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 $\lim_{x\to\infty}c (x)=0$, 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 $$ s(x+2k\pi)=s(x),\ \ s(x+\pi/2)=c(x),\ \ \ c(x+2k\pi)=c(x). $$

If you are given a function $f:A\to B$ then you are actually given a rule using which given any member of $x\in A$ you can find its image $f(x) \in B$. The function $\sin x$ defined by equation $$\sin x = x - \frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\dots$$ also gives a rule in a direct manner to calculate $\sin x$ given $x$.

Consider the other definition of $\sin x$ given using ratio of specific sides of a right angled triangle. Do you see how we can calculate $\sin x$ given $x$ according to this definition? The fundamental challenge is that given a value $x$ say $x=1$, how do we get to a proper right triangle for this $x=1$ on the basis of which the desired ratio of sides of this triangle can be calculated? What to do when $x=1000$ or $x=\sqrt{2}$? At least there should be a way in theory to get a right angled triangle based on $x$. Further we need to deal with the case when $x=0$.

When you ponder over the questions raised in last paragraph you realize that the definition of $\sin x$ based on triangles is grossly inadequate and does not get to the essence of the real nature of function $\sin x$. When looked at this way these functions are called trigonometric functions (trigon being another word for triangle). The only advantage of this approach is the ease with which it can be taught to young students of age 13-14 years and moreover the students can be accustomed to practical applications like heights and distances. In fact on a deeper level this approach is equivalent to exploiting the idea of similarity of triangles in a systematic fashion and is thus nothing more than an extension of elementary Euclidean geometry.

The real geometric definition of $\sin x$ does not use triangles, but rather circles and the functions involved are called circular functions. In this approach we can get the value of $\sin x$ given $x$ (at least in theory). Unfortunately this approach requires the use of calculus/analysis for its justification, but luckily the presentation itself does not require any of the powers of calculus. I have presented such an approach (along with justification based on calculus) in another answer. Using this approach we can get the derivative formulas $(\sin x) '=\cos x, (\cos x)' =-\sin x$ from which the Taylor series for these functions can be derived easily and we reach the definition based on infinite series.