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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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Hint: If $x+y = \pi$, then $y = \pi - x$, so you've to minimize $\cos(\pi - x)\sin x$ and $\cos(\pi - x) = -\cos x$ ... – martini Jul 19 '12 at 17:07
In $\,\Lambda_x\,$ you need a minus sign in the right side, as $\,(\cos x)'=-\sin x\,$ – DonAntonio Jul 19 '12 at 17:07
Oops, and also the right side of your $\,\Lambda_y\,$ as a minus sign that doesn't belong there...I think you have the signs of the derivatives of sine and cosine switched. – DonAntonio Jul 19 '12 at 17:10
A better conclusion to obtain from the Lagrange step is that $\lambda = \cos x \cos y = - \sin x \sin y$, which gives $\cos x \cos y + \sin x \sin y = \cos (x-y) = 0$. In such problems, the multipliers are often better at suggesting a form for the solution, rather than the solution itself. – copper.hat Jul 19 '12 at 17:42
@DonAntonio Thanks, fixed now :) – jClark94 Jul 19 '12 at 18:16
up vote 9 down vote accepted

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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Thanks, I'd tried something similar, but made some stupid mistakes. I'll try that now – jClark94 Jul 19 '12 at 18:13
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 $\Bbb R$. – Cameron Buie Jul 19 '12 at 18:56

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