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I'm taking a single variable calculus course and sine series is defined as

$$ \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k + 1)!}. $$

Sine is also defined as opposite length over hypotenuse length for a right angled triangle.

So $$\frac{\text{opposite}}{\text{hypotenuse}} = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k + 1)!}? $$

Is there a concept I'm not understanding here?

Austin Mohr
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blue-sky
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  • no, you're correct. try plugging in values of $x \in [0, \pi]$ into the series, and also working out the sine of a right angled triangle with angle $x$. – mdave16 Apr 12 '17 at 11:44
  • See my answer to a similar (duplicate?) question, which shows Chaikovsky's construction of the terms of the power series as lengths of involutes related to the unit-hypotenuse right triangle. – Blue Feb 01 '20 at 04:09

2 Answers2

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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.

Martin Argerami
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  • I think your argument for the existence of $\beta (=\pi/2)$ is slightly flawed. You say “Suppose that $c(x)>0$ for all $x$; this would imply that $s(x)$ is always increasing and bounded, so $\lim\limits_{x\to\infty}c(x)=0$, but this would contradict…” This would be correct if the limit actually existed in the first place; but it is conceivable (and of course what actually happens) that the limit doesn’t exist, so we don’t get a contradiction. I’m sure that a correct argument must also invoke the fact that $c’=-s$ (but you only seemed to have used $s’=c$). – peek-a-boo Mar 24 '24 at 05:33
  • @peek-a-boo: the contradiction arises on the assumption that $c(x)>0$ for all positive $x$. This leads to $s$ being increasing and bounded, so the limit exists. The existence of the limit implies that the derivative converges to zero, and one gets the contradiction. I have written this with more care somewhere else, I'll try to dig it up tomorrow. – Martin Argerami Mar 24 '24 at 05:42
  • Perhaps, but I still don’t see it immediately. Because you could imagine keeping $c(x)$ positive, but say on a very small interval $(n-\epsilon,n+\epsilon)$ make $c(x)$ very large and positive, but then for the remainder of the interval, make it very small and positive (then make it peak again at $x=n+1$), so $\limsup\limits_{x\to\infty} c(x)=\infty$ and $\liminf\limits_{x\to\infty}c(x)=0$. One way I see of arguing this away is that $c’=-s$ which would then be negative on $(0,\infty)$, so $c$ is itself decreasing and hence has a limit (prohibiting my above type of counterexample). – peek-a-boo Mar 24 '24 at 05:45
  • That said, I much prefer your proof. – Martin Argerami Mar 24 '24 at 05:45
  • @peek-a-boo: at that stage in my argument, we already know that $-1\leq c(x),s (x)\leq 1$. – Martin Argerami Mar 24 '24 at 05:47
  • ok fine, then make $c(x)$ be equal to $0.5$ on a small interval. Then we’d still have $\limsup c(x)>0$. My point is that we need some extra information about monotonicity of $c$ (and that we do get from the other defining equation $c’=-s$). – peek-a-boo Mar 24 '24 at 05:48
  • I see what you mean. I'll think about it. – Martin Argerami Mar 24 '24 at 05:50
  • I have added the argument I had. In the end, it's pretty similar to yours. Regarding your previous comment, I never meant to say that the argument worked without $c'=-s$. And thanks for your remarks, feedback is scarce and very much appreciated. – Martin Argerami Mar 24 '24 at 11:02
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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.

Paramanand Singh
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