# Fundamental Mathematics/Trigonometry

## Angles

When two lines AB and AC cuts at one point A , angle is formed at point A between AB and AC . Angle is denoted as $\angle A$  . Angle is measured in Degree (o) or Radian (Rad) .

$1^{o}=({\frac {360}{2\pi }})^{o}$
$1Rad=({\frac {2\pi }{360}})Rad$

$\theta ^{c}=\theta ^{\circ }\times {\frac {\pi }{180}}$

$\phi ^{\circ }=\phi ^{c}\times {\frac {180}{\pi }}$

## Trigonometric functions

### Fundamental trigonometric functions

Relation of right triangle sides are defined as trigonometric functions. They are:

 Trignometric Function Definition Mathematical Formula Graphs Sine Ratio of opposite side over hypotnuse $\sin A={\frac {a}{h}}$  Cosine Ratio of adjacent side over hypotnuse $\cos A={\frac {b}{h}}$  Tangent Ratio of opposite over adjacent $\tan A={\frac {a}{b}}$  Cotangent Ratio of adjacent over opposite $\cot A={\frac {b}{a}}$  Secant Ratio of 1 over opposite $\sec A={\frac {h}{b}}$  Cosecant Ratio of 1 over adjacent $\csc A={\frac {h}{a}}$  ### Chracteristics

 Periodic $\sin(x)=\sin(x+2k\pi )\,$ $\cos(x)=\cos(x+2k\pi )\,$ $\tan(x)=\tan(x+k\pi )\,$ Symmetry $\sin(-x)=-\sin(x)\,$ $\cos(-x)=\;\cos(x)\,$ $\tan(-x)=-\tan(x)\,$ Phase shift $\sin(x)=\cos \left({\frac {\pi }{2}}-x\right)$ $\cos(x)=\sin \left({\frac {\pi }{2}}-x\right)$ $\tan(x)=\cot \left({\frac {\pi }{2}}-x\right)$ Complex Power $\cos(x)={\frac {e^{ix}+e^{-ix}}{2}}\;$ $\sin(x)={\frac {e^{ix}-e^{-ix}}{2i}}\;$ với $i^{2}=-1.\,$ $e^{\imath x}=\cos(x)+\imath \sin(x)\,$ và$\imath ={\sqrt {-1}}.\,$ de Moivre's formula:$\cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^{n}\,$ Limit $\lim _{x\rightarrow 0}{\frac {\sin(x)}{x}}=1,$ $\lim _{x\rightarrow 0}{\frac {1-\cos(x)}{x}}=0,$ Derivative ${d \over dx}\cos(x)=-\sin(x)$ ${d \over dx}\sin(x)=\cos(x)$ ## Trigonometric formulas

### Fundamental trigonometry function

$\tan(x)={\frac {\sin(x)}{\cos(x)}}\qquad \operatorname {cotg} (x)={\frac {\cos(x)}{\sin(x)}}={\frac {1}{\tan(x)}}$

### Sum and difference angles

$\sin(x\pm y)=\sin(x)\cos(y)\pm \cos(x)\sin(y)\,$
$\cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)\,$
$\tan(x\pm y)={\frac {\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}}$

### Reduced power

$\cos ^{2}(x)={1+\cos(2x) \over 2}$
$\sin ^{2}(x)={1-\cos(2x) \over 2}$
$\sin ^{2}(x)\cos ^{2}2(x)={1-\cos(4x) \over 4}$
$\sin ^{3}(x)={\frac {23\sin 2(x)-\sin(3x)}{4}}$
$\cos ^{3}(x)={\frac {32\cos(x)+\cos(3x)}{4}}$

### Half angles

$\cos \left({\frac {x}{2}}\right)=\pm \,{\sqrt {\frac {1+\cos(x)}{2}}}$
$\sin \left({\frac {x}{2}}\right)=\pm \,{\sqrt {\frac {1-\cos(x)}{2}}}$
$\tan \left({\frac {x}{2}}\right)={\sin(x/2) \over \cos(x/2)}=\pm \,{\sqrt {1-\cos x \over 1+\cos x}}.\qquad \qquad (1)$
$\tan \left({\frac {x}{2}}\right)=\pm \,{\sqrt {(1-\cos x)(1+\cos x) \over (1+\cos x)(1+\cos x)}}=\pm \,{\sqrt {1-\cos ^{2}x \over (1+\cos x)^{2}}}$
$={\sin x \over 1+\cos x}.$
$\tan \left({\frac {x}{2}}\right)=\pm \,{\sqrt {(1-\cos x)(1-\cos x) \over (1+\cos x)(1-\cos x)}}=\pm \,{\sqrt {(1-\cos x)^{2} \over (1-\cos ^{2}x)}}$
$={1-\cos x \over \sin x}.$
$\tan \left({\frac {x}{2}}\right)={\frac {\sin(x)}{1+\cos(x)}}={\frac {1-\cos(x)}{\sin(x)}}.$
$t=\tan \left({\frac {x}{2}}\right),$
 $\sin(x)={\frac {2t}{1+t^{2}}}$ and $\cos(x)={\frac {1-t^{2}}{1+t^{2}}}$ and $e^{ix}={\frac {1+it}{1-it}}.$ ### Product of trigonomtric functions

$\cos \left(x\right)\cos \left(y\right)={\cos \left(x+y\right)+\cos \left(x-y\right) \over 2}\;$
$\sin \left(x\right)\sin \left(y\right)={\cos \left(x-y\right)-\cos \left(x+y\right) \over 2}\;$
$\sin \left(x\right)\cos \left(y\right)={\sin \left(x-y\right)+\sin \left(x+y\right) \over 2}\;$

### Inverse trigonometric functions

$\arcsin(x)+\arccos(x)=\pi /2\;$
$\arctan(x)+\operatorname {arccot}(x)=\pi /2.\;$
$\arctan(x)+\arctan(1/x)=\left\{{\begin{matrix}\pi /2,&{\mbox{n}}{\acute {\hat {\mbox{e}}}}{\mbox{u}}\ x>0\\-\pi /2,&{\mbox{n}}{\acute {\hat {\mbox{e}}}}{\mbox{u}}\ x<0\end{matrix}}\right..$
$\arctan(x)+\arctan(y)=\arctan \left({\frac {x+y}{1-xy}}\right)\;$
$\arctan(x)-\arctan(y)=\arctan \left({\frac {x-y}{1+xy}}\right)\;$
$\sin(\arccos(x))={\sqrt {1-x^{2}}}\,$
$\cos(\arcsin(x))={\sqrt {1-x^{2}}}\,$
$\sin(\arctan(x))={\frac {x}{\sqrt {1+x^{2}}}}$
$\cos(\arctan(x))={\frac {1}{\sqrt {1+x^{2}}}}$
$\tan(\arcsin(x))={\frac {x}{\sqrt {1-x^{2}}}}$
$\tan(\arccos(x))={\frac {\sqrt {1-x^{2}}}{x}}$

### Complex number

$\cos(x)={\frac {e^{ix}+e^{-ix}}{2}}\;$
$\sin(x)={\frac {e^{ix}-e^{-ix}}{2i}}\;$
$i^{2}=-1.\,$

With

$e^{\imath x}=\cos(x)+\imath \sin(x)\,$

Hence

$\imath ={\sqrt {-1}}.\,$

### Infinite Product

$\sin x=x\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)$
$\sinh x=x\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)$
$\cos x=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}(n-{\frac {1}{2}})^{2}}}\right)$
$\cosh x=\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}(n-{\frac {1}{2}})^{2}}}\right)$
${\frac {\sin x}{x}}=\prod _{n=1}^{\infty }\cos \left({\frac {x}{2^{n}}}\right)$

### Limits

$\lim _{x\rightarrow 0}{\frac {\sin(x)}{x}}=1,$
$\lim _{x\rightarrow 0}{\frac {1-\cos(x)}{x}}=0,$

### Derivatives

${d \over dx}\cos(x)=-\sin(x)$
${d \over dx}\sin(x)=\cos(x)$
${d \over dx}\tan(x)=\sec ^{2}(x)$
${d \over dx}\cot(x)=-\csc ^{2}(x)$
${d \over dx}\sec(x)=\sec(x)\tan(x)$
${d \over dx}\csc(x)=-\csc(x)\cot(x)$
${d \over dx}\arcsin(x)={\frac {1}{\sqrt {1-x^{2}}}}$
${d \over dx}\arctan(x)={\frac {1}{1+x^{2}}}$

### Common Trigonomtric functions

$\sin \left(x+y\right)=\sin x\cos y+\cos x\sin y$
$\sin \left(x-y\right)=\sin x\cos y-\cos x\sin y$
$\cos \left(x+y\right)=\cos x\cos y-\sin x\sin y$
$\cos \left(x-y\right)=\cos x\cos y+\sin x\sin y$
$\sin x+\sin y=2\sin \left({\frac {x+y}{2}}\right)\cos \left({\frac {x-y}{2}}\right)$
$\sin x-\sin y=2\cos \left({\frac {x+y}{2}}\right)\sin \left({\frac {x-y}{2}}\right)$
$\cos x+\cos y=2\cos \left({\frac {x+y}{2}}\right)\cos \left({\frac {x-y}{2}}\right)$
$\cos x-\cos y=-2\sin \left({\frac {x+y}{2}}\right)\sin \left({\frac {x-y}{2}}\right)$
$\tan x+\tan y={\frac {\sin \left(x+y\right)}{\cos x\cos y}}$
$\tan x-\tan y={\frac {\sin \left(x-y\right)}{\cos x\cos y}}$
$\cot x+\cot y={\frac {\sin \left(x+y\right)}{\sin x\sin y}}$
$\cot x-\cot y={\frac {-\sin \left(x-y\right)}{\sin x\sin y}}$

### Double angle trigonometric functions

$\sin(2x)=2\sin(x)\cos(x)\,$
$\cos(2x)=\cos ^{2}(x)-\sin ^{2}(x)=2\cos ^{2}(x)-1=1-2\sin ^{2}(x)\,$
$\tan(2x)={\frac {2\tan(x)}{1-\tan ^{2}(x)}}$

In general,

If Tn is Chebyshev n ordered then

$\cos(nx)=T_{n}(\cos(x)).\,$

De Moivre formula:

$\cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^{n}\,$

hạt nhân Dirichlet Dn(x) sẽ xuất hiện trong các công thức sau:

$1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots +2\cos(nx)\;$
$={\frac {\sin \left(\left(n+{\frac {1}{2}}\right)x\right)}{\sin(x/2)}}\;$
$\sin(nx)=2\sin((n-1)x)\cos(x)-\sin((n-2)x)$
$\cos(nx)=2\cos((n-1)x)\cos(x)-\cos((n-2)x)$

### Triple angle trigonometric functions

$\sin(3x)=3\sin(x)-4\sin ^{3}(x)$
$\cos(3x)=4\cos ^{3}(x)-3\cos(x)$

### Series

${\begin{matrix}\arcsin z&=&z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \\&=&\sum _{n=0}^{\infty }\left({\frac {(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {z^{2n+1}}{(2n+1)}}\end{matrix}}\,\quad \left|z\right|<1$
${\begin{matrix}\arccos z&=&{\frac {\pi }{2}}-\arcsin z\\&=&{\frac {\pi }{2}}-(z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots )\\&=&{\frac {\pi }{2}}-\sum _{n=0}^{\infty }\left({\frac {(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {z^{2n+1}}{(2n+1)}}\end{matrix}}\,\quad \left|z\right|<1$
${\begin{matrix}\arctan z&=&z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots \\&=&\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\end{matrix}}\,\quad \left|z\right|<1$
${\begin{matrix}\operatorname {arccsc} z&=&\arcsin \left(z^{-1}\right)\\&=&z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{-7}}{7}}+\cdots \\&=&\sum _{n=0}^{\infty }\left({\frac {(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {z^{-(2n+1)}}{2n+1}}\end{matrix}}\,\quad \left|z\right|>1$
${\begin{matrix}\operatorname {arcsec} z&=&\arccos \left(z^{-1}\right)\\&=&{\frac {\pi }{2}}-(z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{-7}}{7}}+\cdots )\\&=&{\frac {\pi }{2}}-\sum _{n=0}^{\infty }\left({\frac {(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {z^{-(2n+1)}}{(2n+1)}}\end{matrix}}\,\quad \left|z\right|>1$
${\begin{matrix}\operatorname {arccot} z&=&{\frac {\pi }{2}}-\arctan z\\&=&{\frac {\pi }{2}}-(z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots )\\&=&{\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\end{matrix}}\,\quad \left|z\right|<1$

### Integrals

$\arcsin \left(x\right)=\int _{0}^{x}{\frac {1}{\sqrt {1-z^{2}}}}\,\mathrm {d} z,\quad |x|<1$
$\arccos \left(x\right)=\int _{x}^{1}{\frac {1}{\sqrt {1-z^{2}}}}\,\mathrm {d} z,\quad |x|<1$
$\arctan \left(x\right)=\int _{0}^{x}{\frac {1}{1+z^{2}}}\,\mathrm {d} z,\quad \forall x\in \mathbb {R}$
$\operatorname {arccot} \left(x\right)=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,\mathrm {d} z,\quad z>0$
$\operatorname {arcsec} \left(x\right)=\int _{x}^{1}{\frac {1}{|z|{\sqrt {z^{2}-1}}}}\,\mathrm {d} z,\quad x>1$
$\operatorname {arccsc} \left(x\right)=\int _{x}^{\infty }{\frac {-1}{|z|{\sqrt {z^{2}-1}}}}\,\mathrm {d} z,\quad x>1$

### Complex trigonometric functions

$\arcsin(z)=-i\log \left(i\left(z+{\sqrt {1-z^{2}}}\right)\right)$
$\arccos(z)=-i\log \left(z+{\sqrt {z^{2}-1}}\right)$
$\arctan(z)={\frac {i}{2}}\log \left({\frac {1-iz}{1+iz}}\right)$