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

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




$egingroup$ Are $a,b$ arbitrary, or are you assuming that $binriz15.combbZ$ is an integer? $endgroup$
No, và there"s a precise reason.

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First, the geometric definition of $cos$ talks about angles, & the product of two angles doesn"t make sense.

Moreover, when you view the cosine as an exponential complex function, as you know $$cosx= frac e^i x + e^-i x2 $$ 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.



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.

Not really, but I suppose this works:$$cos ab=Re<(cos(b)+isin(b))^a>$$

You can get the above equation by taking the real part of de Moivre"s formula:$$cos n heta +isin n heta=(cos( heta)+isin( heta))^n ,$$


For general $a$ & $b$, 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 $2pi$, so adding $2pi$ to lớn $b$ does not change any of these functions. But adding $2pi$ to $b$ can change $cos (ab)$ - for instance, if $a=1/2$, if sends $cos (ab)$ khổng lồ $-cos(ab)$. Only if $a$ is an integer can we avoid this problem.


As many experts already noted here, an argument of cos(⋯) is an angle, and a sensible briz15.comematical structure on angles is the one of abelian-groups under “+”. We can showroom and subtract angles, as well as multiply them by integers. To lớn some extent we can multiply angles by rational numbers, i.e. Solve equations like$$ qx = pa,quad x,a ext are angles, p,qinriz15.combb Z, q≠0.$$If $a$ is specified modulo 2π radians, then such solutions, placed on the trigonometric circle, will khung 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 & an irrational number $t$. If $a$ is specified modulo 2π radians (that is a typical condition), then possible values of $ta$ will khung a dense subset of the trigonometric circle, & 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 và this page for some partial cases.

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