How do you evaluate $ \sin 360{}^\circ $ .

Hint: We will use the circular function of the trigonometry find the value of $ \sin 360{}^\circ $ . We will also use the formula $ \sin \left( 2\pi +\theta \right)=\sin \left( 360{}^\circ +\theta \right)=\sin \theta $ and we will first find the value of $ \theta $ as $ \theta $ must lies in the range of $ \left[ 0,\dfrac{\pi }{2} \right] $ because we generally do not know the value of trigonometric function for angle greater than $ \dfrac{\pi }{2} $ .

Complete step by step answer:
 We will first recall the concept of the circular function of the trigonometric. All trigonometric functions can also be seen on the unit circle. The unit circle is divided into quadrants by the Cartesian coordinates, so the signs of each circular function can be determined by the value of t. If the value of t has a positive value for both x and y, then it lies in Quadrant I. In Quadrant II, only the value of y is positive, and x has a negative value, so sin t and cosecant t will be positive. All other functions will be negative. In Quadrant III, both coordinates for x and y are negative. This means that tangents and their reciprocal cotangents will be positive. In Quadrant 4, x is positive and y is negative, so cosines and their reciprocal, secants, will be positive.
Since we know that the whole angle of the circle is equal to $ 360{}^\circ $ and when we move in an anticlockwise direction the angle is counted as positive and when we move in clockwise direction angle is counted as negative.
So, to represent $ 360{}^\circ $ we have to move a whole circle in an anticlockwise direction, so the angle will lie on the x-axis so we can say that it is equivalent to $ 0{}^\circ $ and we know that sine is positive in the first quadrant and negative in the fourth quadrant.
So, we can write $ \sin 360{}^\circ =\sin 0{}^\circ $ .
And, since from standard trigonometric table we know that $ \sin 0{}^\circ =0 $
 $ \Rightarrow \sin 360{}^\circ =\sin 0{}^\circ =0 $
This is our required solution.

Note:
We can also solve above question alternatively by using trigonometric formula $ \sin \left( 2\pi +\theta \right)=\sin \left( 360{}^\circ +\theta \right)=\sin \theta $ . When we put $ \theta =0{}^\circ $ , we will get:
 $ \Rightarrow \sin \left( 2\pi +0{}^\circ \right)=\sin \left( 360{}^\circ +0{}^\circ \right)=\sin 0{}^\circ $
 $ \Rightarrow \sin \left( 360{}^\circ \right)=0 $