$$ \arctan\left(\frac{x}{\sqrt{a^2-x^2}}\right)$$
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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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
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Think geometrically. What is the $\tan(\phi)$ ?
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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
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$\begingroup$ @Iuli I used Mathematica. The command can be found here. $\endgroup$ – Sasha Aug 23 '12 at 16:07
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$\begingroup$ So $\arctan\dfrac{x}{\sqrt{a^2-x^2}}=\arccos\dfrac x a$. $\endgroup$ – Michael Hardy Aug 23 '12 at 20:46
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1$\begingroup$ @MichaelHardy I think it should be $\arcsin \frac{x}{a}$, instead. $\endgroup$ – Sasha Aug 23 '12 at 21:01
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$\begingroup$ @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
