# Thread: Is it really equal? Why?

1. ## Is it really equal? Why?

$\displaystyle \displaystyle{ \cos\left\{ tan^{-1}\left(\frac{1}{R}\sqrt{\frac{L}{C}}\right) \right\}\stackrel{?}{=}\frac{1}{\sqrt{1+\frac{L}{R ^2C}}} }$

Even though I think it is ir-relevant: $\displaystyle R, L, C$ are resistance, inductance and capacitance, respectively.

2. Originally Posted by courteous
$\displaystyle \displaystyle{ \cos\left\{ tan^{-1}\left(\frac{1}{R}\sqrt{\frac{L}{C}}\right) \right\}\stackrel{?}{=}\frac{1}{\sqrt{1+\frac{L}{R ^2C}}} }$

Even though I think it is ir-relevant: $\displaystyle R, L, C$ are resistance, inductance and capacitance, respectively.
consider a right triangle ...

$\displaystyle \tan{y} = \frac{opp}{adj}$

$\displaystyle \tan^{-1}\left(\frac{opp}{adj}\right) = y$

$\displaystyle opp = \frac{1}{R} \sqrt{\frac{L}{C}}$

$\displaystyle adj = 1$

hypotenuse = $\displaystyle \sqrt{1^2 + \left(\frac{1}{R} \sqrt{\frac{L}{C}}\right)^2}$

$\displaystyle \cos{y} = \frac{adj}{hyp} = \frac{1}{\sqrt{1 + \frac{L}{R^2 C}}}$