I had this crazy idea trying to prove the Pythagorean trigonometric identity;$$\sin^2x+\cos^2x=1$$by squaring the infinite Taylor series of $\sin x$ and $\cos x$. But it came out quite beautiful, involving also a combinatorics identitie.

The proof:

$$\sin x=\frac{x}{1}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\\\\\sin^2x=x^2-x^4\left (\frac{1}{1!3!}+\frac{1}{3!1!}\right )+x^6\left (\frac{1}{1!5!}+\frac{1}{3!3!}+\frac{1}{5!1!}\right )-...\\\\\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\\\\\cos^2x\!=\!1\!-\!x^2\left(\!\frac{1}{0!2!}\!+\!\frac{1}{2!0!}\!\right)\!+\!x^4\left(\!\frac{1}{0!4!}\!+\!\frac{1}{2!2!}\!+\!\frac{1}{4!0!}\!\right)\!-\!x^6\left(\!\frac{1}{0!6!}\!+\!\frac{1}{2!4!}\!+\!\frac{1}{4!2!}\!+\!\frac{1}{6!0!}\!\right)\!+...$$We should have shown that the series for both $\sin x$ and $\cos x$ converge absolutely (since we changed the arrangement), but it's obvious since the absolute value of all terms of $\sin x+\cos x$ add up to $e^x$.$$\sin^2x+\cos^2x=\\=1-x^2\left(\frac{1}{0!2!}-\frac{1}{1!1!}+\frac{1}{2!0!}\right)+x^4\left(\frac{1}{0!4!}-\frac{1}{1!3!}+\frac{1}{2!2!}-\frac{1}{3!1!}+\frac{1}{4!0!}\right)-x^6\left(\frac{1}{0!6!}-\frac{1}{1!5!}+\frac{1}{2!4!}-\frac{1}{3!3!}+\frac{1}{4!2!}-\frac{1}{5!1!}+\frac{1}{6!0!}\right)+...=\\\\=1+\sum_{n=1}^{\infty}(-1)^nx^{2n}\sum_{k=0}^{2n}\frac{(-1)^k\binom{2n}{k}}{(2n)!}$$ Since we can show easily that $\sum_{i=0}^n(-1)^i\binom{n}{i}=0$ by expanding $(1-1)^n$ using Binom's formula. So:$$\sin^2x+\cos^2x=1-0+0-0+...=1$$

I think it's beautiful. I just wanted to ask, do Taylor's series of those functions depend on this identity? Because if they do, the proof will be circular.


Nice proof! To address your question,

Do Taylor's series of those functions depend on this identity? Because if they do, the proof will be circular.

You are right to be worried about circularity! However, which concepts or theorems depend on which others is a matter of certain flexibility. Often, we take one thing to be the definition of a concept, and then have to prove the other properties as theorems -- but we could alternately have used some other property as a definition, and then the original definition would have to be a theorem.

Specific to your case, I have seen definitions of $\sin$ and $\cos$ where we start by defining $\sin$ using arclength, then we define $\cos x$ to satisfy $\cos^2 x + \sin^2 x = 1$. If we take this approach, certainly, there is nothing to prove. However, this is not the only possible approach! It is also common to define $\sin$ and $\cos$ using their Taylor series. Under this approach, you have given a very nice proof that $\sin^2 x + \cos^2 x = 1$. Your proof could also be valid if we define $\sin$ and $\cos$ to be a basis of functions satisfying $f''(x) = -f(x)$.

In summary, it depends on what you define $\sin$ and $\cos$ to be; however, your proof is not necessarily circular. And it is a nice example of deriving a result about some functions from their Taylor series.

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    $\begingroup$ Thanks for this answer, but what about the usual definition of $sin$ and $cos$, as the proportion of sides of a triangle, that of course also doesn't depend on $\sin^2x+\cos^2x=1$, am I right? $\endgroup$ – 76david76 Sep 20 '16 at 22:17
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    $\begingroup$ @76david76: Are $\sin$ and $\cos$ usually defined that way? That's often the way they're introduced, but then what is $\sin 3\pi/2$? $\endgroup$ – Brian Tung Sep 20 '16 at 22:31
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    $\begingroup$ It can be an expansion of the first definition. Is there really no way to define these functions out of the unit circle? $\endgroup$ – 76david76 Sep 20 '16 at 22:49
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    $\begingroup$ @76david76 Most functions in analysis are not defined using geometry because we would first need a rigorous definition of points, lines, and so on. It is much easier to define them using taylor series. However, yes, you can develop these functions using the unit circle. When I said "we start by defining sinsin using arclength", that is referring to how we let $\sin \theta$ be the $y$-coordinate of the point on the unit circle such that the arc length from $(1,0)$ is $2 \pi y$. If you define them using the unit circle, $\cos^2 \theta + \sin62 \theta$ is immediate like I said. $\endgroup$ – 6005 Sep 20 '16 at 23:11

No, they do not depend on the Pythagoren identity. Using the differential equations that define $\sin$ and $\cos$

  • $\cos(0) = 1$
  • $\sin(0) = 0$
  • $\cos' = -\sin$
  • $\sin' = \cos$

it's very easy to show that $\forall x \in \mathbb{R}\cos(x) = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i}}{(2x)!}$ and $\forall x \in \mathbb{R}\sin(x) = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i + 1}}{(2x + 1)!}$.


You are just showing that the Pythagorean theorem is a consequence of some property of the (complex) exponential function, there is no circularity in such argument. For instance, we may define, for any $z\in\mathbb{C}$, $$ f(z)=e^{z}=\sum_{n\geq 0}\frac{z^n}{n!} \tag{1}$$ and prove through a combinatorial argument that such a function fulfills $e^{z}\cdot e^{w}=e^{z+w}$.
Since the Taylor coefficients at $0$ of such analytic function are real, we have $f(\bar{z})=\overline{f(z)}$,
hence for any $\rho\in\mathbb{R}$ $$ \left\| e^{i\rho}\right\| = e^{i\rho}\cdot e^{-i\rho} = e^0 = 1. \tag{2}$$ If we define $\cos(\rho)$ and $\sin(\rho)$ as the real/imaginary part of $e^{i\rho}$, we get that $$ \sin(\rho)=\sum_{n\geq 0}\frac{(-1)^n \rho^{2n+1}}{(2n+1)!},\qquad \cos(\rho)=\sum_{n\geq 0}\frac{(-1)^n \rho^{2n}}{(2n)!}\tag{3} $$ and $(2)$ can be read as: $$ \sin^2(\rho)+\cos^2(\rho) = 1.\tag{4}$$

  • $\begingroup$ I was going to upvote this answer except I saw a problem so I downvoted it. It is true that $\forall \rho \in \mathbb{R}\sin^2(\rho) + \cos^2(\rho) = 1$. However, you didn't prove that and that's the reason I downvoted your answer. Without the extra detail of the Pythagorean identity, I would have considered your answer good and upvoted it. It shows that there is a very nice easy way to prove what the Taylor series' of $\sin$ and $\cos$ are that's actually much easier than squaring the Taylor series of $\sin$ then subtracting it from 1 then figuring out which square root of it is $\cos$. $\endgroup$ – Timothy Jun 2 at 23:06
  • $\begingroup$ @Timothy: the Pithagorean theorem is proved in (2). $\endgroup$ – Jack D'Aurizio Jun 3 at 9:03

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