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What is the minimum value of $$ \cos x+\cos y+\cos(x-y). $$ Here $x,y$ are arbitrary real numbers. Mathematica gives (with NMinimize) $-3/2$. But I don't know if this is correct and if so, how to prove it.

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The local max or min of a 2-variable function comes where both partial derivatives are 0. So if we say that

$$ z = \cos x+\cos y+\cos(x-y) $$ then $$ \frac{\partial z}{\partial x} = \sin x + \sin(x-y) = 0 $$ and $$ \frac{\partial z}{\partial y} = \sin y - \sin(x-y) = 0 $$ adding the two yields

$$ \begin{align} \sin (x) + \sin(y) &= 0 \\ \rightarrow \sin (x) &= -\sin(y) \\ \rightarrow x &= -y \end{align} $$ plugging that into the first partial yields $$ \begin{align} \sin x + \sin(x+x) &= 0 \\ \rightarrow \sin x &= - \sin(2x) \\ \rightarrow x &= -2x + 2\pi\text{*} \\ \rightarrow 3x &= 2\pi \\ \rightarrow x &= \frac{2\pi}{3} \approx 2.1 \end{align} $$ and $$ y = -\frac{2\pi}{3} \approx -2.1 $$

plugging those values into the original equation yields

$$ \begin{align} \mathcal z &= \cos (\frac{2\pi}{3})+\cos (-\frac{2\pi}{3})+\cos(\frac{4\pi}{3})\\ &= -\frac{1}{2} -\frac{1}{2} -\frac{1}{2}\\ &= -\frac{3}{2} \end{align} $$

*x=0 would fit as well but that would be a local maximum. The proof of that is left for the student.

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  • $\begingroup$ Hi Stanley, thanks for your help. There are some little problems though...like how do I know that this is indeed the global minimum? $\endgroup$
    – Alex
    Sep 2 '15 at 23:37
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    $\begingroup$ Why does $\sin x=-\sin y\rightarrow x=-y$? $\endgroup$
    – user84413
    Sep 2 '15 at 23:53
  • $\begingroup$ @user84413 because the $\sin$ function is negatively reflective across the y axis, meaning $\sin(x) = -\sin(-x)$. $\endgroup$
    – D Stanley
    Sep 3 '15 at 0:45
  • $\begingroup$ @Alex technically you would take the second derivatives and plug in the values. if the second derivatives at those values are both positive it is a local minimum. You can also verify it in excel :) $\endgroup$
    – D Stanley
    Sep 3 '15 at 0:48
  • $\begingroup$ But $\sin\frac{4\pi}{3}=-\sin\frac{\pi}{3}$, and $\frac{4\pi}{3}\ne-\frac{\pi}{3}$. $\endgroup$
    – user84413
    Sep 3 '15 at 20:03
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Consider 3 coplanar unit vectors $\hat{\alpha}$ , $\hat\beta$ , $\hat\gamma$

Let $X$ be the angle between $\hat\alpha$ and $\mathrm{\hat\beta}$ and let $Y$ be the angle between $\hat\beta$ and $\hat\gamma$ then it follows that the angle between $\hat\alpha$ and $\hat\gamma$ is $(X+Y)$, hence we can say that:-

$$\hat\gamma\cdot\hat\alpha + \hat\alpha\cdot\hat\beta + \hat\beta\cdot\hat\gamma= \cos{X} + \cos{Y} + \cos{(X+Y)}$$

$$ {(\hat\alpha + \hat\beta + \hat\gamma)}^{2} \ge 0\\ \implies\hat\alpha^2 + \hat\beta^2 + \hat\gamma^2 + 2(\hat\gamma\cdot\hat\alpha + \hat\alpha\cdot\hat\beta + \hat\beta\cdot\hat\gamma) \ge 0$$ Since $\hat\alpha^2 + \hat\beta^2 + \hat\gamma^2=3$, we can say that:-$$\hat\gamma\cdot\hat\alpha + \hat\alpha\cdot\hat\beta + \hat\beta\cdot\hat\gamma= \cos{X} + \cos{Y} + \cos{(X+Y)} \ge \frac{-3}{2}$$

This result can be extended to the given equation by replacing $Y$ with $(-Y)$ since cos is an even function.

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Since cosine is even, the problem is equivalent to

Minimize $\cos(-x)+\cos(y)+\cos(x-y)$.

which is equivalent to

Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c=0$.

Since cosine is periodic with period $2\pi$, this is equivalent to

Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c\equiv 0\pmod{2\pi}$.

If two or more of $\cos a$, $\cos b$, $\cos c$ are positive, say $\cos a>0$ and $\cos b>0$, then replacing $a$ and $b$ with $a+\pi$ and $b+\pi$ will preserve the constraint $a+b+c\equiv 0\pmod{2\pi}$ and negate $\cos a$ and $\cos b$, reducing the sum. So we can assume that at most one of the three is positive. We thus reduce the problem to:

Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c\equiv 0\pmod{2\pi}$, with $a,b\in[\frac\pi2,\frac{3\pi}2]$ and $c\in[0,2\pi)$.

And now we solve:

\begin{align*} \cos a + \cos b + \cos c &= \cos a + \cos b + \cos (2n\pi - a - b) \\ &= \cos a + \cos b + \cos(a+b) \\ &= \cos a + \cos b + 2\cos^2\big(\tfrac{a+b}2\big) - 1 \\ &\ge 2\cos\big(\tfrac{a+b}2\big) + 2\cos^2\big(\tfrac{a+b}2\big) - 1 &&\text{(cosine is convex on $[\tfrac\pi2,\tfrac{3\pi}2]$)} \\ &= 2\big(\cos\big(\tfrac{a+b}2\big) + \tfrac12\big)^2 - \tfrac32 \\ &\ge -\tfrac32 \end{align*} with equality when $a=b$ and $\cos\big(\tfrac{a+b}2\big) = -\frac12$; for example, $a=b=c=\frac{2\pi}3$.

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Let $$\displaystyle z=\cos x+\cos y+\cos(x-y) = 2\cos\left(\frac{x+y}{2}\right)\cdot \cos\left(\frac{x-y}{2}\right)+2\cos^2 \left(\frac{x-y}{2}\right)-1$$

so we get $$\displaystyle 2\cos^2 \left(\frac{x-y}{2}\right)-2\cos\left(\frac{x+y}{2}\right)\cdot \cos \left(\frac{x-y}{2}\right)-(1+z) =0$$

Now Let $$\displaystyle \cos\left(\frac{x-y}{2}\right) = t\;,$$ and $$\displaystyle \bullet \; -1\le \cos \left(\frac{x\pm y}{2}\right)\le 1$$

So we get $$\displaystyle 2t^2-2\cos \left(\frac{x+y}{2}\right)-(1+z) =0\;,$$ Now for real roots, its $\bf{Discriminant\geq 0}$

So we get $$\displaystyle 4\cos \left(\frac{x+y}{2}\right)+8(1+z)\geq 0$$

So we get $$\displaystyle \cos \left(\frac{x+y}{2}\right)+2(1+z)\geq 0\Rightarrow 2(1+z)\geq \cos \left(\frac{x+y}{2}\right)\geq -1$$

So we get $$\displaystyle (1+z)\geq -\frac{1}{2}\Rightarrow z\geq -\frac{3}{2}$$

and equality hold when $\displaystyle \cos \left(\frac{x+y}{2}\right) = -1=\cos \pi\Rightarrow x+y=2\pi$

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