Location of negative angle on unit circle

Hi!

What quadrant and location is the following angle?

$\displaystyle -\frac{11}{12}$

$\displaystyle -\frac{11}{12} = 2\pi -\frac{11\pi}{12}$

$\displaystyle \frac{13}{12} \implies$ Six and a half revolutions, stopping at $\displaystyle \pi$.

$\displaystyle \pi - \frac{\pi}{12} = \frac{11\pi}{12}$

Quadrant II.

Re: Location of negative angle on unit circle

On coordinate axes remember that for angle x , 0<x<pi/2 , x is in I quadrant; pi/2<x<pi , x is in II quadrant; pi<x<3pi /2, x is in III quadrant; and 3pi/2<x<2pi , x is in IV quadrant;

Another thing we need to remember is that angle is measured positive when measure anticlockwise and negative when measured clockwise.

Even when an angle is given as negative we can change into positive, e.g., angle x = - 32/7 pi = -6pi + 10pi/7 since even multiples ( positive or negative ) of pi take the terminating line to coincide with the positive x axis we are at the starting point.

Thus we will have x = - 32/7 pi = -6pi + 10pi/7 the same as angle x = 10pi/7

It is just as simple as that

Re: Location of negative angle on unit circle

I'm sorry, I don't fully understand what you're saying.

For $\displaystyle -\frac{11\pi}{12}$ I changed it to a positive, $\displaystyle \frac{13\pi}{12}$.

Re: Location of negative angle on unit circle

Quote:

Originally Posted by

**Unreal** Hi!

What quadrant and location is the following angle?

$\displaystyle -\frac{11}{12}$

$\displaystyle -\frac{11}{12} = 2\pi -\frac{11\pi}{12}$

$\displaystyle \frac{13}{12} \implies$ Six and a half revolutions, stopping at $\displaystyle \pi$.

$\displaystyle \pi - \frac{\pi}{12} = \frac{11\pi}{12}$

Quadrant II.

You need to edit this post, as I expect you have left off $\displaystyle \displaystyle \begin{align*} \pi \end{align*}$ from a few of your numbers. The $\displaystyle \displaystyle \begin{align*} \pi \end{align*}$ is incredibly important and without it your solution would change dramatically.

Re: Location of negative angle on unit circle

Post edited

Quote:

Originally Posted by

**unreal** hi!

What quadrant and location is the following angle?

$\displaystyle -\frac{11\pi}{12}$

$\displaystyle -\frac{11\pi}{12} = 2\pi -\frac{11\pi}{12}$

$\displaystyle \frac{13\pi}{12} \implies$ six and a half revolutions, stopping at $\displaystyle \pi$.

$\displaystyle \pi - \frac{\pi}{12} = \frac{11\pi}{12}$

quadrant ii.

Re: Location of negative angle on unit circle

After the first response, the edit option was no longer available.

Thanks anyway.

Re: Location of negative angle on unit circle

A couple of things, NEVER write down that two numbers are equal when they are clearly not. $\displaystyle \displaystyle \begin{align*} -\frac{11\pi}{12} \end{align*}$ is NOT equal to $\displaystyle \displaystyle \begin{align*} 2\pi - \frac{11\pi}{12} \end{align*}$. What you should be saying is that by going through a revolution of the unit circle, the point on the circle which makes an angle of $\displaystyle \displaystyle \begin{align*} -\frac{11\pi}{12} \end{align*}$ is also made by an angle of $\displaystyle \displaystyle \begin{align*} 2\pi - \frac{11\pi}{12} = \frac{13\pi}{12} \end{align*}$.

I have no idea what you have done after this. $\displaystyle \displaystyle \begin{align*} \frac{13\pi}{12} \end{align*}$ IS your angle. Notice that it is $\displaystyle \displaystyle \begin{align*} \pi + \frac{\pi}{12} \end{align*}$, which means it is positioned in the THIRD quadrant at an angle of $\displaystyle \displaystyle \begin{align*} \frac{\pi}{12} \end{align*}$ below the x-axis.

Re: Location of negative angle on unit circle

For - 11pi/12 i would like to change it to -2pi + 13pi/12

Thus we have the angle -11pi/12 the same as angle 13pi/12. since -2pi brings us back to the starting line.

Now 13pi/12 = pi + pi/12 that clearly indicates that the angle is in the third quadrant.