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

Hint: Try to prove $\cos \left( -a \right)=\cos a$ and $\cos \left( {{360}^{\circ }}-a \right)=\cos a$ by ASTC rule by considering proper quadrants and sign conventions. For $\cos \left( {{360}^{\circ }}-a \right)$, consider the ${{4}^{th}}$ quadrant and for $\cos \left( -a \right)$ ${{1}^{st}}$ and ${{4}^{th}}$ quadrant are to be considered according to the ASTC rule.

Complete step by step answer:
ASTC rule: We have different trigonometric functions like $\sin ,\cos ,\tan $etc. ASTC stands for all, sin, tan, cos. This rule indicates the positivity of a particular trigonometric function on a particular quadrant as per the following table. For even multipliers of angle ${{90}^{\circ }}$, the function remains the same. But for an odd multiplier of angle ${{90}^{\circ }}$ the values change accordingly.

Now let’s consider our question
As we know $\cos \left( -a \right)=\cos a$……….(1) (as cos is positive in ${{1}^{st}}$ and ${{4}^{th}}$ quadrant)
For $\cos \left( {{360}^{\circ }}-a \right)$,
The angle $\left( {{360}^{\circ }}-a \right)$ falls in $4th$ quadrant. Because each quadrant is taken as ${{90}^{\circ }}$ so, 4 quadrants together form ${{360}^{\circ }}$.
Hence, $\cos \left( {{360}^{\circ }}-a \right)=\cos a$……….(2) (with a positive sign because it’s in $4th$ quadrant according to the ASTC rule)
From (1) and (2) we get,
$\cos \left( -a \right)=\cos \left( {{360}^{\circ }}-a \right)=\cos a$
Hence proved.

Note:
ASTC rule should be strictly followed for getting the exact value with proper sign convention. For angles that are an odd multiplier of ${{90}^{\circ }}$, the value of sin becomes cos and vice-versa, tan becomes cot and vice-versa, sec becomes cosec and vice-versa. But the sign convention will be according to sin, tan and sec respectively. Beside ASTC rule, $\left( {{90}^{\circ }}+\theta \right)$ and $\left( {{90}^{\circ }}-\theta \right)$ formulae can also be used.