I am about to start university in October, to study computer science, and have been asked by my university to complete a number of problem sheets. I have become stuck on the following question, and therefore would appreciate any help possible.
The numbers $x$ and $y$ are subject to the constraints $x+y=\pi$. Find the values of $x$ and $y$ for which $\cos x\sin y$ takes its minimum value.
Using this question as a starting point towards a solution, I have the following steps attempted so far.
\begin{align*} \Lambda(x, y, \lambda)&=\cos x\sin y+\lambda(x+y-\pi)\\ \frac{\partial\Lambda}{\partial x}&=-\sin x\sin y+\lambda=0\\ \frac{\partial\Lambda}{\partial y}&=\cos x\cos y+\lambda=0\\ \frac{\partial\Lambda}{\partial\lambda}&=x+y-\pi=0\\ x&=\cos^{-1}\left(\frac{\lambda}{\sin y}\right)\\ y&=\cos^{-1}\left(\frac{-\lambda}{\cos x}\right)\\ \cos^{-1}\left(\frac{\lambda}{\sin y}\right)+\cos^{-1}\left(\frac{-\lambda}{\cos x}\right)&=\pi\\ \end{align*}
Unfortunately, such maths is way beyond my abilities, as I have only studied A-Level Maths and Further Maths, and I am working from the first answer in the referenced question and the Wikipedia articles on Lagrange Multipliers and Partial Derivatives. I'm unsure of the correct tags to apply, so any help there would also be wonderful.
Edit: After receiving a number of hints, this is part of my solution, however, I'm not sure on how to properly phrase the last bit of the question with respect to properly solving the inequality or expressing values for $y$. \begin{align*} \sin x\cos y&=\sin x\cos(\pi-x)\\ &=-\sin x\cos x\\ &=\sin x\cos x\\ &=\frac12\sin2x\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac12\sin2x\right)&=\cos2x\\ \cos2x&=0\Rightarrow x=\frac12\left(n\pi-\frac\pi2\right),n\in\mathbb{Z}\\ \frac{\mathrm{d}}{\mathrm{d}x}\cos2x&=-2\sin2x\\ -2\sin2x>0&\Rightarrow\sin2x<0\\ \end{align*}