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How can an angle be negative?

Writer Andrew Mclaughlin
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How can be angle be negative like sine(-60) , cosine(-50) ? Which quadrant do they fall if we have the negative angles ? I dont see any negative angles in full circle of 360

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4 Answers

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Angles increase positively by increasing counter clockwise. So a negative angle is one that starts in a clockwise direction. 60 is the angle 60 degrees above the x-axis so -60 is the angle 60 degrees below the x-axis.

Angle measures are considered cyclic and any angle $x$ is equal to $x \pm 360$. So $-60$ is the same thing as $300$.

In particular 180 = -180. Also convenient are -90 = 270. And of course 0 = 360.

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Believe it or not you will encounter situations where you need to consider angles of 720 degrees. These are simply 0.

Sometimes you'll have to do something like sin(x + y) or cos(x - y) and have things like x = 135 and y = 243. It's useful to know that 125 + 243 = 368 = 8 and 135 - 243 = -108 make sense. (Actually -108 = 108 clockwise is far more intuitive than 252 degrees is.)

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Hint: $$ \sin (330°)=\sin (-30°) $$ In other words: the positive angles are counterclockwise, the negative angles are clockwise.

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For a negative angle, you simply need to rotate in the other direction (i.e., clockwise instead of anti-clockwise).

This means that for an angle of $-40^\circ$, you end up at the same point (and with the same values of $\sin$ and $\cos$ as an angle of $320^\circ$.

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Negative angles are connected to an orientation on the circle, just as negative abscissae are connected to a direction of travel on a straight line: once you've decided what is the positive direction of travel (from left to right, or from right to left), the straight line becomes an (oriented) axis, and points on this axis can have a positive or negative abscissa.

Similarly, once you've decided what is a positive rotation on a circle (clockwise or counterclockwise), it becomes an oriented (or trigonometric) circle. Remember an angle, in a sense is but a curvilinear abscissa on a circle with radius $1$.

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