Four numbers and four equations

Let $a = \sin (x), b = \cos (x), c = \sin (y), d = \cos (y)$ . Then $ac + bd = \sin(x)\sin(y) + \cos(x)\cos(y) = \cos (x - y) = 0$ We have that $x - y = 0 ; x = y + 90(2n + 1)$ where $n$ is a non- negative integer. We then have $ab + cd = \sin(x)\cos(x) + \sin(y)\cos(y) = \frac{\sin(2x) + \sin(2y)}{2} = \frac{\sin(180(2n + 1) + 2y) + \sin(2y)}{2}$ . We have that $\sin(180(2n + 1) + 2y) = \sin(180(2n + 1))\cos(2y) + \cos(180(2n + 1))\sin(2y) = -\sin(2y)$ , since $\sin(180k) = 0$ and \cos(180(2k + 1)) = -1) where $k$ is a non- negative integer( these facts are immediately obvious when we look at the unit circle). So, $\frac{\sin(180(2n + 1) + 2y) + \sin(2y)}{2} = \frac{-\sin(2y) + \sin(2y)}{2} = 0$ . Therefore, the only - and therefore maximum - value of $ab + cd$ is $0$