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I'm studying Calculus but I forgot to do this part. The formula for getting theta is Arctan(y/x), and I am solving a problem where x=2 and y=-2. My notes shows that Arctan(-1), with some adjustments, approximates to -pi/4. How so? I need to know how to get the theta equivalent of it so I can graph it in a unit circle.

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In fact, it is not an approximation, as Michael's answer shows. – Cameron Buie Sep 30 '13 at 11:26
You can already graph $-\pi/4$ in the unit circle; it is $\pi/4$ radians clockwise from the $x$ axis. In polar coordinates, $\theta$ can have any real number as its value, including negative ones. – Carl Mummert Sep 30 '13 at 11:35

Since $\sin(-\pi/4)=-\sqrt{2}/2$ and $\cos(-\pi/4)=\sqrt{2}/2$, we have $\tan(-\pi/4)=-1$.


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You don't need to sign your answers. Your name links to your profile after everything you write on the StackExchange network. – kahen Sep 30 '13 at 11:37

The general issue is that the tangent function has a period of 180 degrees, but there are 360 degrees in a circle. Therefore the arctangent function cannot, in general, tell the correct angle when you convert from cartesian coordinates to polar coordinates, because for any point $(x,y)$ the point $(-x,-y)$ will lead to the same value of arctangent when you use the formula $\theta = \arctan(y/x)$.

So, when converting from cartesian to polar, you have to use additional knowledge of which quadrant the point is in, which allows you to "fix" the value of arctangent to give the correct angle. You can do this by adding 180 degrees ($\pi$ radians) to the angle from the formula if it is in the opposite quadrant from the one you want.

In computer programming, there is a two-argument form of arctangent, called atan2, which exists precisely because of this problem.

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x=2 and y=-2

So $\theta$ lies in fourth quadrant

its value is $ -\alpha$

where $\alpha = arc tan (y/x)$ {avoid negative sign)

$\alpha = arc tan (2/2)$

$\alpha = \frac {\pi}{4}$

so $\theta = - \alpha$

$\theta = - \frac {\pi}{4}$


first quadrant, $\theta = \alpha$

second quadrant,$\theta= \pi - \alpha$

third quadrant,$\theta = -(\pi - \alpha)$

fourth quadrant,$\theta = - \alpha$

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