I have been struggling to solve the following problem which seems to be some kind of generalized trigonometric Pythagorean identity:
Let $x_{ij}$ $(1 \leq i \leq m$, $1\leq j \leq n)$ be real numbers, then prove that:
$$\prod_{j=1}^{n}\left(1-\prod_{i=1}^{m}\sin^2(x_{ij})\right) + \prod_{i=1}^{m}\left(1-\prod_{j=1}^{n}\cos^2(x_{ij})\right) \geq 1$$
I have tried the following induction:
Prove for $n=1$ and any $m$ and then do an induction on $n$. The base case is immediate using $1-\cos^2(x_{ij}) = \sin^2(x_{ij})$. However I have not managed to pull off the inductive step.
If anyone has a contribution it would be much apreciated.
Best Answer
Since $\sin^2x_{ij}$ is an arbitrary number in $[0,1]$, and $\cos^2x_{ij}=1-\sin^2x_{ij}$, let's rewrite your expression, with $p_{ij}\in[0,1]$:
$$\prod_{j=1}^n(1-\prod_{i=1}^m p_{ij})+\prod_{i=1}^m(1-\prod_{j=1}^n(1-p_{ij}))$$
Now, consider a matrix of independent events $E_{ij}$ on a given probability space, where the probability of $E_{ij}$ occurring is $p_{ij}$.
The first product is the probability of the event $A$: "in every column $j$, at least one event $E_{ij}$ does not occur".
The second product is the probability of the event $B$: "in every row $i$, at least one event $E_{ij}$ is occurring".
Now, we have $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
If $A$ is false, it means that there is a column where all events occur, but then $B$ must be true. Therefore, $P(A\cup B)=1$. And finally, we may write:
$$P(A)+P(B)-1=P(A)+P(B)-P(A\cup B)=P(A\cap B)\in[0,1]$$
Hence $P(A)+P(B)\ge1$, which is exactly your inequality.