Find the minimum of $\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$

Let $0<x,y<\frac {\pi}{2}$ such that $\sin (x+y)=\frac 23$, then find the minimum of

$$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$$

A) $\frac 23$

B) $\frac 43$

C) $\frac 89$

D) $\frac {16}{9}$

E) $\frac{32}{27}$

My attempts:

I think that the all possible answers are wrong. Because, by Am-Gm inequality we have

$$\frac{\sin x}{\cos y}+\frac{\cos y}{\sin x}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}≥2+2=4.$$

But, Wolfram Alpha gives us a different result : The Global Minimum doesn't exist. However, ​the local minimum must be $6.$

But, still the problem is not solved. Because Wolfram's graph shows that the minimum can be less than $6$.

I also tried

Let $$\sin x=a,\cos y=b,\cos x=c,\sin y=d$$ with

$$ab+cd=\frac 23≥2\sqrt{abcd}\implies abcd≤\frac 19\\ a^2+c^2=b^2+d^2=1 $$

then I need

$$\min \left(\frac ab+\frac ba+\frac cd+\frac dc\right)$$

But, I can't do anything from here. Finally, I attach the graph drawn by WA.

Solution 1:

Simplifying gives: $$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}=\frac{\sin^2 x+\cos^2y}{\sin x\cos y}+\frac {\cos^2 x+\sin^2y}{\sin y\cos x}=\frac{\sin(x+y)\sin(x-y)+1}{\sin x\cos y}+\frac {1-\sin(x+y)\sin(x-y)}{\sin y\cos x}=\frac {8}{3\sin2x\sin 2y}+\frac{2\sin (x-y)}{3}\left(\frac{1}{\sin x\cos y}-\frac{1}{\sin y\cos x}\right)=\frac {8}{3\sin2x\sin 2y}-\frac 83\frac{\sin^2(x-y)}{\sin2x\sin 2y}=\frac{8\cos^2(x-y)}{3\sin 2x\sin2y}=\frac{16\cos^2(x-y)}{3(\cos 2(x-y)-\cos 2(x+y))}=\frac{16}{3(2-10/9\sec^2(x-y))}$$

The denominator is maximum when $\sec^2(x-y)$ is minimum, which happens when $x=y$, in which case the quantity has value $6$. So the required global minimum should be $6$.

Edit: The edit shows that the given expression (let's call it $f(x,y)$ for brevity) can't take any value smaller than $6$. To see that, let's first note (Refer Note) that $\color{blue}{ 2-\frac {10}9\sec^2(x-y)\gt 0 \text{ for all } (x,y)\in (0,\frac \pi 2)\times (0,\frac \pi 2)}$.

Suppose on the contrary that there exist some $a$ and $b$ in $(0,\frac \pi 2)$ such that $f(a,b)<6$. It follows that $$\frac{16}{3(2-10/9\sec^2(x-y))}\lt 6\implies 8<18-10\sec^2(x-y)\implies \sec^2(x-y)\lt 1 $$ and this is a contradiction. So the assumption that $f$ takes any value less than $6$ is wrong. Hence global minimum value of $f$ under the given conditions is $6$.

Note: If $\sec^2(x-y)\ge \frac{18}{10}$ then $\cos^2(x-y)\le \frac {10}{18}\implies 0\lt \cos (x-y)\le \sqrt {\frac{10}{18}}$ (because $x-y \in (-\frac \pi 2, \frac \pi 2)$ so $\cos $ is +ve) so let's consider two cases: 1) $0\le x-y)\lt \frac \pi 2 \text{ and } 2) -\frac \pi 2\lt x-y \lt 0$.

Case 1: it follows that $x-y\ge \arccos\sqrt {\frac{10}{18}}$ (noting that $\cos$ is decreasing on $[0,\pi/2))$. Given that $x+y=\arcsin \frac 23$, it follows that $y\le 0$ which is not possible.

Case 2: it follows that $x-y\le -\arccos\sqrt {\frac{10}{18}}$ (noting that $\cos$ is increasing on $(-\frac \pi 2, 0))$. Then proceeding as in case 1) results in a contradiction.

This establishes the blue colored part.

Solution 2:

Let $t:=x-y$ and $u:=x+y$. We have

$$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}=2\cos(t)(\csc(2y)+\csc(2x))=\frac{4\sin(u)\cos^2(t)}{\cos^2(t)-\cos^2(u)}.$$ As $u$ is known to be constant, we have a univariate function in $t$. It has a stationary point at $t=0$, giving

$$\sin(x+y)=\sin(2x)=\frac23$$ and

$$2\tan(2x)+2\cot(2x)=6.$$

Solution 3:

You are correct. Let $$f(x,y) = g + \frac{1}{g} + h + \frac{1}{h}$$ where $$g(x,y) = \frac{\sin x}{\cos y}, \quad h(x,y) = \frac{\cos x}{\sin y}.$$ Then since $0 \le x, y \le \pi/2$, we must have $0 \le \sin x, \cos x, \sin y, \cos y \ge 1$, hence $g, h \ge 0$. Then AM-GM proves $g + 1/g \ge 2$, as well as $h + 1/h \ge 2$, thus $f \ge 4$, and this is without the added constraint $\sin (x+y) = \frac{2}{3}$. None of the answer choices is greater than $4$, so all are incorrect.

But also be advised that Wolfram Alpha is also misleading, because we have established that it is impossible for $f$ to be negative. Thus there must be a lower bound.

The actual minimum is $6$. To see this, let $c_1 = \arcsin \frac{2}{3} \approx 0.729728$, and $c_2 = \pi - c_1 \approx 2.41186$. These comprise the two possible values for $x + y$ in the square $[0,\pi/2]^2$. Then the transformation $u = x+y$, $v = x-y$ gives $$f = 8\frac{\sin u \cos^2 v}{\cos 2v - \cos 2u} = 4 \frac{\sin u \cos^2 v}{ \cos^2 v + \sin^2 u - 1}$$ Then since $\sin u = 2/3$, $$f = \frac{24\cos^2 v}{-5 + 9 \cos^2 v} = \frac{8}{3} + \frac{\frac{40}{27}}{t - \frac{5}{9}}$$ where $t = \cos^2 v$. Consider the range of $t$. Since $x + y \in \{c_1, c_2\}$ and $0 \le x, y \le \pi/2$, it follows that $0 \le |x-y| \le c_1$, and in turn, $$\frac{5}{9} \le t \le 1.$$ As $f$ is obviously a decreasing function of $t$ on this interval, the minimum is attained when $t$ is as large as possible, namely $t = 1$, corresponding to $v = 0$, or $x = y \in \{c_1/2, c_2/2\}$, and $$f = \frac{8}{3} + \frac{40/27}{4/9} = 6.$$

Here is a plot of $f(x,y)$ and the two curves for which $\sin (x+y) = \frac{2}{3}$. As you can see, they do have a global minimum, and as $f$ is never negative, neither are the curves.