{t}^{\cancel{2}}} + v_{0} \cdot \frac{v_{1} - v_{0}}{a_{t}} = \frac{v_{1}^2 - 2v_{1}v_{0} + v_{0}^2}{2a_{t}} + \frac{v_{0}v_{1} - v_{0}^2}{a_{t}} = \frac{v_{1}^2 - \cancel{2v_{1}v_{0}} + \cancel{v_{0}^2} + \cancel{2v_{1}v_{0}} - \cancel{2}v_{0}^2}{2a_{t}} = \frac{v_{1}^2 - v_{0}^2}{2a_{t}}$
 
 **Doba jízdy**
 

KFY FYI1/Priklad12.md

@@ -22,7 +22,7 @@ pro optické rozhraní 2 platí
 ### Výpočet
 
 vztah mezi úhly $\gamma$ a $\beta$ - viz. pravoúhlý trojúhelník
-- $\beta = \frac{\pi}{2} - \gamma \implies \sin \beta = \sin (\frac{\pi}{2}) = \cos \gamma$
+- $\beta = \frac{\pi}{2} - \gamma \implies \sin \beta = \sin \left(\frac{\pi}{2} - \gamma\right) = \cos \gamma$
 
 určení numerické apertury
 - $\sin \alpha_{m} = n_{1} \cdot \sin \beta = n_{1} \cdot \cos \gamma =$