Why is $\sin(d\Phi) = d\Phi$ where $d\Phi$ is very small?

Just draw the diagram!

What does $\sin x$ mean? it's the ratio of the opposite side to the hypotenuse in a triangle.

Now, let's draw a triangle with a small angle $x$ inside the unit circle:

$\quad\quad\quad$

Now clearly, when the angle becomes really small, the opposite side is approximately the arc length. In radians, the arc length in a unit circle is exactly the angle $x$, and so we have for small angles:

$$\sin x = \frac{\text{opposite}}{\text{hypot}} = \frac{\text{opposite}}{1}\approx \frac{x}{1} = x$$

If you are familiar with Taylor series you know that the series of $\sin(x)$ expanded at $0$ is:

$$\sin{(x)} = x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots$$

Then, if $x$ is very small you can neglect all term of order greater than one getting:

$$\sin{(x)} \approx x$$

You can also show this result using basic trigonometry but this approach seems easier to me.