Minimum of product of three trigonometric function

Question

If $x,y,z$ are real numbers with $x\geq y\geq z \geq 15^\circ$ and $x+y+z=90^\circ$, then find the range of $\cos x\cos y\cos z$.

I tried Jensen inequality and AM-GM inequality:

$$3(\cos x\cos y\cos z)^{\frac{1}{3}}\leq \cos x+\cos y+\cos z\leq 3 \cos\bigg(\frac{x+y+z}{3}\bigg)=\frac{3\sqrt{3}}{2},$$ so $$ \cos x\cos y\cos z\leq \frac{3\sqrt{3}}{8}.$$

Equality holds when $x=y=z=30^\circ$.

Also,$$ \cos x\cos y\cos z=\frac{1}{4}\bigg[\sin(2x)+\sin(2y)+\sin(2x+2y)\bigg].$$

I am unable to find the minimum of $\cos x\cos y\cos z$.

Answer 1

Note that $x\geqslant y\geqslant z\geqslant15°$ and $x + y + z = 90°$ implies $2y + z\leqslant90°$ and $15°\leqslant z\leqslant30°$, then\begin{align*} \cos x\cos y\cos z &= \frac12(\cos(x + y) + \cos(x - y))\cos z\\ &= \frac12(\cos(90° - z) + \cos(90° - 2y - z))\cos z\\ &= \frac12(\sin z + \sin(2y + z))\cos z \geqslant \frac12(\sin z + \sin 3z)\cos z\\ &= \sin 2z\cos z\cdot\cos z = 2\sin z\cos^3 z. \end{align*} Now define $f(z) = \sin z\cos^3 z$. For $15°\leqslant z\leqslant30°$,\begin{align*} f'(z) &= \cos^4 z - 3\sin^2 z\cos^2 z = (\cos^2 z - 3\sin^2 z)\cos^2 z\\ &= (1 - 4\sin^2 z)\cos^2 z \geqslant 0, \end{align*} thus denoting $z_0 = 15°$, then\begin{align*} &\mathrel{\phantom{=}}{} \cos x\cos y\cos z \geqslant 2\sin z\cos^3 z \geqslant 2\sin z_0\cos^3 z_0\\ &= \sin 2z_0 \cos^2 z_0 = \sin 2z_0 \cdot \frac12(\cos 2z_0 + 1)\\ &= \frac12 \cdot \frac12\left( \frac{\sqrt3}2 + 1 \right) = \frac{2 + \sqrt3}8. \end{align*} Because for $(x, y, z) = (60°, 15°, 15°)$, $\cos x\cos y\cos z = \dfrac{2 + \sqrt3}8$, then the minimum is indeed $\dfrac{2 + \sqrt3}8$.

Answer 2

If I have understood your question correctly, it is to minimize $\cos x\cdot\cos y\cdot\cos z$ with condition $x\ge y\ge z \ge 15$. If I have understood it incorrectly, please write it.

The answer for what I have understood is: $$(x, y, z) = (60, 15, 15)\\\cos x\cdot\cos y\cdot\cos z=\frac{1}{4}+\frac{\sqrt{3}}{8}\approx0.4665$$

As I could not do it algebraically, I decided to create a C++ script with your limitations, which lead to this result. If you are curious about it, here it is:

#include <iostream>
#include <cmath>
#define PI 3.1415926538

using namespace std;

float randomBetween(float minVal, float maxVal){
    return (minVal + (rand()/float(RAND_MAX)) * (maxVal - minVal));
}

int main(){
    cout << "Start! \n" << endl;
    float minimumResult = 1;
    while(1 == 1){
        float x = randomBetween(15.0, 60.0);
        float y = randomBetween(15.0, (75.0 - x));
            float z = 90.0 - x - y;
            float testValue = cos(PI/180.0*x)*cos(PI/180.0*y)*cos(PI/180.0*z);
            if(testValue < minimumResult){
                minimumResult = testValue;
                cout << "(" << x << ", " << y << ", " << z << ") \t" << testValue << endl;
        }
    }
}

Answer 3

The least value of function $$f(x,y) = \cos x\cos y \cos \left(\dfrac\pi2 - x-y\right),$$ or $$f(x,y) = \cos x\cos y \sin(x+y),$$ can be achieved in the stationary points of $f(x)$ or in the bounds of the domain.

The stationary points can be defined from the system $f'_x = f'_y = 0,$ or \begin{cases} -\sin x \cos y \sin(x+y) + \cos x\cos y\cos(x+y)=0,\\ -\cos x \sin y \sin(x+y) + \cos x\cos y\cos(x+y)=0. \end{cases} Subtraction leads to equation \begin{align} \sin(x-y)\sin(x+y)=0. \end{align} Taking in account the issue conditions, this means $$y=x,\quad -\sin x \sin 2x + \cos x \cos 2x = 0,\quad x \in[15^\circ, 37.5^\circ],$$ $$\cos 3x =0,\quad y=x,\quad x \in[15^\circ, 37.5^\circ],$$ \begin{align} x=y=z = 30^\circ,\quad f(30^\circ, 30^\circ) = \dfrac{3\sqrt3}8. \end{align}

The task is symmetric on $x, y, z,$ so WLOG it is sufficient to consider the bounds $z=15^\circ,\ x+y = 75^\circ,$ and $z= 60^\circ,\ x=y = 15^\circ.$

The first bound gives $$g(x) = f(x,75^\circ-x) = \cos15^\circ\cos x\cos (75^\circ-x),\quad x\in[15^\circ,\ 60^\circ],$$ $$g(x) = \dfrac12\cos15^\circ(\cos 37.5^\circ + \cos(37.5^\circ - 2x)),$$ with the least value $$g_m = g(15^\circ) = g(60^\circ) = \cos^2 15^\circ\cos 60^\circ = \dfrac14(1+\cos30^\circ),$$ $$g_m = \dfrac{2+\sqrt3}8 < \dfrac{2\sqrt3+\sqrt3}8 = f(30^\circ, 30^\circ).$$

The second bound gives the same value $g_m.$

Thus, in the issue constraints, $$\boxed{\min \{\cos x\cos y\cos z\} = \dfrac{2-\sqrt3}8\approx 0.46656}$$ at $$\begin{pmatrix}x\\y\\z\end{pmatrix}\in \left\{ \begin{pmatrix}15^\circ\\15^\circ\\60^\circ\end{pmatrix}, \begin{pmatrix}15^\circ\\60^\circ\\15^\circ\end{pmatrix}, \begin{pmatrix}60^\circ\\15^\circ\\15^\circ\end{pmatrix}\right\} $$