how to find minimum value of $\cos(a)+\cos(b)+\cos(a+b)$?

Question

How to find the minimum value of the above expression $\cos{a} + \cos{b} + \cos(a+b)$?

I need analytical method (not using contradiction).

Answer 1

By C-S $$\cos{a}+\cos{b}+\cos(a+b)=\cos{a}(1+\cos{b})-\sin{a}\sin{b}+\cos{b}\geq$$ $$\geq\cos{b}-\sqrt{((1+\cos{b})^2+\sin^2b)(\cos^2a+\sin^2a)}=$$ $$=\cos{b}-2|\cos\frac{b}{2}|=2|\cos\frac{b}{2}|^2-1-2|\cos\frac{b}{2}|\geq-\frac{3}{2}.$$ The equality occurs for $|\cos\frac{a}{2}|=\frac{1}{2}$, which says that $-\frac{3}{2}$ is a minimal value.

Done!

Answer 2

defining the function $$f(a,b)=\cos(a)+\cos(b)+\cos(a+b)$$ then we get the partial derivatives with respect to $a,b$ as. $$f_a=-\sin(a)-\sin(a+b)$$ $$f_b=-\sin(b)-\sin(a+b)$$ solve the system $$f_a=0$$ and $$f_b=0$$

Answer 3

We seek critical points of the function $f(x,y)=\cos x + \cos y + \cos(x+y)$. We can restrict our attention to the region $[-\pi,\pi)\times[-\pi,\pi)$, and by symmetry, we can assume $x\leq y$. We calculate partial derivatives:

$$f_x(x,y) = -\sin x - \sin(x+y)$$ $$f_y(x,y) = -\sin y - \sin(x+y)$$

For these to simultaneously equal $0$, we must have $\sin x = \sin y = - \sin(x+y)$. This is certainly satisfied if both $x$ and $y$ are multiples of $\pi$, and also for $x=y=\pm\frac{2\pi}{3}$. Thus, we examine the points:

$(-\pi,-\pi), (-\pi,0), (0,0),(-\frac{2\pi}{3},-\frac{2\pi}{3}),(\frac{2\pi}{3},\frac{2\pi}{3})$.

Plugged into our original function, these yield the respective values:

$-1, -1, 3, -\frac32,-\frac32$

The minimum value we see there is $-\frac32$, so we have a solution, $a=b=\pm\frac{2\pi}{3}$

Answer 4

Since cosine is an even function, therefore we can modify the given expression as: $$\text{minimize} \quad \cos(-x)+\cos(-y)+\cos(x+y).$$ Said differently, $$\text{minimize} \quad \cos(a)+\cos(b)+\cos(c) \quad \text{such that} \quad a+b+c=0.$$ Likewise using the periodicity of cosine we can also have $$\text{minimize} \quad \cos(a)+\cos(b)+\cos(c) \quad \text{such that} \quad a+b+c=2k\pi, \text{where } k \in \mathbb{Z}.$$

Observe that if $\cos a>0$ and $\cos b>0$, then by replacing $a \to a+\pi$ and $b \to b+\pi$ we can still satisfy $a+b+c=2k\pi$ but reduce the value of the expression because $\cos(a+\pi)<0$ and $\cos(b+\pi)<0$.

So without the loss of generality, we can assume that two cosine terms are negative and only one might be positive. In essence, we can confine $a,b \in \left[\frac{\pi}{2},\frac{3\pi}{2}\right]$ and $c \in [0,2\pi]$.

Now we can consider this modified problem and using convexity of cosine on $\left[\frac\pi2,\frac{3\pi}{2}\right]$, we get

\begin{align*} \cos a + \cos b + \cos c &= \cos a + \cos b + \cos (2k\pi - a - b) \\ &= \cos a + \cos b + \cos(a+b) \\ &= \cos a + \cos b + 2\cos^2\left(\frac{a+b}{2}\right) - 1 \\ &\ge 2\cos\left(\frac{a+b}{2}\right) + 2\cos^2\left(\frac{a+b}{2}\right) - 1\\ &= 2\left(\cos\left(\frac{a+b}{2}\right) + \frac{1}{2}\right)^2 - \frac{3}{2} \\ &\geq -\frac{3}{2} \end{align*}