Question

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

Answer 1

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

Quadrant	Positive function
$1^{st}$	All
$2^{nd}$	sin and cosec
$3^{rd}$	tan and cot
$4^{th}$	cos and sec

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