Why is $\sin(d\Phi) = d\Phi$ where $d\Phi$ is very small? I haven't touched Physics and Math (especially continuous Math) for a long time, so please bear with me.
In essence, I'm going over a few Physics lectures, one which tries to calculate the Force exerted by uniform magnetic field on a semi circular current carrying wire.
The mathematics that puzzles me is this, that:
$$
\sin(d \phi) \thickapprox d\phi
$$
where $d\phi$ is very small. Link to video.
 A: 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:
$$\sin x = \frac{\text{opposite}}{\text{hypot}} = 
\frac{\text{opposite}}{1}\approx \frac{x}{1} = x$$
A: 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.
A: You can give a linear approximation for $\sin$ near $0$ based on this formula: $$f(x)\approx f(x_0)+(x-x_0)f'(x_0),$$ and using the fact that: $\sin^\prime=\cos$, you get: $$\eqalign{\sin x&\approx \sin0+(x-0)\cos(0)= x.}$$
So when $x$ is very small, you have that $\sin x\sim x.$

What this intuitively means, is that when you observe closely the graph of the curve $\color{darkmagenta}{\sin x}$ near $0$, it starts to resemble a line, and this line is described by $y=\color{darkblue}x$.
$\phantom{XXX}$
A: I think one way to think of it is $\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$. Which means that as $x$ becomes very small, the ratio goes to one, i.e, $\sin x$ can be approximated by $x$.
A: Substituting $x$ for $d \Phi$...
I would say $\sin x \approx x$ when $x \approx 0$ because...


*

*$\sin x$ is a smoothly varying function with no discontinuities.

*$\sin x = 0$ when $x = 0$

*The gradient of $\sin x$ is equal to the gradient of $x$ when $x = 0$

*The second order derivative of $\sin x$ is $-\sin x$, which is $0$ when $x=0$


On point 3, the derivative of $\sin x$ is $\cos x$ which evaluates to $1$ when $x=0$, and the derivative of $x$ is $1$ (at all points).
This is closely related to the Taylor series argument.
A: As already was said Taylor's theorem answer this question in a precise way.
$\sin(d\Phi)=d\Phi+O((d\Phi)^3).$
I don't see another better explanation than this one.
