Find ${{\theta }_{n}}$ for minimum $\sum\limits_{n=1}^{N}{\sin {{\theta }_{n}}}$ if $\sum\limits_{n=1}^{N}{\cos {{\theta }_{n}}}=A$ Is there any analytical solution for minimizing $\sum\limits_{n=1}^{N}{\sin {{\theta }_{n}}}$ ?
subject to
 $\sum\limits_{n=1}^{N}{\cos {{\theta }_{n}}}=A$ 
& 
$0<{{\theta }_{n}}<{\pi/2}$
 A: 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*}
A: 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.

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

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. $\square$

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}

