2
$\begingroup$

I have found this statement somewhere, however, I dont really understand it.

Could someone explain me where does $2$ before $\operatorname{tg}(x/2)$ come from?

$\frac {dx}{\cos^2(\frac{x}{2})} = 2d(\operatorname{tg}(\frac{x}{2})) $

  • 0
    $\displaystyle{sec^2 x = \frac{1}{cos^2 x}}$2012-03-15

1 Answers 1

5

Hint: (Edited to make the spoiler spoil less things!)

  • $d(\tan x)=\sec^2 x \rm dx $
  • Chain Rule
  • $\operatorname{tg}(x)$ is an archaic name for $\tan x$

$d(\tan \dfrac x 2)=\dfrac 1 2\sec^2\dfrac x 2 \mathrm{d}x \implies \dfrac{\mathrm{d}x}{\cos^2 \dfrac{x}{2}}=2\mathrm{d}(\tan \dfrac{x}{2})$

  • 0
    $\mathrm{tg}$ is also the notation used in Russia and the Balkan countries, if memory serves.2012-08-01