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.
 A: 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:
$\forall \varepsilon >0\ \exists n_0 > 0$ so that if $n > n_0$ then $|a_n - l| < \varepsilon$.
Taking $l = 1$ and $n_0 = 5$ then regardless of how $\varepsilon$ is, the definition holds, so indeed, $$\lim_{n\to\infty} \cos (n!) = 1$$
A: The argument you gave is entirely correct. In $n \ge 6$, then
$$ 
$\cos(n!) = 
\cos\big( n \times (n-1) \times (n-2)\times\ldots\times 8 \times 7\times 720\big) = 
\cos(360) = 1
$$
So, for all $n$ greater than $6$, we have $\cos(n!)=1$, and this certainly implies that $\cos(n!)$ tends to $1$ as $n$ tends to infinity. The mathematical way of saying this is
$$
\lim_{n \to \infty} \cos(n!) = 1
$$
But, as others have said, it doesn't make sense to simply substitute $n=\infty$ in the expression $\cos(n!)$, because $\cos(x)$ only makes sense when $x$ is a number, and $\infty$ is not a number (or, not one of the right type, anyway).
A: 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 $\cos(\infty!)=\cos(\sqrt{2\pi})\approx0.99904$
