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Let $m,n$ be natural numbers and let $\alpha_{ij}$ be reals. For each $i=1,2,...,m$; $j=1,2,...,n$. Show that:

$\displaystyle \prod_{j=1}^{n}\left(1-\prod_{i=1}^{m}\cos^2 \alpha_{ij}\right)+ \prod_{i=1}^{m}\left(1-\prod_{j=1}^{n}\sin^2 \alpha_{ij}\right)\ge 1$

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  • $\begingroup$ english please? I have no idea what sean or $sen^{2}$ mean $\endgroup$
    – K Split X
    Aug 1, 2017 at 1:54
  • $\begingroup$ $\mathrm{sen}(\cdot)$ is simply $\mathrm{sin}(\cdot)$, but yes this site uses English. $\endgroup$
    – Eff
    Aug 1, 2017 at 1:58
  • $\begingroup$ @Eff oh i see. strange because in the title s/he wrote $\sin$, $\endgroup$
    – K Split X
    Aug 1, 2017 at 1:59
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    $\begingroup$ Looks like some probabilistic interpretation might help here. $\endgroup$
    – Aryabhata
    Aug 1, 2017 at 2:09

1 Answer 1

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Let $X_{ij}$ be independent random variables with $$P(X_{ij}=0)=\cos^2\alpha_{ij},$$ $$P(X_{ij}=1)=\sin^2\alpha_{ij}.$$ Let $A$ be the event $(\exists i)(\forall j)(X_{ij}=1)$ and $B$ the event $(\exists j)(\forall i)(X_{ij}=0)$. These are mutually exclusive, so $$ P(A\cap B)=0. $$ We have $$\begin{eqnarray*} P(A)&=&1-P(\neg A)\\ &=&1-P\left((\forall i)(\exists j)(X_{ij}=0)\right)\\ &=&1-\prod_iP\left((\exists j)(X_{ij}=0)\right)\\ &=&1-\prod_i\left(1-P((\forall j)(X_{ij}=1))\right)\\ &=&1-\prod_i\left(1-\prod_j P(X_{ij}=1)\right)\\ &=&1-\prod_i\left(1-\prod_j\sin^2\alpha_{ij}\right). \end{eqnarray*}$$ Similarly $$ P(B)=1-\prod_j\left(1-\prod_i\cos^2\alpha_{ij}\right). $$ Hence $$ \prod_j\left(1-\prod_i\cos^2\alpha_{ij}\right)+\prod_i\left(1-\prod_j\sin^2\alpha_{ij}\right) =2-P(A)-P(B). $$ But $1\geq P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)$, so the RHS is $\geq1$.

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    $\begingroup$ Very nice explanation. :-) (+1) $\endgroup$
    – user90369
    Aug 1, 2017 at 8:26

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