Can cos(n!) in degrees tend to one if n>6? does cos(n!)  in degrees tend to 1.
consider cos(n!)=cos(n*...*6!),6!=720=360*2.So this is like rotating on the plane n*...7*2 times so cos(n!)=1,When n>6 .Does this proof hold even when n tends to infinity.please give a reply which suits the mind of a beginner.
Indeed $\cos (n!) = 1$ for $n \geq 6$ since $n! \equiv 0 \pmod{360}$. But $n$ must be a finite number, $\cos \infty$ makes no sense, but we can say that $\displaystyle\lim_{n\to\infty} \cos (n!) = 1$:

We say that the sequence $\{a_n\}$ tends to the finite limit $l$ as $n$ goes to infinity if:

$\forall \varepsilon &gt;0\ \exists n_0 &gt; 0$ so that if $n &gt; n_0$ then $|a_n - l| &lt; \varepsilon$.

Taking $l = 1$ and $n_0 = 5$ then regardless of how $\varepsilon$ is, the definition holds, so indeed, $$\lim_{n\to\infty} \cos (n!) = 1$$
Using Riemann zeta function,infinite factorial can be regularized $$\prod_{n=1}^\infty\lambda_n=\exp(-\zeta'_{\lambda}(0))\\\zeta_\lambda(s)=\sum_{n=1}^\infty \lambda_n^{-s}\\\zeta_\lambda'(s)=-\sum_{n=1}^\infty\lambda_n^{-s}\log(\lambda_n)$$
Plugging in $\lambda_n=n$ we get that
$$\prod_{n=1}^\infty n=\infty!=\sqrt{2\pi}$$
NOTE: $\zeta'(0)=-\frac{1}{2}\ln({2\pi})$ which can be derived directly from the Wallis formula

So $\cos(\infty!)=\cos(\sqrt{2\pi})\approx0.99904$