Question

Simplify the expression:
\[\large 4030!!\left(2^{\frac{\cos(4030\pi)-1}{4}-2015}\right)\left(\pi^{\frac{\sin^2(2015\pi)}{2}}\right)\]
where \(n!!\) denotes the double factorial, defined for integers \(n\ge-1\) by
\[n!!=\begin{cases}n\times(n-2)\times\cdots\times5\times3\times1&\text{for odd }n\\[1ex]
n\times(n-2)\times\cdots\times6\times4\times2&\text{for even }n\\[1ex]
0&n=-1,0\end{cases}\]

Answer 1

$$2^{\frac{\cos(4030\pi)-1}4-2015}=2^{\frac{\cos(0)-1}4-2015}=2^{-2015}=\frac1{2^{2015}}$$

similarly $$\pi^{\frac{\sin^2(2015\pi)}2}=\pi^{\sin^2(\pi)/2}=\pi^0=1$$
and also: $$(2n)!!=2^n\cdot n$$ so $$4030!!=2^{2015}\cdot 2015!$$ giving the result $$2015!$$

Answer 2

Correct, as usual! A simple "contest math" question