Answer
See solution
Work Step by Step
The given equation is a linear combination of these two functions: $\cos(\omega t)$ and $\sin(\omega t)$.
It contains the zero vector:
Regardless of the values of the functions, their sum will be 0 if the constants $c_1$ and $c_2$ are 0.
It is closed under vector addition:
Let us take $A=a_1\cos(\omega t)+a_2\sin(\omega t)$ and $B=b_1\cos(\omega t)+b_2\sin(\omega t)$
$A+B=a_1\cos(\omega t)+a_2\sin(\omega t)+b_1\cos(\omega t)+$$b_2\sin(\omega t)=(a_1+b_1)\cos(\omega t)+(a_2+b_2)\sin(\omega t)$
It is closed under scalar multiplication:
Let us take $A=a_1\cos(\omega t)+a_2\sin(\omega t)$. $bA=a_1b\cos(\omega t)+a_2b\sin(\omega t)$
Because it fulfills these three conditions, it is indeed a vector space
