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

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.

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    Set $x = a \sin(\phi)$, and simplify the argument of the $\arctan$. Are you assuming $02012-08-23

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

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    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 :)2012-08-23
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    @Iuli I used [_Mathematica_](http://www.wolfram.com/mathematica/). The command can be found [here](https://docs.google.com/document/d/19Is-ofRbRPBMNOI6kdbAOA9XR-uouUqEzEcy7OcIUIM/edit).2012-08-23
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    So $\arctan\dfrac{x}{\sqrt{a^2-x^2}}=\arccos\dfrac x a$.2012-08-23
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    @MichaelHardy I think it should be $\arcsin \frac{x}{a}$, instead.2012-08-23
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    @Sasha : You're right: I'm accustomed to pictures in which the "opposite" side is vertical and the "adjacent" side horizontal.2012-08-23