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)=?\$ \$egingroup\$ Are \$a,b\$ arbitrary, or are you assuming that \$binriz15.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,qinriz15.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.

## Conclusion

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.