Relationship Between Sine as a Series and Sine in Triangles

Define $$s (x)=\sum_{k=0}^\infty (-1)^k\frac {x^{2k+1}}{(2k+1)!}. $$ This is a power series with infinite radius of convergence; in particular, we are allowed to differentiate and to take antiderivatives term by term. Denote its derivative by $c (x)=s' (x) $. An easy computation shows that $c'(x)=-s (x) $. So $c (x)c'(x)+s (x)s' (x)=0$. This is the derivative of $c (x)^2+s (x)^2$, which is thus constant; using that $s (0)=0$ and $c (0)=1$, we obtain $$\tag {1}c (x)^2+s (x)^2=c(0)^2+s(0)^2=1.$$ One can also get from the series that $$s (-x)=-s (x),\ \ \ \ c (-x)=c (x), $$ and that$^1$ $$\tag {2}s (x+y)=s (x)c (y)+s (y)c (x), $$ $$c (x+y)=c (x)c (y)-s (x)s (y).$$ From $(1) $ we get that $-1\leq s (x)\leq1$ for all $x $. Suppose that $c(x) >0$ for all $x $; this would imply that $s (x) $ is always increasing and bounded, so $\lim_{x\to\infty}c (x)=0$ (details at the bottom), but this would contradict $$c (2x)=c (x)^2-s (x)^2=2c (x)^2-1.$$ Similarly, we cannot have $c(x)<0$ for all $x$. So there exists $\beta>0$ with $c (\beta)=0$; by $(1) $ $s (\beta)=1$, and by $(2)$ $s (2\beta)=0$. We can take the least such $\beta>0$: since $c(0)=1$, we are guaranteed that $c(x)>0$ on $[0,\beta)$. Now from $(1)$ and $s(-x)=-s(x)$ we get that the range of $s (x) $ is $[-1,1] $. Thus $(c (t),s (t)) $, $0\leq t \leq 2\beta $, parametrizes the upper unit semi circle, and $2\beta $ is the number we usually call $\pi $. The length of the arc corresponding to the parameter running from $0$ to $x $, is $$ \int_0^x\sqrt {c'(t)^2+s' (t)^2}\,dt=\int_0^x1\,dt=x. $$ So the angle (in radians) corresponding to the arc joining $(1,0) $ and $(c (x),s (x)) $ is $x $.

Now take your square triangle with angle $x $, opposite $b $, and hypotenuse $h $. Fit it in the circle of radius $h $, with the angle $x $ lying at the origin, and the adjacent side on the positive horizontal axis. By the above, the vertex touching the circle has coordinates $(hc (x),hs (x)) $. So $b=hs (x) $, and $$s (x)=\frac bh=\sin x.$$

1. One can prove $(2)$ by direct manipulation of the series, but here is another method. Fix $y$, and define $f(x)=s(x+y)$. Then $$\tag{3} f''(x)=-f(x),\ \ \ \ f(0)=s(y),\ \ f'(0)=c(y). $$ It is well-known that the initial-value-problem above has a unique solution; posing that $f(x)=\sum_{k=0}^\infty a_kx^k$, one readily obtains that $$ a_{2k}=\frac{(-1)^ks(y)}{(2k)!},\ \ \ a_{2k+1}=\frac{(-1)^kc(y)}{(2k+1)!}. $$ It follows immediately that $$ s(x+y)=f(x)=c(x)s(y)+s(x)c(y). $$

For the sake of completeness, let us show here that both functions $s(x)$ and $c(s)$ have period $4\beta$ (i.e., $2\pi$). We first go by induction: first, by $(2)$ $$ s(2\beta)=2s(\beta)\,c(\beta)=0. $$Now assume that $s(2(k-1)\beta)=0$. Then, by $(2)$, $$ s(2k\beta)=s(2(k-1)\beta+2\beta)=s(2(k-1)\beta)\,c(2\beta)+c(2(k-1)\beta)\,s(2\beta)=0. $$ Also from $(2)$, $$ c(2\beta)=c(\beta+\beta)=c(\beta)^2-s(\beta)^2=0^2-1^2=-1, $$ and $$c(4\beta)=c(2\beta)^2-s(2\beta)^2=(-1)^2-0^2=1.$$Then by induction we obtain $$c(2k\beta)=(-1)^k.$$ Now $$ s(x+4k\beta)=s(x)c(4k\beta)+c(x)s(4k\beta)=s(x)c(4k\beta)=s(x). $$ We can obtain the periodicity of $c(x)$ in the same way, or use $$ s(x+\beta)=s(x)c(\beta)+c(x)s(\beta)=0+c(x)=c(x). $$ Then $$ c(x+4k\beta)=s(x+4k\beta+\beta)=s(x+\beta)=c(x). $$ After we write $\pi=2\beta$, the three formulas we found become $$ s(x+2k\pi)=s(x),\ \ s(x+\pi/2)=c(x),\ \ \ c(x+2k\pi)=c(x). $$

Proof that if $c(x)>0$ for all $x$ then $\lim_{x \to\infty}c(x)=0$. Since $s'(x)=c(x)>0$ for all $x$ and $s(x)\leq1$, we get that $s_0=\lim_{x \to\infty}s (x)$ exists. Similarly, from $c'(x)=-s(x)<0$ and $c(x)>0$ we get that $c_0=\lim_{x\to\infty}c(x)$ exists. If $c_0>0$, there exists $x_0$ such that $s(x)>s_0-\frac{c_0}2$ for all $x\geq x_0$. Then $$ s (x_0+1)=s (x _0)+\int_{x _0}^{x _0+1}c\geq s_0-\frac{c_0}2+c_0>s_0, $$ a contradiction. Hence $c_0=0$, as claimed.