Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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



share|cite|improve this question
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

5 Answers 5

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.

share|cite|improve this answer
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.

share|cite|improve this answer
It does not have to be an integer: – Navin May 9 '14 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: $$\cos n\theta +i\sin n\theta=(\cos(\theta)+i\sin(\theta))^n \,$$

share|cite|improve this answer
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.

share|cite|improve this answer
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.


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.