Mathematical Methods

The reader is assumed to have some basic understanding of geometry and algebra.
Topics covered in this section are as follows:

1. Basic Geometry
2. Trigonometry

1. Basic Geometry

• Pythagoras Theorem:



For a right angled triangle, as shown in Figure 1.1, with the hypotenuse \(c\), base \(b\), and height \(a\): $$c^2=a^2+b^2 \hspace{1cm}(1.1)$$ This relation was given by ancient Greek philosopher, Pythagoras upon whom the theorem is named.
This thorem would be used for finding most prominantly the length of vectors, followed by its traditional applications in finding the side length of right-angled triangles and much more.

Here is a simple proof using triangle similarity.




• Proof:

Consider the Figure 1.1 again for \(\triangle BCD\) and \(\triangle ABC\). Clearly, $$ \begin{align} \angle CBD &= \angle CBA =\theta \\ \angle BCD &= \angle BAC =90^{\circ}-\theta \\ \angle BDC &= \angle BCA =90^{\circ} \end{align} $$ This shows that the \(\triangle BCD \sim \triangle ABC, \) by the \(AAA\) similarity criterion. A similar argument for \(\triangle ACD\) and \(\triangle ABC\) would give \(\triangle ACD \sim \triangle ABC. \)

Using similarity, $$\frac{BC}{BD}=\frac{AB}{BC}\implies BD= \frac{BC^2}{AB}\hspace{1cm}(\text{i})$$ and $$\frac{AB}{AC}=\frac{AC}{AD}\implies AD= \frac{AC^2}{AB}\hspace{1cm}(\text{ii})$$ Using,\((\)i\()\) and \((\)ii\()\) and using the fact that \(AD+BD=AB\), we get:

$$ \begin{align} AB=AD+BD &\implies AB = \frac{BC^2}{AB} + \frac{AC^2}{AB} \\ &\implies AB^2 = BC^2+AC^2\\ &\implies c^2 = a^2+b^2 \end{align} $$
,which proves the theorem.




• Distance Formula for 2 and 3 dimensions:

The distance formula is used to calculate the distance between two points in space. Now, the space can be of any dimensions (at least, theoretically) but, for now, we are concerned with only the 2 and 3 dimensional space.

Lets consider them both one by one.




• 2 dimensional case:

For two points in the two dimensional plane, say \(A(x_1,y_1)\) and \(B(x_2,y_2)\), the distance between them is: $$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \hspace{1cm} (1.2)$$



Consider the Figure 1.2 shown. The points \(A\) and \(B\) are shown, with the horizontal distance along the \(x\)-axis as \(x_2-x_1\) and the vertical distance along the \(y\)-axis as \(y_2-y_1\). Further, we apply the Pythagoras Theorem to the \(\triangle ABC\), so as to find the length of side \(AB\) using the sides \(AC\) and \(BC\), as below.

$$ \begin{aligned} &AB^2= AC^2+BC^2 \\ \implies &r^2 = (x_2-x_1)^2+(y_2-y_1)^2 \\ \implies &r = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \end{aligned} $$

This shows the equation \(1.2\).




• 3 dimensional case:

In the three dimensional "normal" space, the distance between two points \(A(x_1,y_1,z_1)\) and \(B(x_2,y_2,z_2)\) is given by: $$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \hspace{1cm}(1.3)$$



Consider the Figure 1.3 shown.

Now, to start with we first project the points \(A\) and \(B\) to the \(x-y\) plane turning them into \(C(x_1,y_1,0)\) and \(D(x_2,y_2,0)\), respectively as shown. Further, we apply the 2 dimensional distance formula (Eq 1.2) for finding the distance \(r'\) to be: $$r'= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ The above equation is easily verified by comparing the top view (Figure 1.4) shown below with the Figure 1.2. We identify that the length of the green and the blue segments are the same i.e. \(r'\). And, the vertical distance of the point \(A\) and \(B\) or the green segment and the point \(B\) is \(\left(z_2-z_1\right)\).



Then, we again employ the Pythagoras theorem, this time for the triangle with two vertices as \(A\) and \(B\) and the green segment as the base. So, we get: $$ \begin{aligned} r &= \sqrt{(r')^2+(z_2-z_1)^2} \\ \implies r&= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \end{aligned} $$ By this, we have proved the distance formula in 3-dimensions.

One special case: A case which arises very frequently, especially when dealing with vectors, is the distance of a point, say \((x,y,z)\) from the origin \((0,0,0)\). Using equation 1.3 ,we get this distance to be: $$ r = \sqrt{x^2+y^2+z^2} \hspace{2cm} (1.4) $$ This is often referred as the length of a vector from origin to the point \((x,y,z)\).

2. Trigonometry

Trigongometry is the study of angles, more specifically the study of angles of a triangle. We start by defining some some relations in between the angle and side length of the right angled triangle, and move towards a general form.

• Some definitions:

Consider the figure shown.



The three sides of the triangle \(ABC\), base \(BC\) has length \(b\), perpendicluar (the side in front of the angle under consideration, here, \(\theta\)) \(AB\) has length \(p\) and the longest side, hypotenuse (the side in front of right angle) \(AC\) has length \(h\).

We define the following:

1. Sine of \(\theta\) :

   Sine of angle \(\theta\) is defined as the ratio of the length of the perpendicular and the hypotenuse and is denoted as: $$ \sin{\theta}=\frac{p}{h} \hspace{1cm} (2.1) $$ Upon rearranging, we get \(p=h \sin{\theta}\). Thus, \(p\) is said to be the projection of \(h\) over the perpendicular side. This is much useful when dealing with vector projection.

2. Cosine of \(\theta\) :

   Cosine of angle \(\theta\) is defined as the ratio of the length of the base and the hypotenuse and is denoted as: $$ \cos{\theta}=\frac{b}{h}\hspace{1cm} (2.2) $$ As before, \(b=h \cos{\theta}\). Thus, \(b\) is the projection of \(h\) on the base.

3. Tangent of \(\theta\) :

   Tangent of angle \(\theta\) is the ratio of sine and cosine of \(\theta\) or, length of the perpendicular and the base. It is denoted as: $$ \tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=\frac{p}{b} \hspace{1cm}(2.3) $$ Now, \(\tan{\theta}\) has an immense significance as it represents the slope (how slant is the line with respect to the \(x\)-axis) of the hypotenuse.

4. Cosecant of \(\theta\) :

   It is defined as the reciprocal of sine. It is denoted as \(\csc{\theta}\) or sometimes \(\text{cosec }\theta\) i.e. $$ \csc{\theta}=\frac{1}{\sin{\theta}}=\frac{h}{p} $$

5. Secant of \(\theta\) :

   It is defined as the reciprocal of cosine. It is denoted as \(\sec{\theta}\) i.e. $$ \sec{\theta}=\frac{1}{\cos{\theta}}=\frac{h}{b} $$

6. Cotangent of \(\theta\) :

   It is defined as the reciprocal of tangent. It is denoted as \(\cot{\theta}\) i.e. $$ \cot{\theta}=\frac{1}{\tan{\theta}}=\frac{b}{p} $$




Table 2.1: Values of trigonmetric functions at some common angles

Function \(/ \theta\)	\(0^{\circ}\)	\(30^{\circ}\)	\(45^{\circ}\)	\(60^{\circ}\)	\(90^{\circ}\)
\(\sin{\theta}\)	\(0\)	\(1/2\)	\(1/\sqrt{2}\)	\(\sqrt{3}/2\)	\(1\)
\(\cos{\theta}\)	\(1\)	\(\sqrt{3}/2\)	\(1/\sqrt{2}\)	\(1/2\)	\(0\)
\(\tan{\theta}\)	\(0\)	\(1/\sqrt{3}\)	\(1\)	\(\sqrt{3}\)	\(\inf \)*
\(\csc{\theta}\)	\(\inf \)*	\(2\)	\(\sqrt{2}\)	\(2/\sqrt{3}\)	\(1\)
\(\sec{\theta}\)	\(1\)	\(2/\sqrt{3}\)	\(\sqrt{2}\)	\(2\)	\(\inf \)*
\(\cot{\theta}\)	\(\inf \)*	\(\sqrt{3}\)	\(1\)	\(1/\sqrt{3}\)	\(0\)
*here, \(\inf\) represents infinity or undefined