Sine and cosine

We define the function ${\displaystyle \cos(x)}$ by requiring that

${\displaystyle \cos ''(x)=-\cos(x),\quad \cos(0)=1}$  and  ${\displaystyle \cos '(0)=0.}$

If you sketch the graph of this function using only this information, you will notice that wherever ${\displaystyle \cos(x)}$ is positive, its slope decreases as ${\displaystyle x}$ increases (that is, its graph curves downward), and wherever ${\displaystyle \cos(x)}$ is negative, its slope increases as ${\displaystyle x}$ increases (that is, its graph curves upward).

Differentiating the first defining equation repeatedly yields

${\displaystyle \cos ^{(n+2)}(x)=-\cos ^{(n)}(x)}$

for all natural numbers ${\displaystyle n.}$ Using the remaining defining equations, we find that ${\displaystyle \cos ^{(k)}(0)}$ equals 1 for k = 0,4,8,12…, –1 for k = 2,6,10,14…, and 0 for odd k. This leads to the following Taylor series:

${\displaystyle \cos(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}=1-{x^{2} \over 2!}+{x^{4} \over 4!}-{x^{6} \over 6!}+\dots .}$

The function ${\displaystyle \sin(x)}$ is similarly defined by requiring that

${\displaystyle \sin ''(x)=-\sin(x),\quad \sin(0)=0,\quad {\hbox{and}}\quad \sin '(0)=1.}$

This leads to the Taylor series

${\displaystyle \sin(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}=x-{x^{3} \over 3!}+{x^{5} \over 5!}-{x^{7} \over 7!}+\dots .}$