trigonometric formulas from seriesTrigonometricFormulasFromSeries2009-02-24 15:13:412009-02-24 16:54:53Derivationdefinitions in trigonometry$\pi$sinecosine% this is the default PlanetMath preamble. as your knowledge
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One may define the sine and the cosine functions for real (and complex) arguments using the power series
\begin{align}
\sin{x} \;=\; x-\frac{x^3}{3!}+\frac{x^5}{5!}-+\ldots,
\end{align}
\begin{align}
\cos{x} \;=\; 1-\frac{x^2}{2!}+\frac{x^4}{4!}-+\ldots,
\end{align}
and using only the properties of power series, easily derive most of the goniometric formulas, without any geometry.\, For example, one gets instantly from (1) and (2) the values
$$\sin0 \;=\; 0, \qquad \cos0 \;=\; 1$$
and the \PMlinkname{parity relations}{EvenoddFunction}
$$\sin(-x) \;=\; -\sin{x}, \qquad \cos(-x) \;=\; \cos{x}.$$
Using the Cauchy multiplication rule for series one can obtain the addition formulas
\begin{align}
\begin{cases}
\sin(x\!+\!y) \;=\; \sin{x}\cos{y}+\cos{x}\sin{y},\\
\cos(x\!+\!y) \;=\; \cos{x}\cos{y}-\sin{x}\sin{y}.
\end{cases}
\end{align}
These produce straightforward many other important formulae, e.g.
\begin{align}
\sin2x \;=\; 2\sin{x}\cos{x}, \qquad \cos2x \;=\; \cos^2x-\sin^2x \qquad (y := x)
\end{align}
and
\begin{align}
\cos^2x+\sin^2x \;=\; 1 \qquad\qquad\qquad (y := -x).
\end{align}
The value\, $\displaystyle\cos\frac{\pi}{2} = 0$,\, as well as the formulae expressing the periodicity of sine and cosine, cannot be directly obtained from the series (1) and (2) --- in fact, one must define the number $\pi$ by using the function properties of the \PMlinkescapetext{cosine series} and its \PMlinkname{derivative series}{PowerSeries}.
The equation
$$\cos{x} \;=\; 0$$
has on the interval \,$(0,\,2)$\, exactly one \PMlinkname{root}{Equation}.\, Actually, as sum of a power series,
$\cos{x}$ is continuous,\, $\cos0 = 1 > 0$\, and\, $\cos2 < 1-\frac{2^2}{2!}+\frac{2^4}{4!} < 0$\, (see \PMlinkname{Leibniz' estimate for alternating series}{LeibnizEstimateForAlternatingSeries}), whence there is at least one root.\, If there were more than one root, then the derivative
$$-\sin{x} \;=\; -x+\frac{x^3}{3!}-+\ldots \;=\; -x(1-\frac{x^2}{3!}+-\ldots)$$
would have at least one zero on the interval; this is impossible, since by Leibniz the parentheses series does not change its sign on the interval:
$$1-\frac{x^2}{3!}+-\ldots \;>\;1-\frac{2^2}{3!} \;>\; 0$$
Accordingly, we can define the number pi to be the least positive solution of the equation\, $\cos{x} = 0$, multiplied by 2.
Thus we have\, $0 < \pi < 4$\, and\, $\cos\frac{\pi}{2} = 0$.\, Furthermore, by (5),
$$\sin\frac{\pi}{2} \;=\; 1,$$
and by (4),
$$\sin\pi \;=\; 0, \qquad \cos\pi \;=\; -1, \qquad \sin2\pi \;=\; 0, \qquad \cos2\pi \;=\; 1.$$
Consequently, the addition formulas (3) yield the periodicities
$$\sin(x\!+\!2\pi) \;=\; \sin{x}, \qquad \cos(x\!+\!2\pi) \;=\; \cos{x}.$$