trigonometric formulas from series

One may define the sine and the cosine functions for real (and complex) arguments using the power series

$\sin x=x-{\displaystyle \frac{x^3}{3!}}+{\displaystyle \frac{x^5}{5!}}-+\mathrm{\dots },$(1)

$\cos x=\mathrm{\hspace{0.33em}1}-{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^4}{4!}}-+\mathrm{\dots },$(2)

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

$$\sin 0=\mathrm{\hspace{0.33em}0},\cos 0=\mathrm{\hspace{0.33em}1}$$

and the parity relations (http://planetmath.org/EvenoddFunction)

$$\sin (-x)=-\sin x,\cos (-x)=\cos x.$$

Using the Cauchy multiplication rule for series one can obtain the addition formulas

$\{\begin{array}{cc}\sin (x+y)=\sin x\cos y+\cos x\sin y,\hfill & \\ \cos (x+y)=\cos x\cos y-\sin x\sin y.\hfill & \end{array}$(3)

These produce straightforward many other important formulae, e.g.

$\sin 2x=\mathrm{\hspace{0.33em}2}\sin x\cos x,\cos 2x={\cos }^{2}x-{\sin }^{2}x\mathit{\hspace{1em}\hspace{1em}}(y=:x)$(4)

and

${\cos }^{2}x+{\sin }^{2}x=\mathrm{\hspace{0.33em}1}\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(y=:-x).$(5)

The value  $\cos {\displaystyle \frac{\pi }{2}}=0$,  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 and its derivative series (http://planetmath.org/PowerSeries).

The equation

$$\cos x=\mathrm{\hspace{0.33em}0}$$

has on the interval  $(0,\mathrm{\hspace{0.17em}2})$  exactly one root (http://planetmath.org/Equation).  Actually, as sum of a power series,$\cos x$ is continuous,  $\cos 0=1>0$  and    (see Leibniz’ estimate for alternating series (http://planetmath.org/LeibnizEstimateForAlternatingSeries)), whence there is at least one root.  If there were more than one root, then the derivative

$$-\sin x=-x+\frac{x^3}{3!}-+\mathrm{\dots }=-x(1-\frac{x^2}{3!}+-\mathrm{\dots })$$

would have at least one zero on the interval; this is impossible, since by Leibniz the series in the parentheses does not change its sign on the interval:

$$1-\frac{x^2}{3!}+-\mathrm{\dots }>\mathrm{\hspace{0.33em}1}-\frac{2^2}{3!}>\mathrm{\hspace{0.33em}0}$$

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    and  $\cos \frac{\pi }{2}=0$.  Furthermore, by (5),

$$\sin \frac{\pi }{2}=\mathrm{\hspace{0.33em}1},$$

and by (4),

$$\sin \pi =\mathrm{\hspace{0.33em}0},\cos \pi =-1,\sin 2\pi =\mathrm{\hspace{0.33em}0},\cos 2\pi =\mathrm{\hspace{0.33em}1}.$$

Consequently, the addition formulas (3) yield the periodicities (http://planetmath.org/PeriodicFunctions)

$$\sin (x+2\pi )=\sin x,\cos (x+2\pi )=\cos x.$$