Finding range of this expression If a line makes  angles $\alpha,\beta,\gamma$ with positive axes,then the range of $\sin\alpha\sin\beta + \sin\beta\sin\gamma +\sin\gamma\sin\alpha$ is? I am a noob in finding range , so please help me from beginning.
 A: If lines makes an angle of $\alpha,\beta,\gamma$ with  positive $x,y$ and $z$ axis, Then $\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma = 1$
So $$\sin^2 \alpha+\sin^2 \beta +\sin^2 \gamma = 2$$
Now Using $$(\sin \alpha+\sin \beta+\sin \gamma)^2 = \sin^2 \alpha+\sin^2 \beta+\sin^2 \gamma+2\sum^{cyc} \sin \alpha \sin \beta$$
So $$2\sum^{cys}\sin \alpha\sin \beta = (\sin \alpha+\sin \beta +\sin \gamma)^2-2$$
Now we have to maximize or minimize $\sin \alpha+\sin \beta+\sin \gamma$ 
subjected to the condition $\alpha+\beta+\gamma = 2\pi$
$\bf{ADDED::}$ For $\max$ Using $\bf{A.M\geq G.M}$
$$\sin^2 \alpha+\sin^2 \beta \geq 2(\sin \alpha \sin \beta)$$
$$\sin^2 \beta+\sin^2 \gamma \geq 2(\sin \beta \sin \gamma)$$
$$\sin^2 \alpha+\sin^2 \gamma\geq 2(\sin \alpha \sin \gamma)$$
Now added all these, We get
So $$\sum^{cyc}\sin \alpha\sin \beta \leq 2$$
A: Take a unit vector $\mathbf u$ parallel to the line: its components will be $(cos\alpha , cos\beta , cos\gamma)$,
with $0 \leqslant \alpha , \; \beta ,\; \gamma \leqslant \pi$, and  $cos^2\alpha + cos^2\beta + cos^2\gamma=1$.
We shall note that the $3$ angles are not independent (there are $2$ degrees of freedom on the unit sphere),
and also that we cannot take two of them as independent variables, since their ranges are interrelated. 
For instance there cannot be two angles null.  

So the best practicable way is to resort to spherical coordinates, e,g.
$$
\begin{gathered}
   - \pi  < \phi  = \text{longitude} \leqslant \pi  \hfill \\
   - \pi /2 \leqslant \theta  = \text{latitude} \leqslant \pi /2 \hfill \\ 
\end{gathered} 
$$
with which we will have
$$
\left\{ \begin{gathered}
  \cos \alpha  = \cos \theta \cos \phi  \hfill \\
  \cos \beta  = \cos \theta \sin \phi  \hfill \\
  \cos \gamma  = \sin \theta  \hfill \\ 
\end{gathered}  \right.\quad \left\{ \begin{gathered}
  \sin \alpha  = \sqrt {1 - \cos ^{\,2} \theta \cos ^{\,2} \phi }  \hfill \\
  \sin \beta  = \sqrt {1 - \cos ^{\,2} \theta \sin ^{\,2} \phi }  \hfill \\
  \sin \gamma  = \sqrt {1 - \sin ^{\,2} \theta }  \hfill \\ 
\end{gathered}  \right.
$$
Note that
$$
\begin{gathered}
  0 \leqslant \sin \alpha  = \cos \left( {\pi /2 - \alpha } \right)\quad \left| {\;0 \leqslant \alpha  \leqslant \pi } \right. \hfill \\
  \cos ^{\,2} \alpha  + \cos ^{\,2} \beta  + \cos ^{\,2} \gamma  = 1\quad  \Rightarrow \quad \sin ^{\,2} \alpha  + \sin ^{\,2} \beta  + \sin ^{\,2} \gamma  = 2 \hfill \\ 
\end{gathered} 
$$
so the sines are non-negative, and we can take the plus sign for the square root.
Therefore we can work to find the range of the function
$$
\begin{gathered}
  f(\theta ,\phi ) = \sin \alpha \sin \beta  + \;\sin \beta \sin \gamma  + \;\sin \gamma \sin \alpha  =  \hfill \\
   = \sqrt {\left( {1 - \cos ^{\,2} \theta \cos ^{\,2} \phi } \right)\left( {1 - \cos ^{\,2} \theta \sin ^{\,2} \phi } \right)}  +  \hfill \\
   + \sqrt {\left( {1 - \cos ^{\,2} \theta \sin ^{\,2} \phi } \right)\left( {1 - \sin ^{\,2} \theta } \right)}  +  \hfill \\
   + \sqrt {\left( {1 - \cos ^{\,2} \theta \cos ^{\,2} \phi } \right)\left( {1 - \sin ^{\,2} \theta } \right)}  \hfill \\ 
\end{gathered} 
$$
or, better, just work with 
$$
g(\theta ,\phi ) = \sin \alpha  + \sin \beta  + \;\sin \gamma 
$$
since
$$
g(\theta ,\phi )^2  = 2 + 2f(\theta ,\phi )
$$
and find that
$$
1 \leqslant f(\theta ,\phi ) \leqslant 2
$$
A: It will be maximum when two angles are 90° and other one is 0°. While it will be minimum when all three angles are 45°
