Editing Trigonometric functions (section)

=== Definition via integration ===
Another way to define the trigonometric functions in analysis is using integration.<ref name="Hardy"/><ref>{{citation|last=Bartle|year=1964|title=Elements of real analysis|publisher=|pages=315–316}}</ref> For a real number <math>t</math>, put
<math display="block">\theta(t) = \int_0^t \frac{d\tau}{1+\tau^2}=\arctan t</math>
where this defines this inverse tangent function. Also, <math>\pi</math> is defined by
<math display="block">\frac12\pi = \int_0^\infty \frac{d\tau}{1+\tau^2}</math>
a definition that goes back to [[Karl Weierstrass]].<ref>{{cite book |last=Weierstrass |first=Karl |author-link=Karl Weierstrass |chapter=Darstellung einer analytischen Function einer complexen Veränderlichen, deren absoluter Betrag zwischen zwei gegebenen Grenzen liegt |trans-chapter=Representation of an analytical function of a complex variable, whose absolute value lies between two given limits |language=de |title=Mathematische Werke |volume=1 |publication-place=Berlin |publisher=Mayer & Müller |year=1841 |publication-date=1894 |pages=51–66 |chapter-url=https://archive.org/details/mathematischewer01weieuoft/page/51/ }}</ref>

On the interval <math>-\pi/2<\theta<\pi/2</math>, the trigonometric functions are defined by inverting the relation <math>\theta = \arctan t</math>. Thus we define the trigonometric functions by
<math display="block">\tan\theta = t,\quad \cos\theta = (1+t^2)^{-1/2},\quad \sin\theta = t(1+t^2)^{-1/2}</math>
where the point <math>(t,\theta)</math> is on the graph of <math>\theta=\arctan t</math> and the positive square root is taken.

This defines the trigonometric functions on <math>(-\pi/2,\pi/2)</math>. The definition can be extended to all real numbers by first observing that, as <math>\theta\to\pi/2</math>, <math>t\to\infty</math>, and so <math>\cos\theta = (1+t^2)^{-1/2}\to 0</math> and <math>\sin\theta = t(1+t^2)^{-1/2}\to 1</math>. Thus <math>\cos\theta</math> and <math>\sin\theta</math> are extended continuously so that <math>\cos(\pi/2)=0,\sin(\pi/2)=1</math>. Now the conditions <math>\cos(\theta+\pi)=-\cos(\theta)</math> and <math>\sin(\theta+\pi)=-\sin(\theta)</math> define the sine and cosine as periodic functions with period <math>2\pi</math>, for all real numbers.

Proving the basic properties of sine and cosine, including the fact that sine and cosine are analytic, one may first establish the addition formulae. First,
<math display="block">\arctan s + \arctan t = \arctan \frac{s+t}{1-st}</math>
holds, provided <math>\arctan s+\arctan t\in(-\pi/2,\pi/2)</math>, since
<math display="block">\arctan s + \arctan t= \int_{-s}^t\frac{d\tau}{1+\tau^2}=\int_0^{\frac{s+t}{1-st}}\frac{d\tau}{1+\tau^2}</math>
after the substitution <math>\tau \to \frac{s+\tau}{1-s\tau}</math>. In particular, the limiting case as <math>s\to\infty</math> gives
<math display="block">\arctan t + \frac{\pi}{2} = \arctan(-1/t),\quad t\in (-\infty,0).</math>
Thus we have
<math display="block">\sin\left(\theta + \frac{\pi}{2}\right) = \frac{-1}{t\sqrt{1+(-1/t)^2}} = \frac{-1}{\sqrt{1+t^2}} = -\cos(\theta)</math>
and
<math display="block">\cos\left(\theta + \frac{\pi}{2}\right) = \frac{1}{\sqrt{1+(-1/t)^2}} = \frac{t}{\sqrt{1+t^2}} = \sin(\theta).</math>
So the sine and cosine functions are related by translation over a quarter period <math>\pi/2</math>.