How do you prove $\sin ( - a) = \sin ({360^ \circ } - a) = - \sin a$?

Hint: In this question we will use the trigonometric identity formula of $\sin (a + b)$ for equating the given term. On doing some simplification we get the final answer.

Formula used: $\sin (a + b) = \sin a\cos b - \cos a\sin b$

Complete step-by-step solution:
We have the term given to us as:
$ \Rightarrow \sin ( - a)$
Now we know that $\sin ( - a)$ can also be written as $\sin ({360^ \circ } - a)$ because addition of ${360^ \circ }$ doesn’t change the final value of the term therefore we can write is as:
$ \Rightarrow \sin ({360^ \circ } - a)$
Now since the above expression is in the format of $\sin (a + b)$ we can expand it using the formula and write it as:
$ \Rightarrow \sin ({360^ \circ })\cos (a) - \sin (a)\cos ({360^ \circ })$
Now we know that the value of $\sin ({360^ \circ }) = 0$ and the value of $\cos ({360^ \circ }) = 1$
On substituting both the values in the expression we get:
$ \Rightarrow 0 \times \cos (a) - \sin (a) \times 1$
On simplifying we get:
$ \Rightarrow - \sin (a)$
Hence Proved.

Note: It is to be remembered which trigonometric functions are positive and negative in what quadrants.
The formula used over here is for $\sin (a + b)$, the other formulas for the sine and cosine should be remembered.
The other identity formula for cosine and tangent should be remembered too and whenever there is a trigonometric proof required, all the terms in the equation should be converted to the basic trigonometric identities of sine and cosine.
There also exist half angle formulas which are an addition to the general angle’s addition-subtraction formulas.
In this question all the value of the angle is given to us in degrees, the symbol of degrees is $ \circ $ and angle can also be represented in radians where $\pi $ is used which is equal to $180$ degrees.
When you add ${180^ \circ }$ to any angle, its position on the graph reverses, and whenever you add ${360^ \circ }$ to any angle, it reaches the same point after a complete rotation.