Find ${{\theta }_{n}}$ for minimum $\sum\limits_{n=1}^{N}{\sin {{\theta }_{n}}}$ if $\sum\limits_{n=1}^{N}{\cos {{\theta }_{n}}}=A$

Question

Is there any analytical solution for minimizing $\sum\limits_{n=1}^{N}{\sin {{\theta }_{n}}}$ ?

subject to $\sum\limits_{n=1}^{N}{\cos {{\theta }_{n}}}=A$

&

$0<{{\theta }_{n}}<{\pi/2}$

Answer 1

This follows a reasoning through Lagrange's multipliers. Observe that $A>0$.

Let $f(\theta_1,\dots,\theta_N)=-\sum_i\sin(\theta_i)$ and let $g(\theta_1,\dots,\theta_N)=\sum_i\cos(\theta_i)$. Then the problem is equivalent to

\begin{align}&\text{maximize }&&f(\theta_1,\dots,\theta_N)\\ &\text{subject to }&&g(\theta_1,\dots,\theta_N)=A\end{align}

We have that $\nabla g=(-\sin(\theta_1),\dots,-\sin(\theta_N))$ and that $\nabla f=(-\cos(\theta_1),\dots,-\cos(\theta_N))$. Hence, the problem is equivalent to solving the following system:

\begin{equation} \left\{\begin{array}{l} \cos(\theta_1)=\lambda\cdot\sin(\theta_1)\\ \dots\\ \cos(\theta_N)=\lambda\cdot\sin(\theta_N)\\ \sum_i\cos(\theta_i)=A\end{array}\right. \end{equation}

We have that $\sin(\theta_i)^2+\cos(\theta_i)^2=(1+\lambda^2)\sin(\theta_i)^2=1$ so that for all $i$ it must be that

$$\sin(\theta_i)=\sqrt{\frac{1}{1+\lambda^2}}\,,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\cos(\theta_i)=\lambda\cdot\sqrt{\frac{1}{1+\lambda^2}}$$

It follows that $A=N\lambda\cdot\sqrt{\frac{1}{1+\lambda^2}}$ must be satisfied. The only solution is hence

$$\lambda=\frac{A}{\sqrt{N^2-A^2}}$$

Now, notice that if the first $n$ equations are satisfied, then

$$\sum_i\cos(\theta_i)=\lambda\sum_i\sin(\theta_i)=-\lambda\cdot f(\theta_1,\dots,\theta_N)$$

Hence, the last equation is equivalent to $f(\theta_1,\dots,\theta_N)=-A/\lambda$ and the value is

$$\sqrt{N^2-A^2}$$

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But is this actually the sought minimum? No. Let's try an example.

If we set $A=N-1$, then our value is $\sqrt{2N-1}$. However, if we take $\theta_1=\theta_2=\dots=\theta_{N-1}=0$ and $\theta_N=\pi/2$, the conditions will be satisfied but our value will be $1$, which is less than $\sqrt{2N-1}$ for $N\geq 2$.

Of course, these exact values of $\theta$ are not in the domain. Nonetheless, if we change them slightly -- $\theta_1=\theta_2=\dots=\theta_{N-1}=\epsilon_0>0$ and $\theta_N$ the angle slightly less than $\pi/2$ that makes the constraint true --, the value of $1$ will also change (increase) very slightly.

This should be an indicator that the minimum cannot be attained. Suppose the $\theta_i$'s lie in $[0,\pi/2]$, so that by compacity there is a minimum to $\sum_i\sin(\theta_i)$.

Claim: No minimum lies in $(0,\pi/2)$.

Proof: Indeed, suppose there is some $k$ such that $\theta_k$ is minimal among nonzero $\theta$'s and $j$ such that $\theta_k<\theta_j<\frac{\pi}{2}$. We will show that by making $\theta_j$ larger and $\theta_k$ smaller (respecting the constraint), the sum of sines becomes smaller.

More precisely, there are real functions $\Theta_i,\Theta_k$ defined on $[0,\epsilon)$ with

\begin{equation}\left\{\begin{array}{l} \Theta_j(0)=\theta_j,\,\,\,\, \Theta_j \text{ increasing}\\ \Theta_k(0)=\theta_k,\,\,\, \Theta_k \text{ decreasing}\\ \cos(\Theta_j(t))+\cos(\Theta_k(t))=\text{ constant} \end{array}\right.\end{equation}

We will choose $\Theta_j,\Theta_k$ smooth, so the last equation is equivalent to

\begin{align}\tag{*}\label{eq1}\sin(\Theta_j(t)){\Theta_j}'(t)+\sin(\Theta_k(t)){\Theta_k}'(t)=0\end{align}

Taking, say, $\Theta_j(t)=\theta_j+t$, the existence and uniqueness theorem for ODE's guarantees that there is a uniquely defined $\Theta_k$ with $\Theta_k(0)=\theta_k$ satifying $\eqref{eq1}$ for $t$ near $0$. Since the sines are positive and ${\Theta_j}'>0$, equation $\eqref{eq1}$ also implies ${\Theta_k}'<0$, so the monotonicity conditions are also satisfied.

On the other hand, substituting $\Theta_j(t)=\theta_j+t$ into equation $\eqref{eq1}$ yields

$$\sin(\theta_j+t)+\sin(\Theta_k(t)){\Theta_k}'(t)=0,$$

so that

\begin{align} &\frac{d}{dt}\Big(\sin(\Theta_j(t))+\sin(\Theta_k(t))\Big)=\cos(\theta_j+t)+\cos(\Theta_k(t)){\Theta_k}'(t)<0\\ \iff &{\Theta_k}'(t)<\frac{-\cos(\theta_j+t)}{\cos(\Theta_k(t))} \iff \frac{-\sin(\theta_j+t)}{\sin(\Theta_k(t))}<\frac{-\cos(\theta_j+t)}{\cos(\Theta_k(t))} \\ \iff &-\sin(\Theta_k(t))\cos(\theta_j+t)+\sin(\theta_j+t)\cos(\Theta_k(t))>0\\ \iff &\sin((\theta_j+t)-\Theta_k(t))>0 \end{align}

For $t=0$, $\Theta_k(t)=\theta_k<\theta_j$, so that indeed the inequalities hold and the sum of sines decreases.

This fails only in one of two cases:

$(1):$ There is no such index $k$, meaning all the $\theta$'s are $0$. This is not a problem: it can only happen when $A=0$, and in this case the only possibility is all $\theta$'s equal $0$ anyway.

$(2):$ There is no such index $j$, which implies one of the following:

$\,\,\,\,(2.1):$ All nonzero $\theta$'s coincide (violating $\theta_k<\theta_j$).

$\,\,\,\,(2.2):$ All $\theta$'s greater than $\theta_k$ are equal to $\pi/2$.

Since $(1)$ and $(2.2)$ already do not lie in $(0,\pi/2)$, we need only show that on $(2.1)$ having all $\theta$'s coincide and be nonzero does not achieve a minimum. We will do so via explicit calculations.

---

Let $m=\lfloor A\rfloor \in \mathbb{Z}$. It's easy to see that at most $m$ of the $\theta$'s may be $0$. For each $k=0,1,\dots,m$, we will calculate the sum $S(k,A)$ of $N$ sines, where $k$ of them are $0$ and the other $N-k$ coincide and are such that the $A$-constraint is satisfied.

We denote the nonzero angle simply by $\theta$. The constraint implies that

$$(N-k)\cos(\theta)=A-k\implies \cos(\theta)=\frac{A-k}{N-k}$$

It follows that that

$$\sin(\theta)=\frac{\sqrt{{(N-k)}^2-{(A-k)}^2}}{N-k}\,,$$

so the sum of sines is

$$S(k,A)=\sqrt{{(N-k)}^2-{(A-k)}^2}=\sqrt{(N^2-A^2)-2k(N-A))}\,.$$

We see readily that $S(k,A)$ is decreasing in $k$, which concludes the proof. $\square$

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Now, what is the actual minimum? We see that $S(m,A)$ is the least of the $S(k,A)$'s, but to find the minimum it suffices to check the values of $(2.2)$. Since we've come all the way here, we might as well do it.

In this last case, there are some $k$ angles $0$, some $l$ angles $\pi/2$, and the rest coincide. The $A$ constraint becomes

$$(N-k-l)\cos(\theta)=A-k\implies \cos(\theta)=\frac{A-k}{N-k-l}$$

Then

$$\sin(\theta)=\frac{\sqrt{{(N-k-l)}^2-{(A-k)}^2}}{N-k-l}\,,$$

and the sum of sines is

\begin{align} S(k,l,A)&=l+\sqrt{{(N-k-l)}^2-{(A-k)}^2}\\ %&=l+\sqrt{(N^2-A^2)-2k(N-A))+l^2-2l(N-k)}\,. \end{align}

Well, is this more or less than $S(k,A)=S(k,0,A)$? To answer this, we will consider $g(l)=S(k,l,A)$ as a real function of $l$. We have that

\begin{align} g'(l)&=1-\frac{N-k-l}{\sqrt{{(N-k-l)}^2-{(A-k)}^2}}\\ &=1-\frac{1}{\sin(\theta)}\leq 0 \end{align}

Therefore, for any given, fixed $k$, we're better off with as high an $l$ as we can afford. The constraints that must be satisfied are

$$\left\{\begin{array}{l} k+l\leq N\\ 0 \leq k \leq \lfloor A \rfloor\\ N-l\geq A\implies l \leq N-A \end{array}\right.$$

Notice that the last two imply the first one. Thus, for each $k=0,1,\dots,\lfloor A \rfloor$, the best $l$ is always $\lfloor N - A \rfloor=N-\lceil A \rceil$, and we have that

\begin{align} S_{\min}(k,A)=S(k,N-\lceil A \rceil, A)&=N-\lceil A \rceil + \sqrt{{(\lceil A \rceil-k)}^2-{(A-k)}^2}\\ &=N-\lceil A \rceil+\sqrt{({\lceil A \rceil}^2-A^2)-2k(\lceil A \rceil-A)} \end{align}

which once again decreases in $k$.

Finally, the minimum is hence when $k$ is maximum. In other words, we're better off taking as many $\theta$'s equal to $0$ and as many $\theta$'s equal to $\pi/2$ as we can. This yields that the minimum is

\begin{align} &N-\lceil A \rceil+ \sqrt{({\lceil A \rceil}^2-A^2)-2\lfloor A \rfloor(\lceil A \rceil-A)}\\ =&N-\lceil A \rceil + \sqrt{{(\lceil A \rceil-\lfloor A \rfloor)}^2-{(A-\lfloor A \rfloor)}^2} \end{align}

Answer 2

Fimpellizieri is right, Lagrange Multipliers is the way forwards. \begin{eqnarray*} \mathcal{L}=\sum_{n=1}^N \sin \theta_n - \lambda \left( \sum_{n=1}^N \cos \theta_n -A\right) \end{eqnarray*} Now partially differentiate this \begin{eqnarray*} \frac{\partial \mathcal{L}}{\partial \theta_n}= \cos \theta_n + \lambda \sin \theta_n \end{eqnarray*} Set this equal to zero and we have \begin{eqnarray*} \lambda = -\tan \theta_n \end{eqnarray*} So Winther is right too, all of the $\theta$ values are equal. The constraint gives \begin{eqnarray*} \cos \theta_n = \frac{A}{N} \end{eqnarray*} and the optimal value is
\begin{eqnarray*} \sum_{n=1}^N \sin \theta_n = \sqrt{N^2-A^2} \end{eqnarray*}

Answer 3

In the plot below, $n=2$ and the plot is over $\theta_{1}$, where $\theta_{2}$ is a solution to $\cos\theta_{1}+\cos\theta_{2}=c$, that is $\theta_{2}=\arccos{(c-\cos\theta_{1})}$. The blue line is for $c=\frac{1}{2}$, the magenta line is for $c=1$ and the yellow line is for $c=\frac{3}{2}$. What the plot shows is that your problem probably has no solution.