Expanding $\sin(ab)$ in terms of $\sin a$ and $\sin b$.

Question

Motivation behind the question: I've wondered since grade 10 that if $\sin(a+b),\sin(a-b)$ have formulas in terms of $\sin(a)$ and $\sin(b)$, then why not $\sin(ab)$. But I couldn't find any such information sadly.

My attempt:-

$$\sin(ab)=\sin\bigg(\frac{(a+b)^2}{4}-\frac{(a-b)^2}{4}\bigg)$$ This can be expanded as $$\sin \bigg(\frac{a+b}{2}\bigg)^2\cos \bigg(\frac{a-b}{2}\bigg)^2-\cos \bigg(\frac{a+b}{2}\bigg)^2\sin \bigg(\frac{a-b}{2}\bigg)^2$$

Now even if I try to expand the squares, I get something in the form of $\sin(a^2)$, and thus I can't run from multiplication.

Any help, or closed form solution will be greatly appreciated.

Answer 1

Your attempt works, but I'm assuming you're in search of a formula which isolates variables $a$ and $b$.

We can derive such a formula for these by De Moivre's Theorem. $$r^n(\cos(nb) + i\sin(nb))^a = r(\cos(ab)+i\sin(ab))^n$$ Hence, rewriting this in terms of $\cos(ab)$ and $\sin(ab)$, we get: $$\cos(ab)+i\sin(ab) = (\cos(b) + i\sin(b))^a$$ Assuming $a$ is a positive integer, we get, by applying the Binomial Theorem, $$(\cos(b) + i\sin(b))^a=[\cos^a(b)+\frac{a(a-1)}{1\cdot2}\cos^{a-2}(b)\sin^2(b)+\frac{a(a-1)(a-2)(a-3)}{1\cdot2\cdot3\cdot4}\cos^{a-4}(b)\sin^4(b)]\dots+i[a\cos^{a-1}(b)\sin(b)-\frac{a(a-1)(a-2)}{1\cdot2\cdot3}\cos^{a-3}(b)\sin^3(b)\dots]$$ Now, equating the real and imaginary parts, we obtain: $$\cos(ab)={a \choose 0}\cos^a(b)-{a \choose 2}\cos^{a-2}(b)\sin^2(b)+{a \choose 4}\cos^{a-4}(b)\sin^4(b)\dots$$ $$\sin(ab)={a \choose 1}\cos^{a-1}(b)\sin(b)-{a \choose 3}\cos^{a-3}(b)\sin^3(b)+{a \choose 5}\cos^{a-5}(b)\sin^5(b)\dots$$

Answer 2

From the complex exponential definition of sine:

$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$

We obtain the formula:

$$\sin(ab) = \frac{e^{iab} - e^{-iab}}{2i}$$ $$= \frac{(e^{ib})^a - (e^{-ib})^a}{2i}$$ $$= \frac{(\cos(b) + i\sin(b))^a - (\cos(-b) + i\sin(-b))^a}{2i}$$ $$= \frac{(\cos(b) + i\sin(b))^a - (\cos(b) - i\sin(b))^a}{2i}$$

If $a$ is an integer, then you can use the Binomial Theorem to expand the numerator, as in @Schrödinger's Cat's answer. For example, with $a = 2$:

$$\sin(2b) = \frac{(\cos(b) + i\sin(b))^2 - (\cos(b) - i\sin(b))^2}{2i}$$ $$= \frac{(\cos(b) + i\sin(b))^2 - (\cos(b) - i\sin(b))^2}{2i}$$ $$= \frac{\cos^2(b) + 2i\cos(b)\sin(b) - \sin^2(b) - (\cos^2(b) - 2i\cos(b)\sin(b) - \sin^2(b))}{2i}$$ $$= \frac{4i\cos(b)\sin(b)}{2i}$$ $$= 2\cos(b)\sin(b)$$

Obtaining the familiar double-angle identity.

Otherwise, I don't think there's a simple formula.