# Is there a formula for $\sin(xy)$

Can we express a trigonometric function for the product of two angles as a function of trigonometric functions of its factors?

For example: Is there a formula for $\sin(xy)$ as a function of $\sin x$ and $\sin y$ or other trigonometric functions of $x$ and $y$.

• If $x$ or $y$ is an integer, then you have Chebyshev Polynomials of the first kind, but I'm not sure if there is a way to do this with $x,y \in \mathbb{R}$ in general. These are a family of recursively defined polynomials with the property that $f_n(\cos(x)) = \cos(nx)$. Then, $\sin(nx)$ can be calculated from $\cos(nx)$ via $\sin^2 + \cos^2 = 1$.
– user242594
Feb 15, 2016 at 7:21
• Some time ago, I asked this question. If there is a formula (let's say a rational fraction in $\cos$ and $\sin$) for $\sin(xy)$, then there would be a formula for $f(x)=\sin(x^2)$, which is not possible because $f$ is not $2π$-periodic. Feb 15, 2016 at 10:06
• This question is related. Feb 15, 2016 at 10:16
• @Watson Perhaps you may check my answer. Feb 17, 2016 at 1:15
• @JeanMarie The answers suggest that no such formula exists, except under some approximations, hovewer, any combination of angles which makes sense physically, eg. x+y, x-y, x/2+y/3 etc. have corresponding formulas, we can take this as a definition for which combinations makes sense, Ideally, angles are dimensionless, that's why terms like sin(wt) exist. May 23, 2021 at 14:41

Unlikely to be explicitly/purely in terms of $$\sin x$$, $$\sin y$$, etc.

But $$\, \sin xy=\sin \left[ \frac{(x+y)^{2}}{4}-\frac{(x-y)^{2}}{4} \right] =\sin \frac{(x+y)^{2}}{4} \cos \frac{(x-y)^{2}}{4}- \cos \frac{(x+y)^{2}}{4} \sin \frac{(x-y)^{2}}{4}$$

or approximately for small $$\, x$$ and $$y$$,

$$\sin xy \approx \sin x \sin y \left( 1+\frac{\sin^{2} x+\sin^{2} y}{6}+\frac{3}{40}\sin^{4} x-\frac{5}{36}\sin^{2} x \sin^{2} y+\frac{3}{40} \sin^{4} y \right)$$

Actually, there is a sort of formula. Let $$T_d^n=\cos\left(\frac{n-d}{2}\pi\right)\frac{2^dn}{n+d}{{\frac{n+d}2}\choose d}$$ then we have $$\cos(n \arccos(x))=\sum_{d=0}^\infty T_d^nx^d \,\,\,\,\,\,\,\,\,\forall |x|<1$$ as both satisfy the differential equation (with a unique solution) $$(1-x^2)f''=xf'-n^2f, f(0)=\cos\left(\frac{n\pi}2\right),f'(0)=n\sin\left(\frac{n\pi}2\right)$$

Then note that $\cos(nx)=\cos(n \arccos(\cos x))$ for $|x|\leq\pi$ (for integer $n$ however the statement is true for all $x$). Therefore we have $$\cos(n x)=\sum_{d=0}^\infty T_d^n\cos^d(x) \,\,\,\,\,\,\,\,\,\forall |x|\leq\pi$$ and by differentiating once and doing some multiplying, we get $$\sin(n x)=\sum_{d=0}^\infty \frac{d\cdot T_d^n}{n}\cos^{d-1}(x)\sin(x) \,\,\,\,\,\,\,\,\,\forall |x|\leq\pi$$

Note: When $n$ is an integer, $T_d^n=0$ for $d>n$ (the binomial coefficient is $0$), so the series will actually be finite. The $\cos\left({\frac{n-d}{2}\pi}\right)$ will also have a repeating pattern of $\{1,0,-1,0\}$, and the formula will be valid for all $x$. These are the so-called Chebyshev Polynomials.

AFAIK, no, not for the general case $x,y \in \mathbb{R}$. But if $y=n \in \mathbb{N}$, we have

$$\sum_{k=0}^n \binom{n}{k}\cos^kx\sin^{n-k}x\sin \left[ \frac{1}{2}(n-k)\pi \right]$$

There's a couple of other formulations of the linked page, but as said, I doubt that this has been extended to cover all reals. But I have been wrong before and would myself also be very interested in finding such a function.

• Thanks for your useful suggestion! Feb 15, 2016 at 8:25
• What does AFAIK mean? Feb 17, 2016 at 0:30
• @SimpleArt "As Far As I Know" :) Feb 17, 2016 at 7:54

If, say, $-\pi/2 \le x \le \pi/2$, and $-\pi/2 \le y \le \pi/2$, then $$\sin(xy) = \sin(\arcsin(\sin(x)) \arcsin(\sin(y)))$$ expresses $\sin(xy)$ as a function of $\sin(x)$ and $\sin(y)$. Other formulas can be used for other intervals.

On the other hand, there is no way to write $\sin(xy)$ as a continuous function of a finite number of functions of the form $\sin(\alpha x)$, $\cos(\alpha x)$, $\sin(\alpha y)$, $\cos(\alpha y)$ , valid for all real $x,y$ (i.e. no continuous function $F$ on $[-1,1]^n$, choices $v_j$ of $\sin(\alpha_j x)$, $\cos(\alpha_j x)$, $\sin(\alpha_j y)$, $\cos(\alpha_j y)$, and constants $\alpha_1 \ldots \alpha_n$, such that $F(v_1,\ldots,v_n) = \sin(xy)$ for all $x,y \in \mathbb R$).

Such a function $F(v_1, \ldots,v_n)$ would be uniformly continuous on $\mathbb R^2$, and $\sin(xy)$ is not.

One can use binomial expansion in combination with the complex extension of trig functions:

$$\cos(xy)=\frac{e^{xyi}+e^{-xyi}}{2}=\frac{a^{xy}+a^{-xy}}2$$

Using $a=e^i$ for simplicity.

We also have:

$$(a+a^{-1})^n=\sum_{i=0}^{\infty}\frac{n!a^{n-i}a^{-i}}{i!(n-i)!}=\sum_{i=0}^{\infty}\frac{n!a^{n-i}}{i!(n-2i)!}$$

Which is obtained by binomial expansion.

We also have:

$$(a+a^{-1})^n=(a^{-1}+a)^n=\sum_{j=0}^{\infty}\frac{n!a^{2j-n}}{j!(n-j)!}$$

And, combining the two, we get:

$$(a+a^{-1})^n=\frac{\sum_{i=0}^{\infty}\frac{n!a^{n-2i}}{i!(n-i)!}+\sum_{j=0}^{\infty}\frac{n!a^{2j-n}}{j!(n-j)!}}2=\frac12\sum_{i=0}^{\infty}\frac{n!}{i!(n-i)!}(a^{n-2i}+a^{-(n-2i)})$$

If we have $\cos(n)=\frac{a^n+a^{-n}}2$, then we have

$$(2\cos(n))^k=(a^n+a^{-n})^k=\sum_{i=0}^{\infty}\frac{k!}{i!(k-i)!}\frac{a^{n(k-2i)}+a^{-n(k-2i)}}2$$

Furthermore, the far right of the last equation can be simplified back into the form of cosine:

$$\sum_{i=0}^{\infty}\frac{k!}{i!(k-i)!}\frac{a^{n(k-i)}+a^{-n(k-i)}}2=\sum_{i=0}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$

Thus, we can see that for $\cos(ny)$, it simply the first of the many terms in $\cos^n(y)$ and we may rewrite the summation formula as:

$$(2\cos(n))^k=\cos(nk)+\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$

And rearranging terms, we get:

$$\cos(nk)=2^k\cos^k(n)-\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$ This becomes explicit formulas for $n=0,1,2,3,\dots$

I note that there is no way by which you may reduce the above formula without the knowledge that $n,k\in\mathbb{Z}$.

Also, it is quite difficult to produce the formulas for, per say, $\cos(10x)$ because as you proceed to do so, you will notice that it requires knowledge of $\cos(8x),\cos(6x),\cos(4x),\dots$, which you can eventually solve, starting with $\cos(2x)$ (it comes out to be the well known double angle formula), using this to find, $\cos(4x)$, use that to find $\cos(6x)$, etc. all the way to $\cos(10x)$.

Notably, this can be easier than Chebyshev Polynomials because it only requires that you know the odd/even formulas less than the one you are trying to solve. (due to $-2i$)

But this is the closest I may give to you for the formula of $\cos(xy)$, $x,y\in\mathbb{R}$.

It is also true for $x,y\in\mathbb{C}$.

As others have noted, this can also be solved in terms of the Chebyshev Polynomial:

$$T_n(\cos(x))=\cos(nx)$$

Trivially,

$$T_0(\cos(x))=1$$

$$T_1(\cos(x))=\cos(x)$$

Through the sum of angles formula, it is derivable that we have:

$$T_n(\cos(x))=2T_{n-1}(\cos(x))-T_{n-2}(\cos(x))$$

A much easier recursive formula for $n\in\mathbb{Z}$.

The formula for $\sin(nk)$ is easily derivable with binomial expansion:

$$\sin(nk)=\frac{e^{-nki}-e^{nki}}{2i}=\frac{c^{nk}-c^{-nk}}2$$

The solution is very similar to the cosine, with the exception that complex numbers will appear more than one may like.

Also, there is no Chebyshev polynomial for sine as far as I have seen. Probably easier to use $\sin(nk)=\cos(nk+\frac12\pi)$

I shall proceed to attempt to explain how to further use my recursive definition.

$$\cos(nk)=2\cos^k(n)-\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$

We also have:

$$\cos(n(k-2j))=2\cos^{k-2j}(n)-\sum_{i=1}^{\infty}\frac{(k-2j)!}{i!(k-2j-i)!}(\cos(n(k-2j-2i)))$$

Combine the above two to get:

$$\cos(nk)=2\cos^k(n)-\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(2\cos^{k-2j}(n)-\sum_{j=1}^{\infty}\frac{(k-2i)!}{j!(k-2i-j)!}(\cos(n(k-2j-2i))))$$

I'm going to call all of the numbers with the factorials $\beta_i$:

$$\cos(nk)=2\cos^k(n)-\sum_{i=1}^{\infty}\beta_i(2\cos^{k-2j}(n)-\sum_{j=1}^{\infty}\beta_j\cos(n(k-2j-2i))))$$

Through a very painful process, you may factor out each and every term this way, I just imagine it isn't so much of a beautiful process.

You will most likely also run into the problem of divergence, which may be fixed using $\cos(k-2i)=\cos(2i-k)$, allowing the terms in the binomial expansion to always have positive exponents so you don't run into $0^{-1}$ or divergence problems.

• Please tell me if I have made any mistakes along the way. Thank you for spotting them, if any, in advance. Feb 17, 2016 at 1:25
• Nice try "Simple Art" and I think I can get your way of thinking which enlightens a way to calculate cosines for a multiple of an angle $\cos(nk)$. Feb 17, 2016 at 9:13
• However there are many mistakes and the recurrence formula given is wrong. For example for $k=1$ gives $\cos n = 2 \cos n$. Also $$(a+{1\over a})^n = \sum_{i\ge 0}\binom{n}{i}a^{2i-n}$$. Finally, I think your way could have been simplified following the "De Moivre" expansion. Feb 17, 2016 at 9:28
• @nickchalkida Thank you for that. I just realized I made a few mistakes and in the process of fixing them Feb 17, 2016 at 12:31
• @nickchalkida fixes made, and if you may, please try simplifying this. Feb 17, 2016 at 12:39