OpenLLM-France/wikimedia huggingface dataset:
# Student Projects/Healthy habits for a healthy lifestyle

Healthy habits for a Healthy Lifestyle
Healthy habit is any behavior that benefits our physical, mental or emotional health. Inculcating them will lead us to live a healthy lifestyle which adds years to our life and prevent long term illness.
1. Early to bed and Early to rise
The famous quote of Benjamin Franklin, “Early to bed and early to rise, makes a man healthy, wealth and wise” signifies the need of sleeping and getting up early.
* It reduces stress and makes a person happier. We can experience refreshed day.
* Any kind of tasks we aim to do can also be done in less time if we do it in the early morning.
2. Personal hygiene
Maintaining personal hygiene is beneficial to have a better health. And also it is a way of life. This also promotes self esteem and to have a healthy physical state. The following are some basic habits which we should follow in order to maintain personal hygiene:
* Brushing our teeth keeps gums and teeth healthy.
* Regular body shower prevents the development and spread of illness and infections.
* Washing hands before and after eating prevents the spread of germs and also keep germs out of our bodies.
3. Drink adequate amount of water
Drinking large amount of water plays vital role in our daily functioning. Doctors suggest drinking at least 3-4 liters of water per day. To prevent dehydration, we have to drink adequate amount of water everyday. The more we intake water, the more we become healthy.
* It regulates our body temperature and bowel movements, removes toxins.
* It boosts metabolic rate and improves health.
* It also delivers oxygen throughout the body.
4. Healthy diet
To lead a healthy lifestyle, good nutrition plays an important role.
A balanced diet is one that gives our body the nutrients it needs to function correctly. It is the diet that helps to maintain overall health.
It includes fruits, vegetables, whole grains, meat and no processed food and sweetened beverages. The benefits are listed below:
5. Regular walk, physical exercise and yoga
Physical exercise is any bodily movement or activity that enhances and maintains overall health and physical fitness.
Along with healthy diet, regular practice of physical exercise and yoga makes us physically strong and healthy.
* It helps an individual to maintain healthy weight.
* It boosts a person's energy levels and strengthens muscles.
* Regular physical activity can relieve us from stress, anxiety, anger and depression.
* It decreases a person’s risk of developing certain diseases such as obesity, type 2 diabetes and high blood pressure.
* Yoga regulates emotions and calms the mind. It also strengthens emotional and mental health.
In order to achieve a healthy lifestyle, remaining consistent is the most important factor. By sticking to the above mentioned practices and following them regularly makes oneself healthy and free from diseases and also keeps us mentally strong.
Therefore, healthy lifestyle should not be just a temporary fix for weight loss or to recover from illness or any disease instead, it should be a way of life.
# Student Projects/How to maintain healthy skin
## Tips about healthy skin:
### Eat healthy diet:

-It will improve our overall health
-Diet contain fruits,green leafy vegetables,grains
-Drink plenty of water
-Avoid junk food

### Sleep

-Plenty of sleep is essential for healthy skin
-The entire body rests and repairs itself
-Six to eight hours of sleep is essential for an individual

### Exercise

-It will help the circulation of blood easier
-Helps to transport oxygen through blood vessels

### Use sunscreen

-Use sunscreen SPF 15 above
-Sunscreen helps skin to protect form sun
-Apply for every two hours

### Stress management

-Due to more stress our skin becomes more sensitive and dull
-Yoga, exercise,gym are the modes of stress management
# Student Projects/How our Indian education system can be improved

Some of the drawbacks of the Indian education system are-
* Upto 85% of the students in india use to memorize the content given in the book and writing the same in the exams.
* Only 18% of the students use understand the concept of the subject and can able to answer in their own words.
* There is only few subjects containing practical contents.
* Not everyone has sufficient need to access the schools and higher studies.
* There are still many single teacher school in the country. Now the changes which are needed are : *Indian government needs to invest large amount of money on the infrastructure of the schools. * Methods of teaching need more changes. Teachers should encourage students to think logically and be creative. * The biographies of successful persons should be a part of the syllabus, so that students can mould their personalities. *Physical education should be given importance. *There is a definite need of revolutionary changes in the Indian education system. *With the effective learning system, India can successfully utilise it's vast human resource.
# Student Projects/Healthy cooking

Cooking is mostly considered as tedious task. But when you start enjoying it is never a tedious task. Now let us look in to healthy, tasty as well as an easy recipe. The name of the dish is “ALOO DAHI CURRY”. Here is the recipe. INGREDIENTS: 1. Potato 2. Curd 3. Coriander powder 4. Turmeric powder 5. Chilli powder 6. Cumin seeds 7. Garam masala 8. Salt 9. Ginger garlic paste 10. Green chilli 11. Curry leaves 12. Coriander leaves
RECIPE: 1. Boil the potato and cut it in to small pieces. 2. Take 2 spoons of ghee in a hot pan, wait for the ghee to become hot and then add one spoon cumin seeds, garam masala and the ginger garlic paste. Sorte it until the smell tickles your taste buds. Add green chilli. 3. And then to the curd add enough coriander, chilli and turmeric powder. Mix it for about 2 minutes so that the masala is blend with curd well. 4. Add the boiled potatoes to the pan and add the masalas and mix it well. 5. Then add the mixed curd to pan and wait until it boils. 6. Atlast add the coriander and curry leaves. 7. Serve it with chapati.
# PlanetPhysics/Direction Cosines

The Direction Cosines define the orientation of a vector with respect to a coordinate reference frame. Each direction cosine is the cosine of the angle between the vector and its corresponding coordinate axis. Let us first look at a two dimensional example in figure 1: \newline \begin{figure}[!hhp]
   \caption{2D - Direction Cosines} \includegraphics[width=\textwidth]{figure1.eps} \end{figure}
The direction cosines of ${\vec {v}}$ are
   $d_{1}=\cos(\theta )$
   $d_{2}=\cos(\phi )$
The x coordinate is given from simple trigonometry by
   $x=v\cos(\theta )$
where v is the magnitude of the vector ${\vec {v}}$ . Similarily, the y coordinate is given by
   $y=v\sin(\theta )$
but we can convert this to a cosine through the trigonometric identity that
   $cos(90-\theta )=\sin(\theta )$
From figure 1 we see that
   $\phi =90^{o}-\theta$
which can be subsitituded into 3 to get
   $y=v\cos(\phi )$
Note that $\phi$ is the angle between the y-axis and ${\vec {v}}$, so our vector ${\vec {v}}$ can be represented in this 2D coordinate frame by
   ${\vec {v}}={v\cos(\theta )}{\hat {x}}+{v\cos(\phi )}{\hat {y}}$
Extending this concept to three dimensions is quite easy, from figure 2 we can define ${\vec {v}}$ with respect t ${\hat {x}},{\hat {y}},{\hat {z}}$ coordinate frame by
   ${\vec {v}}={v\cos(\alpha )}{\hat {x}}+{v\cos(\beta )}{\hat {y}}+{v\cos(\gamma )}{\hat {z}}$
in a more compact form with
   $v_{1}=v\cos(\alpha )$
   $v_{2}=v\cos(\beta )$
   $v_{3}=v\cos(\gamma )$
we get the relation
   ${\vec {v}}={{\vec {v}}_{1}}{\hat {x}}+{{\vec {v}}_{2}}{\hat {y}}+{{\vec {v}}_{3}}{\hat {z}}$
The directional cosines for figure 2 are
   $d_{1}=\cos(\alpha )$
   $d_{2}=\cos(\beta )$
   $d_{3}=\cos(\gamma )$
An important property of the direction cosines is that
   ${\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}=1$
One important application is to use the direction cosines to define a coordinate system with reference to another. This can be accompished by defining the location of each coordinate axis unit vector with respect to the 'parent'. Once these nine direction cosines are determined (3 for each unit vector), than a transformation matrix exists to carry out coordinate transformations between the child frame and the parent frame.
# PlanetPhysics/Dot Product

The dot product or scalar product is defined as
$\mathbf {A} \cdot \mathbf {B} =A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$
{\mathbf Geometric interpretation}
Using a geometric interpretation allows us to find the angle between two vectors because
$\mathbf {A} \cdot \mathbf {B} =\left|\mathbf {A} \right|\left|\mathbf {B} \right|\cos \theta$
It is also useful to note that if the two vectors are perpindicular, their dot product is zero since
$\cos \left(90^{o}\right)=0$
# PlanetPhysics/Divergence
## Divergence

The divergence of a vector field is defined as
$\nabla \cdot {\mathbf {V} }={\frac {\partial V_{x}}{\partial x}}+{\frac {\partial V_{y}}{\partial y}}+{\frac {\partial V_{z}}{\partial z}}$
This is easily seen from the definition of the dot product and that of the del operator $\mathbf {A} \cdot \mathbf {B} =A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$ $\nabla ={\frac {\partial }{\partial x}}{\mathbf {\hat {i}} }+{\frac {\partial }{\partial y}}{\mathbf {\hat {j}} }+{\frac {\partial }{\partial z}}{\mathbf {\hat {z}} }$
carrying out the dot product with ${\mathbf {V} }$ then gives (1).

### Physical Meaning

(this section is a work in progress)
Building physical intuition about the divergence of a vector field can be gained by considering the flow of a fluid. One of the most simple vector fields is a uniform velocity field shown in below figure.
\begin{figure} \caption{Uniform Flow} \includegraphics[scale=1]{UniformFlow.eps} \end{figure}
Mathematically, this field would be
${\mathbf {V} }=5{\mathbf {\hat {i}} }$
The divergence is then
$\nabla \cdot {\mathbf {V} }={\frac {\partial }{\partial x}}5=0$
Source/Sink flow field ( div > 0 / div < 0)
\begin{figure} \caption{Positive Divergence} \includegraphics[scale=1]{PositiveDivergence.eps} \end{figure}
\begin{figure} \caption{Negative Divergence} \includegraphics[scale=1]{NegativeDivergence.eps} \end{figure}
Circular flow with zero divergence
\begin{figure} \caption{Circular Flow} \includegraphics[scale=1]{CircularFlow.eps} \end{figure}

### Coordinate Systems

Cartesian Coordinates
$\nabla \cdot {\mathbf {V} }={\frac {\partial V_{x}}{\partial x}}+{\frac {\partial V_{y}}{\partial y}}+{\frac {\partial V_{z}}{\partial z}}$
Cylindrical Coordinates
$\nabla \cdot {\mathbf {V} }={\frac {1}{r}}{\frac {\partial }{\partial r}}(rV_{r})+{\frac {1}{r}}{\frac {\partial V_{\theta }}{\partial \theta }}+{\frac {\partial V_{z}}{\partial z}}$
Spherical Coordinates
$\nabla \cdot {\mathbf {V} }={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}(r^{2}V_{r})+{\frac {1}{rsin\theta }}{\frac {\partial }{\partial \theta }}(V_{\theta }sin\theta )+{\frac {1}{rsin\theta }}{\frac {\partial V_{\phi }}{\partial \phi }}$
# PlanetPhysics/Direction Cosine Matrix to Quaternions

The fourth quaternion can be found through the definition given for the axis of rotation (see axis angle of rotation), $\alpha$
   $q_{4}=+/-{\frac {1}{2}}{\sqrt {A_{11}+A_{22}+A_{33}+1}}$
# PlanetPhysics/Double Groupoid Geometry

\newcommand{\sqdiagram}[9]{Failed to parse (unknown function "\diagram"): {\displaystyle \diagram #1 \rto^{#2} \dto_{#4}& \eqno{\mbox{#9}}} }

### Double Groupoids

The geometry of squares and their compositions leads to a common representation of a double groupoid in the following form:
   Failed to parse (unknown function "\D"): {\displaystyle (1) \D= \vcenter{\xymatrix @=3pc {S \ar @<1ex> [r] ^{s^1} \ar @<-1ex> [r] _{t^1} \ar @<1ex> [d]^{\, t_2} \ar @<-1ex> [d]_{s_2} & H \ar[l] \ar @<1ex> [d]^{\,t} \ar @<-1ex> [d]_s \\ V \ar [u] \ar @<1ex> [r] ^s \ar @<-1ex> [r] _t & M \ar [l] \ar[u] }} }
where $M$ is a set of `points', $H,V$ are `horizontal' and `vertical' groupoids, and $S$ is a set of `squares' with two compositions. The laws for a double groupoid make it also describable as a groupoid internal to the category of groupoids.
Given two groupoids $H,V$ over a set $M$, there is a double groupoid $\Box (H,V)$ with $H,V$ as horizontal and vertical edge groupoids, and squares given by quadruples
   Failed to parse (unknown function "\begin{pmatrix}"): {\displaystyle \begin{pmatrix} & h& \<blockquote><math>-0.9ex] v & & v'\<blockquote><math>-0.9ex]& h'& \end{pmatrix} }
for which we assume always that $h,h'\in H,\,v,v'\in V$ and that the initial and final points of these edges match in $M$ as suggested by the notation, that is for example </math>sh=sv, th=sv', \ldots$,etc.Thecompositionsaretobeinheritedfromthoseof<math>H,V$, that is
   Failed to parse (unknown function "\quadr"): {\displaystyle \quadr{h}{v}{v'}{h'} \circ_1\quadr{h'}{w}{w'}{h''} =\quadr{h}{vw}{v'w'}{h''}, \;\quadr{h}{v}{v'}{h'} \circ_2\quadr{k}{v'}{v''}{k'}=\quadr{hk}{v}{v''}{h'k'} ~. }
This construction is right adjoint to the forgetful functor which takes the double groupoid as above, to the pair of groupoids $H,V$ over $M$. Now given a general double groupoid as above, we can define Failed to parse (unknown function "\quadr"): {\displaystyle S\quadr{h}{v}{v'}{h'}} to be the set of squares with these as horizontal and vertical edges.
This allows us to construct for at least a commutative C*--algebra $A$ a double algebroid (i.e. a set with two algebroid structures)
   Failed to parse (unknown function "\D"): {\displaystyle (2) A\D= \vcenter{\xymatrix @=3pc {AS \ar @<1ex> [r] ^{s^1} \ar @<-1ex> [r] _{t^1} \ar @<1ex> [d]^{\, t_2} \ar @<-1ex> [d]_{s_2} & AH \ar[l] \ar @<1ex> [d]^{\,t} \ar @<-1ex> [d]_s \\ AV \ar [u] \ar @<1ex> [r] ^s \ar @<-1ex> [r] _t & M \ar [l] \ar[u] }} }
for which
   Failed to parse (unknown function "\quadr"): {\displaystyle AS\quadr{h}{v}{v'}{h'} }
is the free $A$-module on the set of squares with the given boundary. The two compositions are then bilinear in the obvious sense. Alternatively, we can use the convolution construction Failed to parse (unknown function "\D"): {\displaystyle \bar{A}\D} induced by the convolution C*--algebra over $H$ and $V$. These ideas about algebroids need further development in the light of the algebra of crossed modules of algebroids, developed in (Mosa, 1986, Brown and Mosa, 1986) as well as crossed cubes of (C*) algebras following Ellis (1988).
# PlanetPhysics/Direction Cosine Matrix to Euler 323 Angles

Starting with a direction cosine matrix (DCM), we need to determine the three Euler angles. The connection is made by comparing the DCM elements with the combined Euler 323 sequence. It is important to note that the 12 combinations of Euler angles for a given sequence can be found from a given DCM. The DCM matrix is
   $DCM=\left[{\begin{matrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{matrix}}\right]$
The Euler 323 sequence is
   $R_{3}(\psi )R_{2}(\theta )R_{3}(\phi )=\left[{\begin{matrix}c_{\psi }c_{\theta }c_{\phi }-s_{\psi }s_{\phi }&c_{\psi }c_{\theta }s_{\phi }+s_{\psi }c_{\phi }&-c_{\psi }s_{\theta }\\-s_{\psi }c_{\theta }c_{\phi }-c_{\psi }s_{\phi }&-s_{\psi }c_{\theta }s_{\phi }+c_{\psi }c_{\phi }&s_{\psi }s_{\theta }\\s_{\theta }c_{\phi }&s_{\theta }s_{\phi }&c_{\theta }\end{matrix}}\right]$
If we examine the element in coloum 3 row 3, then by inspection
$A_{33}=cos(\theta )$
Solving for $\theta$ yields
   $\theta =cos^{-1}(A_{33})$
Care must now be taken when evaluating the inverse cosine. It is a multivalued function. Analytically, the convention is to choose the principle value such that
$0\leq \theta \leq \pi$
If a numerical program is used, a function acos usually does this for us. The next step is to analyze the ratio ${\frac {A_{32}}{A_{31}}}$. Using these values from the Euler sequence we get
${\frac {A_{32}}{A_{31}}}={\frac {sin(\theta )sin(\phi )}{sin(\theta )cos(\phi )}}$
Rearranging and using the tangent yields
$tan(\phi )={\frac {A_{32}}{A_{31}}}$
Solving the quadrant ambiquity caused by the inverse tangent is done by examining the signs of the numerator and denominator. Denoting y as the numerator and x as the denominator, then the quadrant is chosen by:
quadrant 1 </math> \left[ \begin{matrix} y & x \\ + & + \\ \end{matrix} \right] $soifxandyarebothpositive,then<math>0\leq \phi \leq \pi /2$. Similarily for the other quadrant possibilites
quadrant 2 </math> \left[ \begin{matrix} y & x \\ + & - \\ \end{matrix} \right] $quadrant3$ \left[ \begin{matrix} y & x \\ - & - \\ \end{matrix} \right] $quadrant4$ \left[ \begin{matrix} y & x \\ - & + \\ \end{matrix} \right] $Ofcourse,itismuchsimpliertouseacalculatorornumericalprogramthatusestheatan2functionwhichwillchoosetheprinciplevalue.InasimilarfashionwecanfindthefinalEuleranglebylookingattheratio<math>{\frac {A_{23}}{A_{13}}}$ which gives the relation ${\frac {A_{23}}{A_{13}}}={\frac {sin(\psi )sin(\theta )}{-cos(\psi )sin(\theta )}}$
Canceling terms and rearrangeing gives us
$\psi =tan^{-1}({\frac {A_{23}}{-A_{13}}})$
which uses the same method to resolove quadrant ambiguity as above. To summarize, we will give the formulas for the conversion from a direction cosine matrix to the Euler 232 angles in Matlab syntax. Be careful how you implement this in other numerical programs. In Matlab it goes atan2(y,x) and in Mathematica it is ArcTan[x, y].
$\phi =atan2(A_{32},A_{31})$ $\theta =acos(A_{33})$ $\psi =atan2(A_{23},-A_{13})$
# PlanetPhysics/Doctoral Theses Supervised by a Grothendieck
### Doctoral theses supervised by Alexander Grothendieck

(original list is in French)

#### An incomplete list of doctoral theses directed by A. Grothendieck

* Schemes of reductive groups, by M. Demazure
[1] Sch\'emas en groupes r\'eductifs; short version of the thesis by M. Demazure, published in Bull. SMF. M\'em {\mathbf 93} (1965), 369-413. (Demazure's thesis is published "in totto" in SGA 3.
* Derived categories in $l$-adic cohomology.
[2] Cat\'egories d\'eriv\'ees en cohomologie l'-adique, (1969), thesis of J.-P. Jouanolou.
* Ample sheaves over group schemes and homogeneous spaces, thesis of Michel
Raynaud, published in Springer Lecture Notes {\mathbf119} (1970). [3] Faisceaux amples sur les sch\'emas en groupes et espaces homog\'enes; thesis of Michel Raynaud.
* Lefschetz theorems in coherent cohomology nad etale cohomology;
[4] Th\`eor\'emes de Lefschetz en cohomologie coh\'erente et en cohomologie \'etale; thesis of Mich\'ele Raynaud, Bull. SMF M\'em. 41 (1976).
* Complexe cotangent et d\'eformations, Springer Lecture Notes {\mathbf 239} (1971); thesis of Luc
Illusie.
* (On) Tannakian Categories , Thesis by Saavedra Rivano, published in Springer
Lecture Notes , 265 (1972). [6] Cat\'egories Tannakiennes, Springer Lecture Notes 265 (1972); thesis of Saavedra Rivano.
* \htmladdnormallink{non-Abelian {http://planetphysics.us/encyclopedia/AbelianCategory3.html} Cohomology}, Thesis by J. Giraud.
[7] Cohomologie non ab\'elienne, Grundlehren Math. Wiss. 179 (1972), Springer; thesis of J. Giraud.
* Ring topos and relative schemes, Thesis by Monique Hakim.
[8] Topos annel\'es et sch\'emas relatifs, Ergebnisse Math. \htmladdnormallink{GR {http://planetphysics.us/encyclopedia/SR.html}.} {\mathbf 64} (1972), Springer; thesis of Monique Hakim.
* Cohomologie cristalline, Springer Lecture Notes 407 (1974); thesis of Pierre Berthelot.
* Derived Categories and Abelian Categories, Thesis by Jean-Louis Verdier.,
[10] Des cat\'egories d\'eriv\'ees des cat\'egories ab\'eliennes; thesis of Jean-Louis Verdier, later published in Ast\'erisque {\mathbf 239} (1996), SMF (G. Maltsiniotis, ed.)
* Grothendieck Categories , Thesis of Hong Suan Xsin.
[11] Gr-Categories; thesis of Hong Suan Xsin.
* Isotopic study of embeddings of a topological $1$-complex into a complex surface; thesis
of Yves Ladegaillerie, short version published in Topology {\mathbf 23} no. 3(1984), 303-311.
* Jacobiennes g\'en\'eralis\'ees globales relatives; thesis of Carlos Contou-Carr\'ere, short version
published in the Grothendieck Festschrift , vol. II (1990).
# PlanetPhysics/Dot Product Example

Examples involving the dot product:
(1) Calculate $\mathbf {A} \cdot \mathbf {B}$ with
$\mathbf {A} =5\mathbf {\hat {i}} +2\mathbf {\hat {j}} -\mathbf {\hat {k}}$ \\ $\mathbf {B} =3\mathbf {\hat {i}} -3\mathbf {\hat {j}} +3\mathbf {\hat {k}}$ \\
answer:
$\mathbf {A} \cdot \mathbf {B} =(5)(3)+(2)(-3)+(-1)(+3)$ \\ $\mathbf {A} \cdot \mathbf {B} =15-6-3=6$ \\
(2) Find the angle between the above vectors.
answer:
We know their dot product, so we just need to calculate their magnitudes $\left|\mathbf {A} \right|={\sqrt {A_{x}^{2}+A_{y}^{2}+A_{z}^{2}}}={\sqrt {5^{2}+2^{2}+(-1)^{2}}}$ \\ $\left|\mathbf {A} \right|={\sqrt {25+4+1}}={\sqrt {30}}=5.48$ \\ \\
$\left|\mathbf {B} \right|={\sqrt {3^{2}+(-3)^{2}+(3)^{2}}}$ \\ $\left|\mathbf {B} \right|={\sqrt {9+9+9}}={\sqrt {27}}=5.2$ \\ \\
Finally
$\mathbf {A} \cdot \mathbf {B} =\left|\mathbf {A} \right|\left|\mathbf {B} \right|\cos \theta$
$\theta =cos^{-1}\left({\frac {\mathbf {A} \cdot \mathbf {B} }{\left|\mathbf {A} \right|\left|\mathbf {B} \right|}}\right)$ \\
$\theta =cos^{-1}\left({\frac {6}{(5.48)(5.2)}}\right)$ \\ $\theta =cos^{-1}(0.21)=77.9^{o}$
# PlanetPhysics/Direction Cosine Matrix to Euler 321 Angles

Starting with a direction cosine matrix (DCM), we need to determine the three Euler angles. The connection is made by comparing the DCM elements with the combined Euler 321 sequence. It is important to note that the 12 combinations of Euler angles for a given sequence can be found from a given DCM. The DCM matrix is
   $DCM=\left[{\begin{matrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{matrix}}\right]$
The Euler 321 sequence is
   $R_{1}(\psi )R_{2}(\theta )R_{3}(\phi )=\left[{\begin{matrix}c_{\theta }c_{\phi }&c_{\theta }s_{\phi }&-s_{\theta }\\-c_{\psi }s_{\phi }+s_{\psi }s_{\theta }c_{\phi }&c_{\psi }c_{\phi }+s_{\psi }s_{\theta }s_{\phi }&s_{\psi }c_{\theta }\\s_{\psi }s_{\phi }+c_{\psi }s_{\theta }c_{\phi }&-s_{\psi }c_{\phi }+c_{\psi }s_{\theta }s_{\phi }&c_{\psi }c_{\theta }\end{matrix}}\right]$
If we examine the element in column 3 row 1, then by inspection
$A_{13}=-sin(\theta )$
Solving for $\theta$ yields
   $\theta =sin^{-1}(-A_{13})$
Care must now be taken when evaluating the inverse sine. It is a multivalued function, which will have values of $\theta$ and $\pi -\theta$. Analytically, the convention is to choose the principle value such that
$-\pi /2\leq \theta \leq \pi /2$
If a numerical program is used, a function arcsine usually does this for us. The next step is to analyze the ratio ${\frac {A_{12}}{A_{11}}}$. Using these values from the Euler sequence we get
${\frac {A_{12}}{A_{11}}}={\frac {sin(\phi )cos(\theta )}{cos(\phi )cos(\theta )}}$
Rearranging the minus sign and using the tangent yields
$tan(\phi )={\frac {A_{12}}{A_{11}}}$
Solving the quadrant ambiguity caused by the inverse tangent is done by examining the signs of the numerator and denominator. Denoting y as the numerator and x as the denominator, then the quadrant is chosen by:
quadrant 1 $\left[{\begin{matrix}y&x\\+&+\\\end{matrix}}\right]$
so if x and y are both positive, then $0\leq \phi \leq \pi /2$. Similarly for the other quadrant possibilities
quadrant 2 $\left[{\begin{matrix}y&x\\+&-\\\end{matrix}}\right]$
quadrant 3 $\left[{\begin{matrix}y&x\\-&-\\\end{matrix}}\right]$
quadrant 4 $\left[{\begin{matrix}y&x\\-&+\\\end{matrix}}\right]$
Of course, it is much simpler to use a calculator or numerical program that uses the atan2 function which will choose the principle value. In a similar fashion we can find the final Euler angle by looking at the ratio ${\frac {A_{23}}{A_{33}}}$ which gives the relation ${\frac {A_{23}}{A_{33}}}={\frac {sin(\psi )cos(\theta )}{cos(\psi )cos(\theta )}}$
Canceling terms and rearranging gives us
$\psi =tan^{-1}({\frac {A_{23}}{A_{33}}})$
which uses the same method to resolve quadrant ambiguity as above. To summarize, we will give the formulas for the conversion from a direction cosine matrix to the Euler 313 angles in Matlab syntax. Be careful how you implement this in other numerical programs. In Matlab it goes atan2(y,x) and in Mathematica it is Arctan[x, y].
$\phi =atan2(A_{12},A_{11})$ $\theta =asin(-A_{13})$ $\psi =atan2(A_{23},A_{33})$