\documentclass[12pt,fleqn]{article}
\setlength{\oddsidemargin}{0.0cm}
\setlength{\evensidemargin}{0.0cm}
\setlength{\topmargin}{-10mm}
\advance\headheight by 2pt
\setlength{\footskip}{8mm}
\setlength{\textheight}{245mm}
\advance\textheight by-\headheight
\advance\textheight by-\headsep
\setlength{\textwidth}{170mm}
\def\baselinestrech{1.0}
\pagenumbering{arabic}
\newcommand{\ds}{\displaystyle}
\newcommand{\tg}{\mathop{\mathrm{tg}}\nolimits}
\newcommand{\ctg}{\mathop{\mathrm{ctg}}\nolimits}
\newcommand{\arctg}{\mathop{\mathrm{arctg}}\nolimits}
\newcommand{\arcctg}{\mathop{\mathrm{arcctg}}\nolimits}
\newcommand{\cosec}{\mathop{\mathrm{cosec}}\nolimits}
\newcommand{\arcsec}{\mathop{\mathrm{arcsec}}\nolimits}
\newcommand{\arccosec}{\mathop{\mathrm{arccosec}}\nolimits}
\renewcommand{\theenumi}{\arabic{enumi}}
\renewcommand{\theenumii}{\arabic{enumii}}

\begin{document}

\begin{center}
\bf Formule trigonometrice
\end{center}

\begin{enumerate}

\item $\ds\sin \alpha =\frac{a}{c}; \quad
      \cos \alpha =\frac{b}{c}; \quad
      \tg \alpha =\frac{a}{b}; \quad
      \ctg \alpha =\frac{b}{a};$\\
      ($a,\ b$ - catetele, $c$ - ipotenuza triunghiului dreptunghic,
      $\alpha $ - unghiul, opus catetei $a$).
\item $\ds\tg \alpha =\frac{\sin \alpha}{\cos \alpha};
      \quad \ctg \alpha =\frac{\cos \alpha}{\sin \alpha}. $
\item $\ds\tg\alpha\ctg\alpha =1. $
\item $\ds \sin \left(\frac{\pi}{2}\pm \alpha \right) =
       \cos \alpha; \quad \sin (\pi \pm \alpha )=\mp \sin \alpha. $
\item $\ds \cos \left( \frac{\pi}{2} \pm \alpha \right) =
         \mp \sin \alpha; \quad
       \cos (\pi \pm \alpha)=-\cos \alpha. $
\item $\ds \tg \left( \frac{\pi}{2} \pm \alpha \right) =
         \mp \ctg \alpha;
       \quad \ctg \left( \frac{\pi}{2} \pm \alpha \right)=\mp \tg \alpha. $
\item $\ds \sec \left( \frac{\pi}{2} \pm \alpha \right) =
         \mp \cosec \alpha;
       \quad \cosec \left( \frac{\pi}{2} \pm \alpha \right) = \sec \alpha. $
\item $\ds \sin^2 \alpha + \cos^2 \alpha =1. $
\item $\ds 1+\tg^2 \alpha= \sec^2 \alpha. $
\item $\ds 1+\ctg^2 \alpha= \cosec^2 \alpha. $
\item $\ds \sin (\alpha \pm \beta) =
         \sin \alpha \cos \beta \pm \sin \beta \cos \alpha. $
\item $\ds \cos (\alpha \pm \beta) =
       \cos \alpha \cos \beta \mp \sin \alpha \sin \beta. $

\footnotetext[0]{\quad Copyright$\copyright$1999 ONG TCV
    Scoala Virtuala a Tanarului Matematician \quad http://math.ournet.md}

\item $\ds \tg (\alpha \pm \beta) =
       \frac{\tg \alpha \pm \tg \beta}{1 \mp \tg \alpha \tg \beta}. $
\item $\ds \ctg (\alpha \pm \beta) =
       \frac{\ctg \alpha \ctg \beta \mp 1}{\ctg \beta \pm \ctg \alpha}. $
\item $\ds\sin 2\alpha =2\sin\alpha\cos\alpha.$
\item $\ds\cos 2\alpha =\cos^2\alpha - \sin^2\alpha.$
\item $\ds\tg 2\alpha =\frac{2\tg \alpha}{1-\tg^2\alpha}.$
\item $\ds\ctg 2\alpha =\frac{\ctg^2 \alpha -1}{2\ctg\alpha}.$
\item $\ds\sin 3\alpha =3\sin\alpha -4\sin^3\alpha.$
\item $\ds\cos 3\alpha =4\cos^3\alpha -3\cos\alpha.$
\item $\ds \left| \sin \frac{\alpha}{2} \right|=
       \sqrt{\frac{1-\cos \alpha}{2}}.$
\item $\ds \left| \cos \frac{\alpha}{2} \right| =
       \sqrt{\frac{1+\cos \alpha}{2}}.$
\item $\ds \left| \tg \frac{\alpha}{2} \right| =
       \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}.$
\item $\ds \tg \frac{\alpha}{2}=
       \frac{\sin \alpha}{1+\cos \alpha}=
       \frac{1-\cos \alpha}{\sin \alpha}.$
\item $\ds \left| \ctg \frac{\alpha}{2} \right| =
       \sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}.$
\item $\ds \ctg \frac{\alpha}{2}=
       \frac{\sin \alpha}{1- \cos \alpha}=
       \frac{1+\cos \alpha}{\sin \alpha}.$
\item $\ds 1+\cos \alpha = 2\cos^2 \frac{\alpha}{2}.$
\item $\ds 1-\cos \alpha = 2\sin^2 \frac{\alpha}{2}.$
\item $\ds \sin \alpha \pm \sin \beta =2\sin \frac{\alpha \pm \beta}{2}
         \cos \frac{\alpha \mp \beta}{2}.$
\item $\ds \cos \alpha + \cos \beta=2\cos \frac{\alpha +\beta}{2}
         \cos \frac{\alpha-\beta}{2}.$
\item $\ds \cos \alpha -\cos \beta=-2\sin \frac{\alpha+\beta}{2}
         \sin \frac{\alpha-\beta}{2}.$

\footnotetext[0]{\quad Copyright$\copyright$1999 ONG TCV
    Scoala Virtuala a Tanarului Matematician \quad http://math.ournet.md}

\item $\ds \tg \alpha \pm \tg \beta =
         \frac{\sin (\alpha \pm \beta)}{\cos \alpha \cos \beta}.$
\item $\ds \ctg \alpha \pm \ctg \beta=
         \frac{\sin (\beta \pm \alpha)}{\sin \alpha \sin \beta}.$
\item $\ds \sin \alpha \sin \beta =
         \frac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta)].$
\item $\ds \sin \alpha \cos \beta =
         \frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)].$
\item $\ds \cos \alpha \cos \beta =
         \frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)].$
\item Ecuatii trigonometrice elementare:
      $$
      \left.
      \begin{array}{l}
      \sin x=a,\ |a|\leq 1; \ x=(-1)^n\arcsin a+\pi n;\\[2mm]
      \cos x=a,\ |a|\leq 1; \ x=\pm \arccos a+2\pi n;\\[2mm]
      \tg x=a,\ x=\arctg a+\pi n;\\[2mm]
      \ctg x=a,\ x=\arcctg a+\pi n
      \end{array}
      \right\}\
      n \in \bf{Z}.
      $$
\item $\ds \arcsin x+\arccos x=\frac{\pi}{2}, \quad |x|\leq 1.$
\item $\ds \arctg x+\arcctg x=\frac{\pi}{2}.$
\item $\ds \arcsec x+\arccosec x=\frac{\pi}{2},\quad |x|\geq 1.$
\item $\ds \sin (\arcsin x)=x,\quad x \in [-1;+1].$
\item $\ds \arcsin (\sin x)=x,\quad
         x \in \left[-\frac{\pi}{2};\frac{\pi}{2} \right].$
\item $\ds \cos (\arccos x)=x,\quad x \in [-1;+1].$
\item $\ds \arccos (\cos x)=x,\quad
         x \in [0;\pi].$
\item $\ds \tg (\arctg x)=x,\quad x \in \bf{R}.$
\item $\ds \arctg (\tg x)=x,\quad
         x \in \left( -\frac{\pi}{2}; \frac{\pi}{2} \right).$

\footnotetext[0]{\quad Copyright$\copyright$1999 ONG TCV
    Scoala Virtuala a Tanarului Matematician \quad http://math.ournet.md}

\item $\ds \ctg (\arcctg x)=x,\quad x \in \bf{R}.$
\item $\ds \arcctg (\ctg x)=x, \quad x \in (0;\pi).$
\item $\ds \arcsin x= \arccos \sqrt{1-x^2}=\arctg \frac{x}{\sqrt{1-x^2}}=
         \arcctg \frac{\sqrt{1-x^2}}{x}, \quad 0<x<1.$
\item $\ds \arccos x= \arcsin \sqrt{1-x^2}=\arctg \frac{\sqrt{1-x^2}}{x}=
         \arcctg \frac{x}{\sqrt{1-x^2}}, \quad 0<x<1.$
\item $\ds \arctg x=\arcsin \frac{x}{\sqrt{1+x^2}}=
         \arccos \frac{1}{\sqrt{1+x^2}}=\arcctg \frac{1}{x},
         \quad  0<x<+\infty.$
\item $\ds \arcctg x=\arcsin \frac{1}{\sqrt{1+x^2}}=
         \arccos \frac{x}{\sqrt{1+x^2}}=\arctg\frac{1}{x},
         \quad 0<x<+\infty.$
\item $\arcsin x+\arcsin y =
      \left[
      \begin{array}{ll}
        \arcsin (x\sqrt{1-y^2}+y\sqrt{1-x^2}), &
           \textrm{daca } xy\leq 0 \textrm{ sau } x^2+y^2\leq 1; \\[2mm]
        \pi-\arcsin (x\sqrt{1-y^2}+y\sqrt{1-x^2}), &
           \textrm{daca } x>0,\ y>0 \ \textrm{si } x^2+y^2>1;\\[2mm]
        -\pi-\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2}), &
           \textrm{daca } x<0, y<0 \ \textrm{si } x^2+y^2>1.
      \end{array}
      \right.$
\item $ \arcsin x-\arcsin y =
      \left[
      \begin{array}{ll}
        \arcsin(x\sqrt{1-y^2}-y\sqrt{1-x^2}), &
          \textrm{daca } xy\geq 0 \ \textrm{sau } x^2+y^2\leq 1;\\[2mm]
        \pi-\arcsin (x\sqrt{1-y^2}-y\sqrt{1-x^2}), &
          \textrm{daca } x>0, y<0 \ \textrm{si } x^2+y^2>1;\\[2mm]
        -\pi-\arcsin (x\sqrt{1-y^2}-y\sqrt{1-x^2}), &
          \textrm{daca } x<0, y>0 \ \textrm{si } x^2+y^2>1.
      \end{array}
      \right.$
\item $ \arccos x+\arccos y =
      \left[
      \begin{array}{ll}
        \arccos (xy-\sqrt{(1-x^2)(1-y^2)}), &
          \textrm{daca } x+y\geq 0; \\[2mm]
        2\pi-\arccos (xy-\sqrt{(1-x^2)(1-y^2)}), &
          \textrm{daca } x+y<0.
      \end{array}
      \right.$
\item $ \arccos x-\arccos y =
      \left[
      \begin{array}{ll}
        -\arccos (xy+\sqrt{(1-x^2)(1-y^2)}), &
          \textrm{daca } x\geq y; \\[2mm]
        \arccos (xy+\sqrt{(1-x^2)(1-y^2)}), &
          \textrm{daca } x<y.
      \end{array}
      \right.$

\footnotetext[0]{\quad Copyright$\copyright$1999 ONG TCV
    Scoala Virtuala a Tanarului Matematician \quad http://math.ournet.md}

\item $ \arctg x +\arctg y =
      \left[
      \begin{array}{ll}
        \ds\arctg \frac{x+y}{1-xy}, & \textrm{daca } xy<1;\\[4mm]
        \ds\pi+\arctg \frac{x+y}{1-xy}, & \textrm{daca } x>0 \ \textrm{si } xy>1;\\[4mm]
        \ds-\pi+\arctg \frac{x+y}{1-xy}, & \textrm{daca } x<0 \ \textrm{si } xy>1.
      \end{array}
      \right.$
\item $ \arctg x -\arctg y =
      \left[
      \begin{array}{ll}
        \ds\arctg \frac{x-y}{1+xy}, & \textrm{daca } xy>-1;\\[4mm]
        \ds\pi+\arctg \frac{x-y}{1+xy}, & \textrm{daca } x>0 \ \textrm{si } xy<-1;\\[4mm]
        \ds-\pi+\arctg \frac{x-y}{1+xy}, & \textrm{daca } x<0 \ \textrm{si } xy<-1.
      \end{array}
      \right.$
\item $ 2\arcsin x=
      \left[
      \begin{array}{ll}
        \arcsin(2x\sqrt{1-x^2}), &
          \ds\textrm{daca } |x|\leq \frac{\sqrt2}{2};\\[3mm]
        \pi-\arcsin(2x\sqrt{1-x^2}), &
          \ds\textrm{daca } \frac{\sqrt2}{2}<x\leq 1;\\[3mm]
        -\pi-\arcsin(2x\sqrt{1-x^2}), &
          \ds\textrm{daca } -1\leq x< -\frac{\sqrt2}{2}.
      \end{array}
      \right.$
\item $ 2\arccos x =
      \left[
      \begin{array}{ll}
        \arccos (2x^2-1) & \textrm{cand } 0\leq x\leq 1;\\[2mm]
        2\pi-\arccos (2x^2-1) & \textrm{cand } -1\leq x<0.
      \end{array}
      \right.$
\item $ 2\arctg x=
      \left[
      \begin{array}{ll}
        \ds\arctg \frac{2x}{1-x^2}, & \textrm{daca } |x|<1;\\[4mm]
        \ds\pi+\arctg \frac{2x}{1-x^2}, & \textrm{daca } x>1;\\[4mm]
        \ds-\pi+\arctg \frac{2x}{1-x^2}, & \textrm{daca } x<-1.
      \end{array}
      \right.$
\item $\ds \frac{1}{2} \arcsin x =
      \left[
      \begin{array}{ll}
        \ds\arcsin \sqrt{\frac{1-\sqrt{1-x^2}}{2}}, & \textrm{daca } 0\leq x\leq 1;\\[4mm]
        \ds-\arcsin \sqrt{\frac{1-\sqrt{1-x^2}}{2}}, & \textrm{daca } -1\leq x<0.
      \end{array}
      \right.$
\item $\ds \frac{1}{2} \arccos x=\arccos \sqrt{\frac{1+x}{2}},\
         \textrm{daca } -1\leq x\leq 1.$
\item $\ds \frac{1}{2}\arctg x=
      \left[
      \begin{array}{l}
        \ds\arctg \frac{\sqrt{1+x^2}-1}{x},\ \textrm{daca } x\neq 0;\\[4mm]
        0,\ \textrm{daca } x=0.
      \end{array}
      \right.$
\end{enumerate}

\footnotetext[0]{\quad Copyright$\copyright$1999 ONG TCV
    Scoala Virtuala a Tanarului Matematician \quad http://math.ournet.md}


\end{document}
