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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 \(n\ge1\) by
\[n!!=\begin{cases}n\times(n2)\times\cdots\times5\times3\times1&\text{for odd }n\\[1ex]
n\times(n2)\times\cdots\times6\times4\times2&\text{for even }n\\[1ex]
0&n=1,0\end{cases}\]
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 \(n\ge1\) by \[n!!=\begin{cases}n\times(n2)\times\cdots\times5\times3\times1&\text{for odd }n\\[1ex] n\times(n2)\times\cdots\times6\times4\times2&\text{for even }n\\[1ex] 0&n=1,0\end{cases}\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$2^{\frac{\cos(4030\pi)1}42015}=2^{\frac{\cos(0)1}42015}=2^{2015}=\frac1{2^{2015}}$$ 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 $$2015!$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correct, as usual! A simple "contest math" question
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