Formulas in Pure Math 2

Formulas in Pure Math 2

Module 2

Table of Contents

Vectors : slide 3

Coordinate geometry : slide 5

Parametric representation : slide 6

Trigonometry : slide 7

Vectors

 $\cdot magnitude\ =\sqrt{a^2+b^2+c^2}$  for a vector  $ai+bj+ck$  .
*  $unit\ vector=\frac{vector}{magnitude}$  
*  Given position vectors  $\overrightarrow{OA\ }and\ \overrightarrow{OB},\ general\ vector\ \ \overrightarrow{AB}\ =\overrightarrow{OB}\ -\overrightarrow{OA}$  
* Perpendicular vectors  $\overline{a}\ \cdot\ \overline{b}\ =0$  
*  $\overline{a}\cdot\overline{b}=\left|a\right|\cdot\left|b\right|\cdot\cos\theta$  WHERE theta is the angle between the 2 vectors
*  $vector\ equation\ of\ a\ line\ :\ \overline{r}=\overline{a}+t\overline{b}$  where a is a vector point on the line and b is the direction vector of the line 
* Parametric equation of a line :
 $x=x_1+tl$  
 $y=y_1+tm$  
 $z=z_1+tn$  
*Cartesian equation of a line 
 $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$  

* 2 vectors parallel, their cross product gives a vector perpendicular to the vectors

 $\overline{a}\cdot\overline{b}=\left(_{a_{2_{a3}}}^{a_1}\right)\cdot\left(_{b_{2_{b_3}}}^{b_1}\right)$  
 $=\left|_{a_3}^{a_2}\ _{b_3}^{b_2}\right|i\ -\left|_{a_3}^{a_1}\ _{b_3}^{b_1}\right|j\ +\left|_{a_2}^{a_1}\ _{b_2}^{b_1}\right|k$  
 $=\left(a_2b_3-a_3b_2\right)i-\left(a_1b_3-a_3b_1\right)j+\left(a_1b_2-a_2b_1\right)k$  
* vector equation of a plane:  $\overline{r}\cdot\overline{n}=d,\ \ \ \ where\ d=\overline{a}\ \cdot\overline{n}$  . a is a point on the plane and n is the normal vector to the plane
* cartesian equation of a plane  $ax+by+cz=d$  
* Distance =  $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$  
* Midpoint =  $\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2},\ \frac{z_1+z_2}{2}\right)$  

Coordinate Geometry

Circles
*Center (a , b) , radius = r : Standard form -  $\left(x-a\right)^2+\left(y-b\right)^2=r^2$  

*Center (-f , -g) , radius = r : General  form -  $x^2+y^2+2fx+2gy+C=0$    $r=\sqrt{f^2+g^2-C}$  
*  $m=\frac{y_2-y_1}{x_2-x_1}$  (grad of line)    * grad of tangent =  $-\frac{1}{grad\ of\ normal}$  
* Equation of line : y = m x + c 
* Distance formula =  $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$  
* Midpoint formula =  $\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$  

Parametric Representation

 $\frac{\left(x-a\right)^2}{r^2}+\ \frac{\left(y-b\right)^2}{r^2}=1\Longrightarrow\ \left(\frac{x-a}{r}\right)^2+\left(\frac{y-b}{r}\right)^2=1$  
 $\frac{x-a}{r}=\cos\ \theta\ \ \ \ \ \ \ \ \ \ \ \frac{y-b}{r}=\sin\theta$  


 $ax^2+bx^2=r^2\Longrightarrow\ \frac{ax^2}{r^2}+\frac{bx^2}{r^2}=1\Longrightarrow\ \left(\frac{ax}{r}\right)^2+\left(\frac{bx}{r}\right)^2$  
 $\frac{ax}{r}=\cos\theta\ \ \ \ \frac{bx}{r}=\sin\theta$  

Trigonometry

 $Tan\ A\ =\ \frac{\sin\ A\ }{\cos\ A}$  
 $\sec\ A\ =\ \frac{1}{\cos\ A}$  
  $\operatorname{cosec}\ A=\ \frac{1}{\sin\ A}$  
 $\cot\ A\ =\frac{1}{Tan\ A}=\ \frac{\cos\ A}{\sin\ A}$  
 $\sin^2A\ +\ \cos\ ^2A\ =\ 1$  
 $Tan\ 2A\ =\ \frac{2TanA}{1-Tan^2A}$  

 $\cos\ 2A\ =\cos^2A-\sin^2A$  

General solution
*  $\sin x=k\Longrightarrow\sin^{-1}\left(k\right)=\propto$   $\left(-1\le k\le1\right)$  

Degrees :  $x=180n+\left(-1\right)^n\left(\propto\right)\ ,\ n\in Z\ \ \ \ Radians\ :\ x=n\pi+\left(-1\right)^n\left(\propto\right),\ n\in Z$  
*  $\cos\ x=k\ \Longrightarrow\cos^{-1}\left(k\right)=\propto$   $\left(-1\le k\le1\right)$  
Degrees :  $x=360n\ \frac{+}{ }\propto,n\in Z\ \ \ \ \ Radians\ :\ 2n\pi\ \frac{+}{ }\propto,\ n\ \in Z$  
*  $\tan x=k\Longrightarrow\ \tan^{-1}\left(k\right)=\propto$  
Degrees :  $x=180n+\propto,\ n\in\ Z\ \ \ \ \ \ Radians\ :\ x=n\pi+\propto,\ n\in Z$  
*  $a\cos\theta+b\sin\theta\ can\ be\ written\ as\ r\sin\left(\theta\ \frac{+}{ }\propto\right)or\ r\cos\ \left(\theta\ \frac{+}{ }\propto\right)\$  where  $R>0\ and\ 0^{\degree}<\propto<90^{\degree}\ or\ 0^c\ <\propto<\frac{\pi^c}{2}$  
 $r=\sqrt{a^2+b^2}$  and  $\propto=\tan^{-1}\left(\frac{b}{a}\right)$ 

 *  $\sin\left(-a\right)=\sin\ A\ \ \$      * $\cos\left(-A\right)=\cos\ A$      * $\tan\left(-A\right)=-\tan\ A$