trigonometric function

#LaTeX

A trigonometric function is a function on the real numbers which relates an angle of a right triangle to a ratio of two side lengths.

\documentclass[tikz,margin={2cm 0cm}]{standalone}
\usepackage{tkz-euclide}
%
\definecolor{oppositecolor}{HTML}{FB464C}
\definecolor{adjacentcolor}{HTML}{45CF6E}
\definecolor{hypotenusecolor}{HTML}{53DFDD}
%
\begin{document}
\begin{tikzpicture}
    \tkzDefPoint(0,0){O}
    \tkzDefPoint(40:3){A}
    % \tkzDrawLine(O,A)
    %
    \tkzCalcLength(O,A)\tkzGetLength{radius}
    \tkzDefPoints{\radius/0/I}
    \tkzDefLine[perpendicular=through A](I,O) \tkzGetPoint{X_t}
    \tkzInterLL(O,I)(X_t,A) \tkzGetPoint{X}
	%
    \tkzPicAngle["$\theta$",draw=black,thin,angle radius=1cm](X,O,A)
    \tkzPicAngle["$\phi$",draw=black,thin,angle radius=0.8cm](O,A,X)
    \tkzMarkRightAngle(A,X,O)
	%
    \tkzDrawSegment[oppositecolor, thick](X,A)
        \tkzLabelSegment[right](X,A){\rotatebox{90}{opposite}}
    \tkzDrawSegment[adjacentcolor, thick](O,X)
        \tkzLabelSegment[below](O,X){adjacent}
     \tkzDrawSegment[hypotenusecolor, thick](O,A)
        \tkzLabelSegment[above=-2.5ex, pos=0.45](O,A){\rotatebox{40}{hypotenuse}}
    %
    \tkzDrawPoints(O,A,X)
\end{tikzpicture}
\end{document}

There are $_{3} P_{2} = 6$ such functions:

\begin{array}{lcl} \sin (θ) = \frac{opposite}{hypotenuse} & \csc (θ) = \frac{hypotenuse}{opposite} = \frac{1}{\sin θ} \\ \cos (θ) = \frac{adjacent}{hypotenuse} & \sec (θ) = \frac{hypotenuse}{adjacent} = \frac{1}{\cos θ} \\ \tan (θ) = \frac{opposite}{adjacent} & \cot (θ) = \frac{adjacent}{opposite} = \frac{1}{\tan θ} \end{array}

The elementary trig functions are the sine function, cosine function, and tangent function.
Their reciprocals are the cosecant function, secant function, and cotangent function, respectively.

The tangent identities are

\begin{array}{ccc} \tan (θ) & = \frac{\sin (θ)}{\cos (θ)} = \frac{\sec (θ)}{\csc (θ)} \\ \cot (θ) & = \frac{\cos (θ)}{\sin (θ)} = \frac{\csc (θ)}{\sec (θ)} \end{array}

Each cofunction can be specified via the complementary angle $(ϕ = \frac{π}{2} - θ)$ through the cofunction identities:

\begin{array}{lllclll} \sin (ϕ) & = \frac{adjacent}{hypotenuse} & = \cos (θ) & \cos (ϕ) & = \frac{opposite}{hypotenuse} & = \sin (θ) \\ \sec (ϕ) & = \frac{hypotenuse}{opposite} & = \csc (θ) & \csc (ϕ) & = \frac{hypotenuse}{adjacent} & = \sec (θ) \\ \tan (ϕ) & = \frac{adjacent}{opposite} & = \cot (θ) & \cot (ϕ) & = \frac{opposite}{adjacent} & = \tan (θ) \end{array}

From the Pythagorean theorem, we derive the Pythagorean identities:

\begin{array}{r} \sin^{2} θ + \cos^{2} θ = 1 \\ \tan^{2} θ + 1 = \sec^{2} θ \\ \cot^{2} θ + 1 = \csc^{2} θ \end{array}

which can be visualized

\documentclass[tikz,margin={2cm 0cm}]{standalone}
\usepackage{tkz-euclide}
%
\definecolor{obspink}{HTML}{FB99CD}
\definecolor{obsred}{HTML}{FB464C}
\definecolor{obsorange}{HTML}{E9973F}
\definecolor{obsyellow}{HTML}{E0DE71}
\definecolor{obsgreen}{HTML}{45CF6E}
\definecolor{obsteal}{HTML}{53DFDD}
\definecolor{obspurple}{HTML}{8A5CF5}
%
\begin{document}
\begin{tikzpicture}
    \tkzDefPoint(0,0){O}
    \tkzDefPoint(40:3){A}
    \tkzDrawLine(O,A)
    %
    \tkzCalcLength(O,A)\tkzGetLength{radius}
    \tkzDefPoints{\radius/0/I, 0/\radius/J}
    \tkzDrawArc[delta=10](O,I)(J)
    \tkzDrawLines(O,I O,J)
	%
    \tkzDefLine[perpendicular=through A](I,O) \tkzGetPoint{X_t}
    \tkzDefLine[perpendicular=through A](O,J) \tkzGetPoint{Y_t}
    %
    \tkzInterLL(O,I)(X_t,A) \tkzGetPoint{X}
    \tkzInterLL(O,J)(Y_t,A) \tkzGetPoint{Y}
	%
    \tkzPicAngle["$\theta$",draw=black,thin,angle radius=1cm](X,O,A)
	%
    \tkzDefLine[tangent at=A](O) \tkzGetPoint{B}
    \tkzInterLL(A,B)(O,J) \tkzGetPoint{C}
    \tkzInterLL(A,B)(O,I) \tkzGetPoint{D}
	%
    \tkzDrawSegment[obsred,thick](X,A)
        \tkzLabelSegment[left,pos=0.4](X,A){$\sin\theta$}
    \tkzDrawSegment[obsgreen,thick](Y,A)
        \tkzLabelSegment[above](Y,A){$\cos\theta$}
    \tkzDrawSegment[obsyellow,thick](O,D)
        \tkzLabelSegment[pos=0.4](O,D){$\sec\theta$}
    \tkzDrawSegment[obspurple,thick](O,C)
        \tkzLabelSegment[left](O,C){$\csc\theta$}
    \tkzDrawSegment[obsteal,thick](A,D)
        \tkzLabelSegment[above right](A,D){$\tan\theta$}
    \tkzDrawSegment[obsorange,thick](C,A)
        \tkzLabelSegment[above right](C,A){$\cot\theta$}
    \tkzDrawSegment[thick](O,A)
        \tkzLabelSegment[above left](O,A){$1$}
    %
    % \tkzDrawPoints(O,A,C,D,X,Y)
\end{tikzpicture}
\end{document}

The cosine and secant functions are even functions, and the sine, cosecant, tangent, and cotangent functions are odd functions. This is expressed by the even/odd identities:

\begin{array}{llcll} \sin (- θ) & = - \sin (θ) & \cos (- θ) & = \cos (θ) \\ \sec (- θ) & = \sec (θ) & \csc (- θ) & = - \csc (θ) \\ \tan (- θ) & = - \tan (θ) & \cot (- θ) & = - \cot (θ) \end{array}

The double angle identities are

\begin{aligned} \sin (2 θ) & = 2 \sin θ \cos θ \\ \cos (2 θ) & = \cos^{2} θ - \sin^{2} θ = 2 \cos^{2} θ - 1 = 1 - 2 \sin^{2} θ \\ \tan (2 θ) & = \frac{2 \tan θ}{1 - \tan^{2} θ} \end{aligned}

The half angle identities are

\begin{aligned} \sin (\frac{θ}{2}) & = \pm \sqrt{\frac{1 - \cos θ}{2}} \\ \cos (\frac{θ}{2}) & = \pm \sqrt{\frac{1 + \cos θ}{2}} \\ \tan (\frac{θ}{2}) & = \pm \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} \end{aligned}

The sum/difference identities are

\begin{aligned} \sin (α \pm β) & = \sin α \cos β \pm \cos α \sin β \\ \cos (α \pm β) & = \cos α \cos β \mp \sin α \sin β \\ \tan (α \pm β) & = \frac{\tan α \pm \tan β}{1 \mp \tan α \tan β} \end{aligned}