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Revision difference : definitions in trigonometry
Version 4 Version 3
\PMlinkescapeword{terms} \PMlinkescapeword{terms}
\PMlinkescapeword{names} \PMlinkescapeword{names}
\PMlinkescapeword{inverses} \PMlinkescapeword{inverses}
\includegraphics{trig.eps} \includegraphics{trig.eps}
\textbf{Informal definitions} \textbf{Informal definitions}
Given a triangle $ABC$ with a signed angle $x$ at $A$ and a Given a triangle $ABC$ with a signed angle $x$ at $A$ and a
right angle at $B$, the ratios right angle at $B$, the ratios
$$\frac{BC}{AC}\qquad \frac{AB}{AC}\qquad \frac{BC}{AB}$$ $$\frac{BC}{AC}\qquad \frac{AB}{AC}\qquad \frac{BC}{AB}$$
are dependent only on the angle $x$, and therefore define functions, are dependant only on the angle $x$, and therefore define functions,
denoted by denoted by
$$\sin x\qquad \cos x\qquad \tan x$$ $$\sin x\qquad \cos x\qquad \tan x$$
respectively, where the names are short for \emph{sine, cosine} and respectively, where the names are short for \emph{sine, cosine} and
\emph{tangent}. Their inverses are rather less important, \emph{tangent}. Their inverses are rather less important,
but also have names: but also have names:
\begin{eqnarray*} \begin{eqnarray*}
\cot x &=& AB/BC = \frac{1}{\tan x} \text{ (cotangent)} \\ \cot x &=& AB/BC = \frac{1}{\tan x} \text{ (cotangent)} \\
\csc x &=& AC/BC = \frac{1}{\sin x} \text{ (cosecant)} \\ \csc x &=& AC/BC = \frac{1}{\sin x} \text{ (cosecant)} \\
\sec x &=& AC/AB = \frac{1}{\cos x} \text{ (secant)} \sec x &=& AC/AB = \frac{1}{\cos x} \text{ (secant)}
\end{eqnarray*} \end{eqnarray*}
From Pythagoras's theorem we have $\cos^2 x+\sin^2 x = 1$ for all (real) $x$. From Pythagoras's theorem we have $\cos^2 x+\sin^2 x = 1$ for all (real) $x$.
Also it is ``clear'' from the diagram at left that functions $\cos$ and $\sin$ Also it is ``clear'' from the diagram at left that functions $\cos$ and $\sin$
are periodic with period $2\pi$. However: are periodic with period $2\pi$. However:
\textbf{Formal definitions} \textbf{Formal definitions}
The above definitions are not fully rigorous, because we have not defined The above definitions are not fully rigorous, because we have not defined
the word \emph{angle}. We will sketch a more rigorous approach. the word \emph{angle}. We will sketch a more rigorous approach.
The power series The power series
$$\sum_{n=0}^\infty\frac{x^n}{n!}$$ $$\sum_{n=0}^\infty\frac{x^n}{n!}$$
converges uniformly on compact subsets of $\mathbb{C}$ and its sum, converges uniformly on compact subsets of $\mathbb{C}$ and its sum,
denoted by $\exp(x)$ or by $e^x$, is therefore an entire function of $x$, denoted by $\exp(x)$ or by $e^x$, is therefore an entire function of $x$,
called the exponential function. called the exponential function.
$f(x)=\exp(x)$ is the unique solution of the boundary value problem $f(x)=\exp(x)$ is the unique solution of the boundary value problem
$$f(0)=1\qquad f'(x)=f(x)$$ $$f(0)=1\qquad f'(x)=f(x)$$
on $\mathbb{R}$. on $\mathbb{R}$.
The sine and cosine functions, for real arguments, are defined in terms The sine and cosine functions, for real arguments, are defined in terms
of $\exp$, simply by of $\exp$, simply by
$$\exp(ix)=\cos x + i(\sin x)\;.$$ $$\exp(ix)=\cos x + i(\sin x)\;.$$
Thus Thus
$$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots$$ $$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots$$
$$\sin x = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots$$ $$\sin x = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots$$
Although it is not self-evident, $\cos$ and $\sin$ are periodic functions on Although it is not self-evident, $\cos$ and $\sin$ are periodic functions on
the real line, and have the same period. That period is denoted by $2\pi$. the real line, and have the same period. That period is denoted by $2\pi$.