1
$\begingroup$

$$ \arctan\left(\frac{x}{\sqrt{a^2-x^2}}\right)$$

Hi, I am not able to solve this problem from last 1 hour. Please help me to solve this question. As I can solve simple inverse trigonometric functions.

$\endgroup$
  • 1
    $\begingroup$ Set $x = a \sin(\phi)$, and simplify the argument of the $\arctan$. Are you assuming $0<x<a$? $\endgroup$ – Sasha Aug 23 '12 at 15:38
4
$\begingroup$

Think geometrically. What is the $\tan(\phi)$ ?

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Please, can you tell me the name of the software which you use to draw these pictures. I try to use GeoGebra and a friend recommend me to use MetaPost but I have some issues to use these software correctly. Thanks :) $\endgroup$ – Iuli Aug 23 '12 at 15:57
  • $\begingroup$ @Iuli I used Mathematica. The command can be found here. $\endgroup$ – Sasha Aug 23 '12 at 16:07
  • $\begingroup$ So $\arctan\dfrac{x}{\sqrt{a^2-x^2}}=\arccos\dfrac x a$. $\endgroup$ – Michael Hardy Aug 23 '12 at 20:46
  • 1
    $\begingroup$ @MichaelHardy I think it should be $\arcsin \frac{x}{a}$, instead. $\endgroup$ – Sasha Aug 23 '12 at 21:01
  • $\begingroup$ @Sasha : You're right: I'm accustomed to pictures in which the "opposite" side is vertical and the "adjacent" side horizontal. $\endgroup$ – Michael Hardy Aug 23 '12 at 21:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.