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 $\cos(ab)$?

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

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

$\cos(ab)=?$

-
Are $a,b$ arbitrary, or are you assuming that $b\in\mathbb{Z}$ is an integer? –  NotNotLogical May 8 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 at 22:41
@Clayton .. but.. do we have any formula for $\cos(a^2)$?? –  Berci May 8 at 22:59
@Berci Well, if $a$ is $2$, we have $cos^22-sin^22$ –  Cole Johnson May 9 at 0:25
@ColeJohnson There is a formula if $a$ is any integer. –  Navin May 9 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.

-
Well, there is one: $$e^{ab}=(e^a)^b=(e^b)^a$$ –  Berci May 8 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 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 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 at 12:23

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.

-
It does not have to be an integer: math.stackexchange.com/a/787186/12438 –  Navin May 9 at 6:00

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: $$\cos n\theta +i\sin n\theta=(\cos(\theta)+i\sin(\theta))^n \,$$

-

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.

-