Express 5 theta in terms of sines of multiple angles.

Given $z = \cos\theta + i \sin\theta$, we know that $z = e^{i\theta}$.
Then $z^{-1} = \cos\theta - i \sin\theta = e^{-i\theta}$.
Subtracting $z^{-1}$ from $z$:

$$z - z^{-1} = (\cos\theta + i \sin\theta) - (\cos\theta - i \sin\theta) = 2i \sin\theta$$

Therefore, $\sin\theta = \frac{z - z^{-1}}{2i}$.

The binomial expansion for $(a-b)^5$ is $a^5 - 5a^4b + 10a^3b^2 - 10a^2b^3 + 5ab^4 - b^5$.
Substitute $a=z$ and $b=z^{-1}$:

$$(z - z^{-1})^5 = z^5 - 5z^4(z^{-1}) + 10z^3(z^{-1})^2 - 10z^2(z^{-1})^3 + 5z(z^{-1})^4 - (z^{-1})^5$$ $$= z^5 - 5z^3 + 10z - 10z^{-1} + 5z^{-3} - z^{-5}$$

Group the terms as $(z^n - z^{-n})$:

$$(z - z^{-1})^5 = (z^5 - z^{-5}) - 5(z^3 - z^{-3}) + 10(z - z^{-1})$$

Using the identity $z^n - z^{-n} = e^{in\theta} - e^{-in\theta} = (\cos(n\theta) + i \sin(n\theta)) - (\cos(n\theta) - i \sin(n\theta)) = 2i \sin(n\theta)$:

$$z^5 - z^{-5} = 2i \sin(5\theta)$$ $$z^3 - z^{-3} = 2i \sin(3\theta)$$ $$z - z^{-1} = 2i \sin\theta$$

Substitute these into the expanded expression:

$$(z - z^{-1})^5 = 2i \sin(5\theta) - 5(2i \sin(3\theta)) + 10(2i \sin\theta)$$ $$= 2i (\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta)$$

$$(2i)^5 = 2^5 \cdot i^5 = 32 \cdot (i^4 \cdot i) = 32 \cdot 1 \cdot i = 32i$$

Now, substitute this and the expanded binomial back into the expression for $\sin^5 \theta$:

$$\sin^5 \theta = \frac{2i (\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta)}{32i}$$ $$\sin^5 \theta = \frac{1}{16} (\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta)$$

The final expression for $\sin^5 \theta$ in terms of sine of multiple angles is:

$$\frac{1}{16} (\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta)$$