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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*}

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    @copper.hat Thanks for the tip, I've never been taught how to use the method, so that wouldn't have occurred. Looks similar to solving linear $2^{\text{nd}}$ order differential equations.2012-07-19

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You've made this unnecessarily complicated. Note that the constraint $x+y=\pi$ means that you can substitute $y=\pi-x$ into your original expression. Thus, you need only minimize the $1$-variable function $\cos(x)\sin(\pi-x)$, which can be achieved through methods of basic calculus.

Edit: One can make this even simpler by using the identities $\sin(\pi-x)=\sin(x)$ and $2\cos(x)\sin(x)=\sin(2x)$.

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    As it turns out, once you know you need to minimize $\frac12\sin(2x)$, you don't even need methods of calculus, since the sine function varies from $-1$ to $1$ inclusive over $\Bbb R$.2012-07-19