Answer
$\cos (\dfrac{3}{4})$
Work Step by Step
Since, $\cos x=1-\dfrac{x^2}{2!}-.....+(-1)^n \dfrac{x^{2n}}{(2n)!}$
Now, $1-\dfrac{3^2}{4^2 (2!)}+\dfrac{3^4}{4^4 (4!)}-\dfrac{3^6}{4^6 (6!)}+....=1-(\dfrac{1}{2!})(\dfrac{3}{4})^2+(\dfrac{1}{4!})(\dfrac{3}{4})^4-...$
Thus,
$1-(\dfrac{1}{2!})(\dfrac{3}{4})^2+(\dfrac{1}{4!})(\dfrac{3}{4})^4-...=\cos (\dfrac{3}{4})$
