anonymous one year ago 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 $n!!=\begin{cases}n\times(n-2)\times\cdots\times5\times3\times1&\text{for odd }n\\[1ex] n\times(n-2)\times\cdots\times6\times4\times2&\text{for even }n\\[1ex] 0&n=-1,0\end{cases}$

1. anonymous

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

2. anonymous

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