Trigonometric identities with multiplication

Question

Why aren't there Trigonometric identities with multiplication inside the function? For example for $\sin(xy)=?$.

Answer 1

Probably this won't be a complete answer but I'll consider only the case where $y\in \Bbb N$ (we can even use $y\in \Bbb Z$ but we can easily note that if $y$ is negative we have $\sin(-yx)=-\sin(yx)$ and $\cos(-xy)=\cos(xy)$ so we can treat only positive values of $y$):

We have those so called "multiplication formulas":

$$\sin(nx)=\sum_{k=0}^n\binom n k\cos^k(x)\sin^{n-k}(x)\sin[\frac 12(n-k)\pi] $$

$$\cos(nx)=\sum_{k=0}^n\binom n k\cos^k(x)\sin^{n-k}(x)\cos[\frac 12(n-k)\pi] $$

(The others for tangent and so on follow from this)

These formulas are pretty easy to memorize if you now about Newton's binomial expansion, which is very similar, and as @frog said can be proven by the identities:

$\sin x = \frac{\mathrm e^{\mathrm ix}-\mathrm{e}^{-\mathrm ix}}{2\mathrm i}$

$\cos x = \frac{\mathrm e^{\mathrm ix}+\mathrm{e}^{-\mathrm ix}}{2\mathrm i}$

Btw these formulas are not used so much because they can be deduced by iterating the addition formulas (which are easier to learn and have more uses):

$\sin(x\pm y)=\sin(x)\cos(y)\pm \cos(x)\sin(y)$

$\cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)$

Answer 2

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:

$$(2\cos(n))^k=\frac12\cos(nk)+\frac12\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$

And rearranging terms, we get:

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

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 $\cos(10x)$.

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 $x,y\in\mathbb{C}$.