Recall the Maclaurin expansion for cos(x), valid for all real x :
[tex]\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}[/tex]
Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives
[tex]\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}[/tex]
The first 3 terms of the series are
[tex]\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}[/tex]
and the general n-th term is as shown in the series.
In case you did mean cos(√(5x)), we would instead end up with
[tex]\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}[/tex]
which amounts to replacing the x with √x in the expansion of cos(√5 x) :
[tex]\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}[/tex]
