6

By Maclaurin series; $\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-....$, and we know that period of $\sin{x}$ is $2π$ because $\sin{(x+2π)} = \sin{x}$. But if we consider only the RHS of the above series then how we can tell that this expression (series) is also of period$=2π$?

Eleven-Eleven
  • 8,806
wasim
  • 61
  • You also need to define $\cos x$ via power series and show that $(\sin x)' = \cos x, (\cos x)' = -\sin x$ and using these show that $\cos x$ is strictly decreasing in $[0, 2]$ and $\cos 0 > 0, \cos 2 < 0$ so that $\cos x$ vanishes only once in $[0, 2]$ and let that root of $\cos x$ be denoted by $\pi/2$. This defines $\pi$. Next we can show via power series (or derivatives) show that $\sin^{2}x + \cos^{2}x = 1, \sin(x + y) = \sin x\cos y + \cos x\sin y$. Using this you can show that $\sin(x + 2\pi) = \sin x$. – Paramanand Singh Dec 17 '13 at 15:19

2 Answers2

1

Consider the exponential function $f(x)=e^{ix}$ for simpler notation, covering both $\sin (x)$ and $\cos (x)$ at once. Let $g(x):=f(x+L)$ where $L=2\pi$. Also denote the series expansions by

$f(x)=\sum_{k=0}^\infty f_k x^k$ and

$g(x)=\sum_{k=0}^\infty g_k x^k$,

respectively. We know that $f_k=i^k/k!$. We find from the binomial theorem that

$g_l=\sum_{k=l}^\infty \frac{i^k}{k!}\binom{k}{l}L^{k-l}=\sum_{r=0}^\infty \frac{i^{r+l}}{r! l!}L^r=\frac{i^l}{l!}\sum_{r=0}^\infty \frac{i^r}{r!}L^r=f_l f(L)$

Hence $g(x)=f(x)\Leftrightarrow \forall l: g_l=f_l \Leftrightarrow f(L)=1$.

The remaining part is to show that $f(L)=e^{i2\pi}=1$.

flonk
  • 2,434
1

Expanding on the comment I have given I first give the proof for $\sin (x + y)$ formula. First we define $$\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots $$ and by differentiation of power series $(\sin x)'= \cos x, (\cos x)' = -\sin x$.

Let us define $$f(z) = 1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + \cdots$$ for all complex $z$. Then by using rule for multiplication of two series we can show easily that $f(z_{1} + z_{2}) = f(z_{1})f(z_{2})$. Now it can be easily seen (by putting $z = ix$ in definition of $f(z)$) that $f(ix) = \cos x + i\sin x$ so that $f(i(x + y)) = f(ix + iy) = f(ix)f(iy)$ and therefore $$\cos(x + y) + i\sin (x + y) = (\cos x + i\sin x)(\cos y + i\sin y)$$ so that $\cos(x + y) = \cos x\cos y - \sin x\sin y$ and $\sin (x + y) = \sin x\cos y + \cos x \sin y$.

Again considering $g(x) = \cos^{2}x + \sin^{2}x$ we can show via differentiation formulas that $g'(x) = 0$ so that $g(x)$ is constant. Clearly since $\sin 0 = 0, \cos 0 = 1$ we have $g(x) = g(0) = 1$. So that proves the fundamental identity $\sin^{2}x + \cos^{2}x = 1$

Using the above formulas and $\cos(\pi/2) = 0$ (note that we have defined $\pi/2$ to be the smallest positive root of $\cos x = 0$), it can be easily shown that $\sin (x + 2\pi) = \sin x$.

Paramanand Singh
  • 87,309