Find maximum/minimum for $\cos(2x) + \cos(y) + \cos(2x+y) $

Question

I have not been able to find the critical points for $\cos(2x) + \cos(y) + \cos(2x+y) $

Answer 1

When you set the partial derivatives equal to $0$, you obtain the two equations \begin{align*} \sin 2x + \sin (2x+y) &= 0 \\ \sin y + \sin (2x+y) &= 0\,. \end{align*} This means that we must have $\sin 2x = \sin y$, and so either $y=x+(2k)\pi$ or $y=(2k+1)\pi-x$ for some integer $k$. Substituting, for example, $y=2x$, you'll find that $\sin 2x = 0$ or $\cos 2x = -1/2$. Continue :)

Answer 2

The largest value cosine takes on is 1. Then for $$\cos(2x)+\cos(y) + \cos(2x+y),$$ the largest possible value it could have would be if all three cosine terms were equal to 1.

Now if the first term is 1, then $x$ must be an integer multiple of $\pi$, and if the second term is 1, then $y$ must be an even integer multiple of $\pi$. If these two conditions hold then the third term must also be 1 (check). This let's one say that the maxima of the function occur at $$\{(k\pi,2l\pi) \in \mathbb{R}^2 \vert k,l \in \mathbb{Z} \}. $$

The case of the minima can be handled similarly. For finding general critical points, i.e. finding the saddle points, we'll need to take some derivatives.

The $x$ and $y$ derivatives are $$ 2\sin(2x) + 2\sin(2x+y)$$ and $$ \sin(y) + \sin(2x+y)$$ respectively. For both expressions to be zero we must have $$\sin(2x) = -\sin(2x+y) = \sin(y).$$ For the first expression to equal the last we must have that

$$\sin(2x) - \sin(y) = 2\sin({2x - y \over 2}) \cos({2x + y\over 2}) = 0$$

where we use a trig identity for the difference of sines. So either $2x = y+2m\pi $ or $ 2x = -y + (2m+1)\pi $ for some integer $m$. In the first case, $$\sin(y) = -\sin(2x + y) = -\sin(2y + 2m\pi) = -\sin(2y) $$ implies $$ \sin(y) + \sin(2y) = 2\sin({y + 2y \over 2})\cos({y - 2y\over 2}) = 2 \sin({3\over 2}y) \cos({1\over 2} y) = 0.$$ Solving $\sin(3y/2) = 0$ and satisfying $2x = y + 2m\pi$ gives the collection of critical points $$\{(k\pi,2l\pi) \in \mathbb{R}^2 \vert k,l \in \mathbb{Z} \}.$$ Solving $\cos(y/2) = 0$ and satisfying $2x = y+ 2m\pi$ gives $$\{({(2k+1)\pi\over 2},(2l+1)\pi) \in \mathbb{R}^2 \vert k,l \in \mathbb{Z} \}.$$ To finish, go back and assume the second possible choice, $2x = -y + (2m+1)\pi$, and solve the equation $\sin(y) = - \sin(2x+y)$ again using the same techniques.

Answer 3

This might be a solution.

Set $X=2x$ and $\cos(X)=(1-t^2)/(1+t^2),\cos(y)=(1-s^2)/(1+s^2)$,$\sin(X)=(2t)/(1+t^2),\cos(y)=(2s)/(1+s^2)$, then the original expression becomes:

$$\cos(X) + \cos(y) + \cos(X+y)=f(s,t)=\frac{3 - s^2 - 4 s t - t^2 - s^2 t^2}{(1+s^2)(1+t^2)}$$

where $$f(s,t)=3 - s^2 - 4 s t - t^2 - s^2 t^2$$

From $f(s,t)=0$, we can solve for $t$:

$$t=\frac{-2 s \pm \sqrt{3 + 6 s^2 - s^4}}{1 + s^2}$$

Notice that: $$\sqrt{3 + 6 s^2 - s^4}=\sqrt{(s^2-(3-2\sqrt{3}))(3+2\sqrt{3}-s^2)}$$

Thus when $3+2\sqrt{3}\ge s^2\ge(3-2\sqrt{3})$ and $s\in \mathbb{R}$, we have real solutions for $t$. Thus we require $3+2\sqrt{3}\ge s^2\ge 0$.