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

(1)

(2)

(3)

(4)

(5)

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 cosine series and its derivative series.

The equation $$\cos{x} \;=\; 0$$ has on the interval $(0,\,2)$ exactly one root. 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 Leibniz' estimate for alternating series), 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 $$\sin(x\!+\!2\pi) \;=\; \sin{x}, \qquad \cos(x\!+\!2\pi) \;=\; \cos{x}.$$