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There is a well known formula for expressing $\sin(\alpha+\beta)$ just using $\sin(\alpha)$ and $\sin(\beta)$. It is enough to replace $\cos$ in the formula $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$ by its equivalent form in terms of $\sin$ function. Thus $\sin(\alpha+\beta)$ is expressible via elementary functions (polynomials, radicals, fractions, ...) and $\sin(\alpha), \sin(\beta)$.

Question 1: What about $\sin(\alpha\cdot\beta)$ and $\sin(\alpha^{\beta})$? Can we express them just using $\sin(\alpha)$ and $\sin(\beta)$ (in any non-trivial sense)? If no, how to prove this fact?

Question 2: Also we can express $\tan(\alpha+\beta)$ in terms of $\tan(\alpha)$ and $\tan(\beta)$. What about $\tan(\alpha\cdot\beta)$ and $\tan(\alpha^{\beta})$?

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  • $\begingroup$ If $\alpha$ is a constant integer, then you can write $\sin(\alpha\beta)$ using only $\sin\beta$ and $\cos\beta$ (see Chebyshev polynomials). But there is no formula for $\sin (\alpha\beta)$ for general $\alpha$ and $\beta$, or for $\sin (\alpha^\beta)$ $\endgroup$
    – Jean-Claude Arbaut
    Commented Nov 4, 2014 at 12:05
  • $\begingroup$ @Jean-ClaudeArbaut How to prove non-existence of such a formula? $\endgroup$
    – user180918
    Commented Nov 4, 2014 at 12:07
  • $\begingroup$ Please help on adding suitable tags for this question. $\endgroup$
    – user180918
    Commented Nov 4, 2014 at 12:10
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    $\begingroup$ For a non integer $\alpha$, you can prove that $\sin(\alpha\beta)$ cannot be written with only algebraic operations, $\sin(\beta)$ and any function of $\alpha$ (call the whole thing $f(\sin\beta,\alpha)$), by considering the periodicity of the functions $\beta\to\sin(\alpha\beta)$ and $\beta\to f(\sin\beta,\alpha)$: the latter is $2\pi$-periodic, not the former. For $\beta\to\sin(\alpha^\beta)$ it's even easier: it's not periodic at all if $\alpha>0$, $\alpha\neq1$. $\endgroup$
    – Jean-Claude Arbaut
    Commented Nov 4, 2014 at 12:12

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The underlying reason why you shouldn't expect any such formulas to exist is that $e^{ix}$ is a homomorphism with respect to addition: $$e^{i(x+y)}=e^{ix}\cdot e^{iy}$$ but not with respect to multiplication, and certainly not with respect to exponentiation. The angle-addition formulae for $\sin$ and $\cos$ come from the homomorphism property: $$e^{i(x+y)}=\cos(x+y)+i\sin(x+y)$$ $$\begin{align*} e^{ix}e^{iy}&=(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))\\ &=\biggl(\cos(x)\cos(y)-\sin(x)\sin(y)\biggr)+i\biggl(\sin(x)\cos(y)+\cos(x)\sin(y)\biggr) \end{align*}$$ (Then just identify the real and imaginary parts.)

Because $e^{ixy}$ is not going to have a relationship with $e^{ix}$ and $e^{iy}$ all the time, perhaps a natural next step would be to relax the question to only being related to one of them, say $e^{ix}$. The only time one could have any hope of a relationship with $e^{ix}$ is if $y$ is an integer, because that is precisely when $$e^{ixy}=(e^{ix})^y$$ is true for all $x$ (see here on complex exponentiation). And, naturally, this is precisely the situation when $\sin(xy)$ and $\cos(xy)$ have actual expressions of the sort you want (see here).

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