Prove that $$\sin n\theta=n\sin \theta-\frac{n(n^2-1)}{3!}\sin^3\theta+\frac{n(n^2-1)(n^2-3^2)}{5!}\sin^5\theta+-\cdots$$

If I am not mistaken, this identity was either proven by Newton or known to him, so if possible I would really like to see the way he approached it, though any solution will suffice.

My brief efforts involved induction on $n$ which failed since I ended up with having to manipulate $\sin( n+1)\theta=\sin( n\theta +\theta)=\sin n\theta \cos\theta+\cos n\theta\sin\theta $, which involves $\cos n\theta$.

I tried the "familiar" method of expansion of $$\sin n\theta=n\theta-\frac{(n\theta)^3}{3!}+\frac{(n\theta)^5}{5!}-+\cdots$$ but this only made it more complicated$$\sin n\theta=n\theta-\frac{(n\theta)^3}{3!}+\frac{(n\theta)^5}{5!}-+\cdots=\\n\Big(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-+\cdots\Big)-\frac{n(n^2-1)}{3!}\Big(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-+\cdots\Big)^3+-\cdots$$ Another attempt would be to use the identity $$\sin n\theta=\frac{1}{2i}(e^{in\theta}-e^{-in\theta})$$ though this was obviously not known to Newton.
In any case, any thougths, ideas are welcome..

A thorough answer has been provided below, but since it involves the use of complex numbers, I deem that the search for another answer, one based solely on the mathematical knowledge up to Newton's time is open.
After a bit more research, it appears that Newton came up with the formula after reading a book by Vieta, but I have been unable to gather further info on whether the formula was known to Vieta as well.

  • $\begingroup$ Wonder if you give the general term. Was a bit worried, since if I put $n = 2 m$, $\theta = \pi/2$ then have $0 = 2m \, (1 - \frac{4m^2-1}{3!} + \frac{(4m^2-1)(4m^2 - 9)}{5!} + \cdots)$ $\endgroup$
    – jim
    Apr 11, 2016 at 16:52
  • $\begingroup$ @jim Well, I found the expression above exactly as it is, but your observation creates a new interesting problem I think! $\endgroup$ Apr 11, 2016 at 17:00
  • $\begingroup$ See this, from Bromwich, “An Introduction to the Theory of Infinite Series” books.google.com/… $\endgroup$
    – Steve Kass
    Apr 13, 2016 at 16:43
  • $\begingroup$ @jim: I have proved the formula for all odd integers $n$ (see my answer). The formula holds for all values of $n$ but the proof for general $n$ is difficult and involves the use of reduction formula for certain integrals. $\endgroup$
    – Paramanand Singh
    Apr 14, 2016 at 7:38
  • $\begingroup$ See some further formulas of this kind at math.stackexchange.com/q/432771/72031 $\endgroup$
    – Paramanand Singh
    Apr 14, 2016 at 7:44

5 Answers 5


This one comes from the classic S.L. Loney's Plane Trigonometry. We have the identity $$1 + 2x\cos t + 2x^{2}\cos 2t + 2x^{3}\cos 3t + \cdots = \frac{1 - x^{2}}{1 - 2x\cos t + x^{2}}\tag{1}$$ This is easily proved by letting $C$ denote the LHS of $(1)$ and $S$ denote $$S = 2x\sin t + 2x^{2}\sin 2t + \cdots$$ so that $$C + iS = 1 + 2xe^{it} + 2x^{2}e^{2it} + \cdots = \frac{1 + xe^{it}}{1 - xe^{it}} = \frac{(1 + xe^{it})(1 - xe^{-it})}{(1 - xe^{it})(1 - xe^{-it})}$$ and thus $$C + iS = \frac{1 + 2ix\sin t - x^{2}}{1 - 2x\cos t + x^{2}}$$ Equating the real part we get the value of sum $C$ as RHS of $(1)$.

It follows that $2\cos nt$ is the coefficient of $x^{n}$ in $\dfrac{1 - x^{2}}{1 - 2x\cos t + x^{2}}$ and thus we have \begin{align} 2\cos nt &= \text{coeff. of }x^{n}\text{ in }\dfrac{1 - x^{2}}{1 - 2x\cos t + x^{2}}\notag\\ &= \text{coeff. of }x^{n}\text{ in }(1 - 2x\cos t + x^{2})^{-1} - \text{coeff. of }x^{n - 2}\text{ in }(1 - 2x\cos t + x^{2})^{-1}\notag\\ &= \text{coeff. of }x^{n}\text{ in }\{1 + x(x - 2\cos t)\}^{-1} - \text{coeff. of }x^{n - 2}\text{ in }\{1 + x(x - 2\cos t)\}^{-1}\notag \end{align} Note that each term in $\{1 + x(x - 2\cos t)\}^{-1}$ is of the form $(-1)^{r}x^{r}(x - 2\cos t)^{r}$ and we need to evaluate the coefficient of $x^{n}$ and $x^{n - 2}$ in this expansion and subtract the two and by the above argument this coefficient will finally be $2\cos nt$.

Let us assume $n$ to be odd positive integer. In this case the first contribution in the desired coefficient comes from the term corresponding to $r = (n - 1)/2$ and in this case we get the coefficient of $x^{n - 2}$ as $$(-1)^{(n - 1)/2}\cdot\frac{n - 1}{2}\cdot(-2\cos t)$$ and in the final calculation this needs to be subtracted and hence the contribution for $r = (n - 1)/2$ is $$-(-1)^{(n - 1)/2}\cdot\frac{n - 1}{2}\cdot(-2\cos t)$$ Similarly we try to find the contributions for $r = (n + 1)/2, (n + 3)/2, \dots$ and finally get \begin{align} 2\cos nt &= (-1)^{(n - 1)/2}\left[-\frac{n - 1}{2}(-2\cos t)\right]\notag\\ &\,\,\,\, + (-1)^{(n + 1)/2}\left[\frac{n + 1}{2}(-2\cos t) - \dfrac{\dfrac{n + 1}{2}\cdot\dfrac{n - 1}{2}\cdot\dfrac{n - 3}{2}}{3!}(-2\cos t)^{3}\right]\notag\\ &\,\,\,\, + (-1)^{(n + 3)/2}\left[\binom{(n + 3)/2}{3}(-2\cos t)^{3} - \binom{(n + 3)/2}{5}(-2\cos t)^{5}\right]\notag\\ &\,\,\,\, + \dots\notag\\ &\,\,\,\, + (2\cos t)^{n}\notag \end{align} Hence by multiplying with $(-1)^{(n - 1)/2}$ we get \begin{align} &(-1)^{(n - 1)/2}(2\cos nt)\notag\\ &\,\,\,\,\,\,\,\,= \cos t\{(n - 1) + (n + 1)\} - \frac{(n + 1)(n - 1)}{3!}\cos^{3}t \{(n - 3) + (n + 3)\} + \cdots\notag \end{align} Hence by dividing by $2$ we get $$(-1)^{(n - 1)/2}\cos nt = n\cos t - \frac{n(n^{2} - 1^{2})}{3!}\cos^{3}t + \frac{n(n^{2} - 1^{2})(n^{2} - 3^{2})}{5!}\cos^{5}t - \cdots$$ Replacing $t$ by $(\pi/2) - \theta$ we get $$\sin n\theta = n\sin \theta - \frac{n(n^{2} - 1^{2})}{3!}\sin^{3}\theta + \frac{n(n^{2} - 1^{2})(n^{2} - 3^{2})}{5!}\sin^{5}\theta - \cdots\tag{2}$$ Note the similarity with the series $$\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots\tag{3}$$ Clearly we can derive $(3)$ from $(2)$ in an intuitive (but non-rigorous) way by putting $x = n\theta$ and keeping $x$ fixed and letting $n \to \infty$ so that $\theta \to 0$. In that case $n\sin \theta = n\sin(x/n) \to x$ as $n \to \infty$ and the formula $(2)$ is transformed into the series in equation $(3)$.

Note: This and many similar expansions of $\sin n\theta, \cos n\theta$ in powers of $\sin \theta, \cos \theta$ are provided in Loney's book and I have not found such elementary proofs anywhere else. I don't know if this classic textbook is studied anywhere or not. Also note that the formula $(2)$ is proved under the assumption that $n$ is an odd positive integer. By changing sign of $n$ it is easily seen that the formula holds for negative odd integers also. The same formula holds for all values of $n$ (the RHS is an infinite series if $n$ is not an integer and the identity should be treated as such in case $n$ is not an integer), but the proof of the identity for general $n$ can not be given via the approach mentioned in the above solution.

Update: From OP's comment it appears that he is interested in a solution which does not use complex numbers at all. Note that the solution presented above uses complex numbers to establish the identity $(1)$. Apart from this no use is made of complex numbers anywhere else.

I provide the following proof of $(1)$ without any use of complex numbers. Consider the expression $$f(n) = x^{n + 1}\cos (n - 1)t - x^{n}\cos nt$$ and then we have \begin{align} f(n) - f(n - 1) &= x^{n + 1}\cos (n - 1)t - x^{n}\cos nt - x^{n}\cos (n - 2)t + x^{n - 1}\cos (n - 1)t\notag\\ &= x^{n - 1}(1 + x^{2})\cos (n - 1)t - 2x^{n}\cos (n - 1)t\cos t\notag\\ &= (1 - 2x\cos t + x^{2})x^{n - 1}\cos (n - 1)t\tag{4} \end{align} Putting $n = 1, 2, \ldots, n$ in $(4)$ and adding the resulting equations we get $$f(n) - f(0) = (1 - 2x\cos t + x^{2})\sum_{k = 0}^{n - 1}x^{k}\cos kt$$ Let us now assume that $|x| < 1$ and then let $n \to \infty$. It is then obvious that $x^{n} \to 0$ and hence $f(n) \to 0$ and thus we have $$\sum_{k = 0}^{\infty}x^{k}\cos kt = -\frac{f(0)}{1 - 2x\cos t + x^{2}} = \frac{1 - x\cos t}{1 - 2x\cos t + x^{2}}$$ The sum in LHS of $(1)$ is given by $$-1 + 2\sum_{k = 0}^{\infty}x^{k}\cos kt = -1 + \frac{2 - 2x\cos t}{1 - 2x\cos t + x^{2}} = \frac{1 - x^{2}}{1 - 2x\cos t + x^{2}}$$ and thus the identity $(1)$ is established for all values of $x, t$ with the constraint $|x| < 1$.

  • $\begingroup$ Thank you for your answer and the very nice book "suggestion"! Since it involves the use of complex numbers it cannot be the one we seek, but again, it is perfectly satisfactory for the problem in general. $\endgroup$ Apr 13, 2016 at 15:29
  • 1
    $\begingroup$ @MathematicianByMistake: The use of complex numbers is only to prove the identity $(1)$. You can prove this identity without using complex numbers but it will be lengthy. $\endgroup$
    – Paramanand Singh
    Apr 14, 2016 at 1:31
  • 2
    $\begingroup$ @MathematicianByMistake: See the updated answer which avoids the use of complex numbers. $\endgroup$
    – Paramanand Singh
    Apr 14, 2016 at 7:27

The series can be retrieved by expressing the Maclaurin expansion of $\sin(nx)$ considered as a function of $\sin x$, avoiding then the use of complex numbers. Moreover, it can be extended to the case of non-integer $n$.

Let $u$ a real number, we introduce the notations $y=\sin x$ et $f(y)=\sin ux$. Successive derivations of $f(y)$ with respect to $x$ give \begin{align*} \cos xf^{\prime}(y)&=u\cos ux\\ -\sin xf^{\prime}(y)+(1-\sin^{2}x)f^{\prime\prime}(y)&=-u^{2}\sin ux % \end{align*} The latter expression can be written as \begin{equation} (1-y^{2})f^{\prime\prime}(y)=yf^{\prime}(y)-u^{2}f \end{equation} Now, successive derivation with respect to $y$ are \begin{align*} (1-y^{2})f^{(3)}(y) & =3yf^{\prime\prime}(y)+(1-u^{2})f^{\prime}(y)\\ (1-y^{2})f^{(4)}(y) & =5yf^{(3)}(y)+(4-u^{2})f^{\prime\prime}(y) \end{align*} This form suggest the recurrence relation: \begin{equation} (1-y^{2})f^{(n+2)}(y)=(2n+1)yf^{(n+1)}(y)+(n^{2}-u^{2})f^{(n)}(y) \end{equation} which is easily established. As $f(0)=0$ and $f^{\prime}(0)=u$, we deduce for $p\geq1$ \begin{align} f^{(2p)}(0)&=0\\ f^{(2p+1)}(0)&=u(1-u^{2})(3^{2}-u^{2})\dots\left[ (2p-1)^{2}-u^{2}\right] \end{align} The Maclaurin expansion of $f(y)$ gives: \begin{equation} \sin ux=u\sum\limits_{p=0}^{\infty}(1^{2}-u^{2})(3^{2}-u^{2})\dots\left[ (2p-1)^{2}-u^{2}\right] \frac{\sin^{2p+1}x}{(2p+1)!} \end{equation} (For $p=0$, the product of the factors in the summation is taken to be equal to 1 by definition).

Additional expansions can be obtained. By taking $y=\sin x$ and $g(y)=\cos ux$, and using the same method, we obtain the same recurrence expression. Now, $g(0)=1$ and $g^{\prime}(0)=0.$ Then \begin{align} g^{(2p+1)}(0)&=0\\ g^{(2p)}(0)&=u^{2}(2^{2}-u^{2})(4^{2}-u^{2})\dots\left[ (2p-2)^{2}-u^{2}\right] \end{align} Thus, \begin{equation} \cos ux=1+\sum\limits_{p=0}^{\infty}(0-u^{2})(2^{2}-u^{2})(4^{2}% -u^{2})\dots\left[ (2p)^{2}-u^{2}\right] \frac{\sin^{2p+2}x}{(2p+2)!} \label{cosux}% \end{equation} By changing $u\to2u$ in the previous expression, we obtain also \begin{equation*} \sin^2ux=\sum\limits_{p=0}^{\infty}(u^{2}-0)(u^{2}-1^{2})(% u^{2}-2^{2})\dots\left( u^{2}-p^{2}\right) \frac{\left( -1 \right)^p2^{2p+1}\sin^{2p+2}x}{(2p+2)!} \end{equation*} From derivation of the above expressions, we can also find the following expansions \begin{align} \frac{\cos ux}{\cos x}&=\sum\limits_{p=0}^{\infty}(1^{2}-u^{2})(3^{2}% -u^{2})\dots\left[ (2p-1)^{2}-u^{2}\right] \frac{\sin^{2p}x}{(2p)!}\\ \frac{\sin ux}{\cos x}&=\frac{1}{u}\sum\limits_{p=0}^{\infty}(0-u^{2}% )(2^{2}-u^{2})(4^{2}-u^{2})\dots\left[ (2p)^{2}-u^{2}\right] \frac{\sin ^{2p+1}x}{(2p+1)!} \label{Dercosux}% \end{align}


Theorem. For $\alpha\in\mathbb{C}$ and $|x|<1$, we have \begin{equation}\label{cos-arcsin-ser}\tag{CS} \cos(\alpha\arcsin x)=\sum_{k=0}^{\infty}\Biggl(\prod_{\ell=1}^{k}\bigl[4(\ell-1)^2-\alpha^2\bigr]\Biggr)\frac{x^{2k}}{(2k)!} \end{equation} and \begin{equation}\label{sin-arcsin-ser}\tag{SS} \sin(\alpha\arcsin x)=\alpha\sum_{k=0}^{\infty} \Biggl(\prod_{\ell=1}^{k}\bigl[(2\ell-1)^2-\alpha^2\bigr]\Biggr) \frac{x^{2k+1}}{(2k+1)!}. \end{equation} Consequently, for $\alpha\in\mathbb{C}$ and $t\in\bigl(-\frac{\pi}2, \frac{\pi}2\bigr)$, we have \begin{equation}\label{cos-arcsin-ser-C}\tag{CT} \cos(\alpha t)=\sum_{k=0}^{\infty}\Biggl(\prod_{\ell=1}^{k}\bigl[4(\ell-1)^2-\alpha^2\bigr]\Biggr)\frac{\sin^{2k}t}{(2k)!} \end{equation} and \begin{equation}\label{sin-arcsin-ser-C}\tag{ST} \sin(\alpha t)=\alpha\sum_{k=0}^{\infty} \Biggl(\prod_{\ell=1}^{k}\bigl[(2\ell-1)^2-\alpha^2\bigr]\Biggr) \frac{\sin^{2k+1}t}{(2k+1)!}. \end{equation}

Proof. Let $f_\alpha(x)=\cos(\alpha\arcsin x)$. Then \begin{gather*} f_\alpha'(x)=-\frac{\alpha}{\sqrt{1-x^2}\,}\sin(\alpha\arcsin x)=-\frac{\alpha}{\sqrt{1-x^2}\,}\sqrt{1-f_\alpha^2(x)}\,\\ \bigl(1-x^2\bigr)\bigl[f_\alpha'(x)\bigr]^2-\alpha^2\bigl[1-f_\alpha^2(x)\bigr]=0,\\ \bigl(1-x^2\bigr)f_\alpha''(x)-xf_\alpha'(x)+\alpha^2f_\alpha(x)=0,\\ \bigl(1-x^2\bigr)f_\alpha^{(3)}(x)-3xf_\alpha''(x)+\bigl(\alpha^2-1\bigr)f_\alpha'(x)=0,\\ \bigl(1-x^2\bigr)f_\alpha^{(4)}(x)-5xf_\alpha^{(3)}(x)+\bigl(\alpha^2-4\bigr)f_\alpha''(x)=0, \end{gather*} and, inductively, \begin{equation}\label{n-deriv-gen+alpha}\tag{P1} \bigl(1-x^2\bigr)f_\alpha^{(k+2)}(x)-(2k+1)xf_\alpha^{(k+1)}(x)+\bigl(\alpha^2-k^2\bigr)f_\alpha^{(k)}(x)=0, \quad k\ge0. \end{equation} Letting $x\to0$ in \eqref{n-deriv-gen+alpha} results in \begin{equation}\label{n-deriv-x=0+alpha}\tag{P2} f_\alpha^{(k+2)}(0)+\bigl(\alpha^2-k^2\bigr)f_\alpha^{(k)}(0)=0, \quad k\ge0. \end{equation} Recusing the relation \eqref{n-deriv-x=0+alpha} and considering $f_\alpha(0)=1$ and $f_\alpha'(0)=0$ arrive at \begin{equation*} f^{(2k)}(0)=\prod_{\ell=1}^{k}\bigl[4(\ell-1)^2-\alpha^2\bigr] \quad\text{and}\quad f^{(2k+1)}(0)=0 \end{equation*} for $k\ge0$. Consequently, the series expansion \eqref{cos-arcsin-ser} is valid.

Let $f_\alpha(x)=\sin(\alpha\arcsin x)$. Then \begin{gather*} f_\alpha'(x)=\frac{\alpha}{\sqrt{1-x^2}\,}\cos(\alpha\arcsin x)=\frac{\alpha}{\sqrt{1-x^2}\,}\sqrt{1-f_\alpha^2(x)}\,\\ \bigl(1-x^2\bigr)\bigl[f_\alpha'(x)\bigr]^2-\alpha^2\bigl[1-f_\alpha^2(x)\bigr]=0,\\ \bigl(1-x^2\bigr)f_\alpha''(x)-xf_\alpha'(x)+\alpha^2f_\alpha(x)=0,\\ \bigl(1-x^2\bigr)f_\alpha^{(3)}(x)-3xf_\alpha''(x)+\bigl(\alpha^2-1\bigr)f_\alpha'(x)=0,\\ \bigl(1-x^2\bigr)f_\alpha^{(4)}(x)-5xf_\alpha^{(3)}(x)+\bigl(\alpha^2-4\bigr)f_\alpha''(x)=0, \end{gather*} and, inductively, \begin{equation}\label{n-deriv-gen+alpha-sin}\tag{P3} \bigl(1-x^2\bigr)f_\alpha^{(k+2)}(x)-(2k+1)xf_\alpha^{(k+1)}(x)+\bigl(\alpha^2-k^2\bigr)f_\alpha^{(k)}(x)=0, \quad k\ge0. \end{equation} Letting $x\to0$ in \eqref{n-deriv-gen+alpha-sin} results in \begin{equation}\label{n-deriv-x=0+alpha-sin}\tag{P4} f_\alpha^{(k+2)}(0)+\bigl(\alpha^2-k^2\bigr)f_\alpha^{(k)}(0)=0, \quad k\ge0. \end{equation} Considering $f_\alpha(0)=0$ and $f_\alpha'(0)=\alpha$ in \eqref{n-deriv-x=0+alpha-sin} and recusing the relation \eqref{n-deriv-x=0+alpha-sin} arrive at \begin{equation*} f^{(2k)}(0)=0 \quad\text{and}\quad f^{(2k+1)}(0)=\alpha \prod_{\ell=1}^{k}\bigl[(2\ell-1)^2-\alpha^2\bigr] \end{equation*} for $k\ge0$. Consequently, the series expansion \eqref{sin-arcsin-ser} is valid.

Replacing $\arcsin x$ by $t$ in \eqref{cos-arcsin-ser} and \eqref{sin-arcsin-ser} yields \eqref{cos-arcsin-ser-C} and \eqref{sin-arcsin-ser-C} immediately.

Remark. The above texts are excerpted from Lemma 3.3 and its proof in the paper [1] listed below. In Remark 18 of the paper [2], several related results are discussed. The preprint [1] has been formally published as the paper [3] below.


  1. F. Qi, Taylor's series expansions for real powers of functions containing squares of inverse (hyperbolic) cosine functions, explicit formulas for special partial Bell polynomials, and series representations for powers of circular constant, arXiv (2021), available online at https://arxiv.org/abs/2110.02749v2.
  2. B.-N. Guo, D. Lim, and F. Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Appl. Anal. Discrete Math. 16 (2022), in press; available online at https://doi.org/10.2298/AADM210401017G.
  3. Feng Qi, Taylor's series expansions for real powers of two functions containing squares of inverse cosine function, closed-form formula for specific partial Bell polynomials, and series representations for real powers of Pi, Demonstratio Mathematica 55 (2022), no. 1, in press; available online at https://doi.org/10.1515/dema-2022-0157.

hint $$\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}$$ then you can use $e^{ix}=\cos(x)+i\sin(x)$ and then use binomial for $(a+b)^n-(a-b)^n$ by using the relation that $\cos(x)=\sqrt{1-\sin^2(x)}$

  • 1
    $\begingroup$ Thank you for the hint! As I mentioned to Paramanand above, we are still looking for Newton's approach.. $\endgroup$ Apr 13, 2016 at 15:30

I suppose you could also make use of the formula $\left(\begin{matrix} \cos n\theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{matrix} \right)=\left(\begin{matrix} \cos \theta & \sin \theta\\ -\sin \theta & \cos \theta \end{matrix} \right)^n$ and substitute $s = \sin \theta, c = \cos \theta = \sqrt{1-s^2}$? Ultimately, though, this is equivalent to the other answers.


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