I am reading Euler's "Introduction to analysis of the infinite", chapter 8, page at the end of page 208, beginning of page 209 and came across his derivation of the power series for $\sin(x)$ and $\cos(x)$. I try hard to follow his steps but fail flat.

He wrote as followed:

$\sin(nz)=\dfrac{n}{1}(\cos(z))^{n-1}\sin(z)+\dfrac{n(n-1)(n-2)}{3!}\cos(z)^{n-3}(\sin(z))^{3}+\dfrac{n(n-1)(n-2)(n-3)(n-4)}{5!}(\cos(z))^{n-5}(\sin(z)^5 .etc$

$\cos(nz)=\cos(z)^{n}+\dfrac{n(n-1)}{2!}(\cos(z))^{n-2}(\sin(z))^{2}+\dfrac{n(n-1)(n-2)(n-3)}{4!}\cos(z)^{n-4}(\sin(z)^4$ .etc

He stated:

Let the arc of z be infinitely small; there will be $\sin(z)=z$ and $\cos(z)=1$; moreover $n$ shall be an infinitely great number, so that the arc $nz$ shall be of finite magnitude, for example $nz=v$; on account of $\sin(z)=z=\dfrac{v}{n}$ there becomes



I have a couple of questions:

  1. What does it mean when he says $z$ be infinitely small and $sin(z)=z$? On the interval $(0,\dfrac{\pi}{2})$, we have the equality $\sin(z)<z$. Besides $-1\leq\sin(z)\leq1$ $\forall z$, does it mean that $z$ approach $0$, because obviously $\sin(z)=z$ only when $z=0$. This makes sense when $\cos(0)=1$, as he stated. Also, what does he mean when he said $n$ is infinitely large?

  2. I've tried to substitute $\sin(z)=z=\dfrac{v}{n}$ and $\cos(z)=1$ but I could not eliminate the whole bunch of $n(n-1)(n-2)$ (for $\sin(nz)$) or $n(n-1)$ (for $\cos(nz)$) or the likes to arrive at the final power series for both functions. Here are what I've got.

$\sin(nz)=\dfrac{n\cdot z}{1}-\dfrac{n(n-1)(n-2)}{3!}z^3+\dfrac{n(n-1)(n-2)(n-3)(n-4)}{5!}z^5+ .etc$

Obviously the first term is $v$, but what about the rest of the terms?

$\cos(nz)=1-\dfrac{n(n-1)}{2!}z^2+\dfrac{n(n-1)(n-2)(n-3)}{4!}z^4- .etc$

Obviously because I don't know what is the value of $n$ so I can't make them varnish. Could you help me here? Thank you so much!

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    $\begingroup$ pdmi.ras.ru/~olegviro/Shchepin/index.html $\endgroup$ Commented Feb 5, 2021 at 3:20
  • $\begingroup$ @ AlexandreEremenko Thanks for your resource, very helpful. But can you answer my question directly? Professor? :) $\endgroup$ Commented Feb 5, 2021 at 3:29
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    $\begingroup$ The approximation sin(z) ~= z is valid for z very small (and hence centered around z=0). It's a very good approximation, the sin(z) function for real z is very nearly a line (try plotting sin(z) and z and they will nearly overlap near z=0) $\endgroup$ Commented Feb 5, 2021 at 12:30
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    $\begingroup$ For the "math rookies" reading the comments, $z$ is always in radians. $sin(degrees)$ doesn't converge to $sin(q) \approxeq q$ very well $\endgroup$ Commented Feb 5, 2021 at 12:38
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    $\begingroup$ I think this belongs to either Math.OF or Math.SE. $\endgroup$ Commented Feb 5, 2021 at 12:45

1 Answer 1


Concerning your question "Obviously the first term is $v$, but what about the rest of the terms?". Since he considers $n$ to be infinitely large he probably has $n(n-1)(n-2)\cdots (n-(k-1))=n^k$ so that $n(n-1)(n-2)\cdots (n-(k-1))z^k=v^k$.


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