I have a series as follows: $$1- \frac{3^2}{4^2 \times 2!} +\frac{3^4}{4^4 \times 4!}-\frac{3^6}{4^6 \times 6!}$$
I understand that this is a different form of a cos(x) Maclaurin series. Replace the x with the 3 and it matches the numerator. I am unsure how to manipulate the cos(x) to get the extra $$4^{2n}$$ in with the factorial in the denominator.
$$ \sum_{i\ge 0}(-)^i\frac{3^{2i}}{4^{2i}(2i)!} $$ is a special case of $$ \cos x = \sum_{i\ge 0}(-)^i\frac{1}{(2i)!}x^{2i}, $$ at $x=3/4$, $$ \sum_{i\ge 0}(-)^i\frac{3^{2i}}{4^{2i}(2i)!} =\cos(3/4). $$