# Is there an identity for cos(ab)?

I know that there is a trig identity for $\cos(a+b)$ and an identity for $\cos(2a)$, but is there an identity for $\cos(ab)$?

$\cos(a+b)=\cos a \cos b -\sin a \sin b$

$\cos(2a)=\cos^2a-\sin^2a$

$\cos(ab)=?$

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Are $a,b$ arbitrary, or are you assuming that $b\in\mathbb{Z}$ is an integer? – NotNotLogical May 8 '14 at 22:40
Might not be too helpful, but you can expand $\cos((a+b)^2)$ and use the identities you have above to get a formula for $\cos(ab)$. – Clayton May 8 '14 at 22:41
@Clayton .. but.. do we have any formula for $\cos(a^2)$?? – Berci May 8 '14 at 22:59
@Berci Well, if $a$ is $2$, we have $cos^22-sin^22$ – Cole Johnson May 9 '14 at 0:25
@ColeJohnson There is a formula if $a$ is any integer. – Navin May 9 '14 at 4:51

No, and there's a precise reason.

First, the geometric definition of $\cos$ talks about angles, and the product of two angles doesn't make sense.

Moreover, when you view the cosine as an exponential complex function, as you know $$\cos{x}= \frac{ e^{i x} + e^{-i x}}{2}$$ you can see that the identities you quoted come from properties of powers, such as $e^{a+b}=e^a e^b$ or $e^{2a} = (e^a)^2$

Since there's no significant formula for $e^{ab}$, there isn't one for the $\cos$ function too.

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Well, there is one: $$e^{ab}=(e^a)^b=(e^b)^a$$ – Berci May 8 '14 at 23:00
Although similar, the formula from $(e^a)^2$ comes from the fact it is $e^a e^a$. As pointed out by someone else in the comments, if $b$ is an integer then a formula does exist (or $a$) – AnalysisStudent0414 May 8 '14 at 23:24
Dimensionally speaking, an angle is a dimensionless quantity; there's no reason that the product of two angles necessarily fails to make sense. (The second half of your answer actually illustrates why this is necessary, because you can't exponentiate dimensioned quantities; if angles weren't dimensionless, then the expression $e^{ix}$ wouldn't make dimensional sense. – Steven Stadnicki May 9 '14 at 4:06
I meant it doesn't have a geometric sense, at least not an immediate one. You can't derive a general formula from geometry, that's what I meant. You're right though – AnalysisStudent0414 May 9 '14 at 12:23
A very interesting “answer” that shows (together with @Berci’s comment) how author’s original thought became ruined when expressed with symbols. $e^{ab} = (e^a)^b$ is indeed a very “significant” identity when b is a rational number. – Incnis Mrsi Nov 9 '14 at 12:52

If $a$ is an integer and $b$ is an angle,

$$\cos(ab) = T_a(\cos b)$$

where $T_n(x)$ is the $n^{th}$ Chebyshev polynomial.

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It does not have to be an integer: math.stackexchange.com/a/787186/12438 – Navin May 9 '14 at 6:00
@Navin Under certain conditions, your answer may not be acceptable because it probably doesn't make things simpler. Given $\cos(1)=a$, write me a formula that solves for $\cos(4.78)$ in terms of $a$. It is not easily done, if it is doable, in your method. – Simple Art Feb 17 at 1:31

Not really, but I suppose this works: $$\cos ab=Re[(\cos(b)+i\sin(b))^a]$$

You can get the above equation by taking the real part of de Moivre's formula: $$\cos n\theta +i\sin n\theta=(\cos(\theta)+i\sin(\theta))^n \,$$

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Expression with real part is correct only for real argument. For complex case you need $\frac{1}{2}(e^{iz}+e^{-iz})$. – Incnis Mrsi Dec 9 '14 at 10:22

For general $a$ and $b$, we cannot write $\cos (ab)$ in terms of the trig functions $\cos a,\sin a, \cos b, \sin b$. This is because the trig functions are periodic with period $2\pi$, so adding $2\pi$ to $b$ does not change any of these functions. But adding $2\pi$ to $b$ can change $\cos (ab)$ - for instance, if $a=1/2$, if sends $\cos (ab)$ to $-\cos(ab)$. Only if $a$ is an integer can we avoid this problem.

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Did you ever hear about multivalued functions? – Incnis Mrsi Nov 9 '14 at 12:42

As many experts already noted here, an argument of cos(⋯) is an angle, and a sensible mathematical structure on angles is the one of under “+”. We can add and subtract angles, as well as multiply them by integers. To some extent we can multiply angles by rational numbers, i.e. solve equations like $$qx = pa,\quad x,a\text{ are angles, }\ p,q\in{\mathbb Z},\ q≠0.$$ If $a$ is specified modulo 2π radians, then such solutions, placed on the trigonometric circle, will form a set of $q$ elements (vertices of a regular $q$-gon), that can be described with an algebraic equation.

There is no “reasonable” multiplication of an angle and an irrational number $t$. If $a$ is specified modulo 2π radians (that is a typical condition), then possible values of $ta$ will form a dense subset of the trigonometric circle, and hence values of trigonometric functions on it will have no use for calculations.

## Conclusion

There are trigonometric identities for products of an angle and a rational number; see other answers and this page for some partial cases.

There are no products of an angle and an irrational number, as well as there are no products of two angles.

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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:

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

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

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