Find the largest and smallest values of the sum of the acute angles that a line through the origin makes with the three coordinate axes.

Let the line through the origin makes angles $\alpha ,\beta \text{and}\gamma$with the three coordinates, respectively.

Then, the sum of angles will be:

$f\left(\alpha ,\beta ,\gamma \right)=\alpha +\beta +\gamma \left(1\right)$

Now, we have:

$$\begin{gathered} {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1 \\ {\cos }^2\gamma =1-{\cos }^2\alpha -{\cos }^2\beta \\ {\cos }^2\gamma ={\sin }^2\alpha -{\cos }^2\beta \\ \cos \gamma =\sqrt{{\sin }^2\alpha -{\cos }^2\beta } \\ \gamma ={\cos }^{-1}\sqrt{{\sin }^2\alpha -{\cos }^2\beta } \left(2\right) \end{gathered}$$

From equations (1) and (2), we get:

$f\left(\alpha ,\beta ,\gamma \right)=\alpha +\beta +{\cos }^{-1}\sqrt{{\sin }^2\alpha -{\cos }^2\beta }$

To find the minimum or maximum angle, we have:

$\frac{\partial f}{\partial \alpha }=0\text{and}\frac{\partial f}{\partial \beta }=0$

For, $\frac{\partial f}{\partial \alpha }=0$:

$$\begin{gathered} \frac{\partial f}{\partial \alpha }=0 \\ 1-\frac{1}{\sqrt{1-{\sin }^2\alpha +{\cos }^2\beta }}\frac{2\sin \alpha \cos \alpha }{2\sqrt{{\sin }^2\alpha -{\cos }^2\beta }}=0 \\ \left({\cos }^2\alpha +{\cos }^2\beta \right)\left({\sin }^2\alpha -{\cos }^2\beta \right)={\sin }^2\alpha {\cos }^2\alpha \\ {\cos }^2\beta \left({\sin }^2\beta -2{\cos }^2\alpha \right)=0 \end{gathered}$$

On further evaluation, we get:

$$\begin{gathered} {\cos }^2\beta =0\text{or}{\sin }^2\beta =2{\cos }^2\alpha \\ \Rightarrow \beta =90°\text{or}{\sin }^2\beta =2{\cos }^2\alpha \end{gathered}$$

Similarly for, $\frac{\partial f}{\partial \beta }=0$:

$$\begin{gathered} \frac{\partial f}{\partial \beta }=0 \\ 1-\frac{1}{\sqrt{1-{\sin }^2\alpha +{\cos }^2\beta }}\frac{-2\cos \beta \left(-\sin \beta \right)}{2\sqrt{{\sin }^2\alpha -{\cos }^2\beta }}=0 \\ \left({\cos }^2\alpha +{\cos }^2\beta \right)\left({\sin }^2\alpha -{\cos }^2\beta \right)={\cos }^2\beta {\sin }^2\beta \end{gathered}$$

As we have $\beta =90°$, we get:

$$\begin{gathered} \left({\cos }^2\alpha +{\cos }^{2}90°\right)\left({\sin }^2\alpha -{\cos }^{2}90°\right)={\cos }^{2}90°{\sin }^{2}90° \\ {\cos }^2\alpha {\sin }^2\alpha =0 \\ 4{\cos }^2\alpha {\sin }^2\alpha =0 \\ {\sin }^{2}2\alpha =0 \end{gathered}$$

This Implies:

$$\begin{gathered} s\mathrm{in}2\alpha =0 \\ 2\alpha =0 \\ \alpha =0° \end{gathered}$$

Now, from equation (2), we get:

$$\begin{gathered} \gamma ={\cos }^{-1}\sqrt{{\sin }^{2}0°-{\cos }^{2}90°} \\ ={\cos }^{-1}0 \\ =90° \end{gathered}$$

Here, the sum of angles will be: $\alpha +\beta +\gamma =0°+90°+90°=180°$.

Now, if ${\sin }^2\beta =2{\cos }^2\alpha$. Then,

$$\begin{gathered} \left({\cos }^2\alpha +{\cos }^2\beta \right)\left({\sin }^2\alpha -{\cos }^2\beta \right)={\cos }^2\beta {\sin }^2\beta \\ {\cos }^2\alpha {\sin }^2\alpha -{\cos }^2\alpha +{\cos }^2\alpha {\sin }^2\beta +{\sin }^2\alpha -{\sin }^2\alpha {\sin }^2\beta =1-{\sin }^2\beta \\ {\cos }^2\alpha {\sin }^2\alpha -{\cos }^2\alpha +{\cos }^2\alpha \left(2{\cos }^2\alpha \right)+{\sin }^2\alpha -{\sin }^2\alpha \left(2{\cos }^2\alpha \right)=1-2{\cos }^2\alpha \\ 2{\cos }^2\alpha ={\sin }^2\alpha \end{gathered}$$

Evaluate as:

$$\begin{gathered} 2{\cos }^2\alpha ={\sin }^2\alpha \\ 2{\cos }^2\alpha =1-{\cos }^2\alpha \\ 3{\cos }^2\alpha =1 \\ \alpha ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{gathered}$$

This Implies:

$$\begin{gathered} {\sin }^2\beta =2{\left(\frac{1}{\sqrt{3}}\right)}^2 \\ 1-{\cos }^2\beta =\frac{2}{3} \\ {\cos }^2\beta =\frac{1}{3} \\ \beta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{gathered}$$

Now, from equation (2), we get:

$$\begin{gathered} \frac{1}{3}+\frac{1}{3}+{\cos }^2\gamma =1 \\ \gamma ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{gathered}$$

Here, the sum of angles will be: $\alpha +\beta +\gamma =3\left({\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)=164.2°$

Hence, the largest and smallest sum of angles are $180°\text{and}164.2°$ respectively.