Differentiation of $\sin x$ using Pythagoras theorem: So, I just tried using Pythagoras theorem to prove:
$\frac{d}{dx}\sin x= \cos x$ but it got me nowhere.
Here is my work.
$$\sin x= p/h.$$; where p and h are lengths of perpendicular and hypotenuse in a right angled triangle.
Differentiating w.r.t $x$, we get
$$\frac{d}{dx}\sin x= \frac{1}{h^2} \left( h\frac{dp}{dx}-p\frac{dh}{dx}\right)\tag{i}$$
We know $h^2=p^2+b^2$. (Where b is the length of the base of the right angled triangle.) Differentiating w.r.t $x$,
$$h\frac{dh}{dx}=p\frac{dp}{dx}$$
(If we put $b$ constant). Substituting in $(i)$,
\begin{align}
\frac{d}{dx} \sin x&= \frac{1}{h^2} \left(h\frac{dp}{dx}-\frac{p^2}{h}\frac{ dp}{dx}\right)\\
       & = \frac{1}{ph^2} \left(b^2\frac{dp}{dx}\right) \tag{ii}
\end{align}
Now, I lost track here because I didn't know how is that equal to $\cos x$. Can anyone help me continue from here?
(Note: I know how to derive this from first principle. I just wanted to try using Pythagoras theorem and trigonometric identities. I kept the base length constant because it appeared to me that if I changed the x, base length won't be effected. But as I can see in the answer, the hypotenuse remains constant. However, can we keep the base constant and end up getting the same result? Maybe not visualising a circle of radius h but just a right angled triangle whose base remains constant w.r.t the reference angle?

Ps: The called duplicate one is out of my understanding. I request a simpler version.
 A: So, the way to think about this is to recognize three important facts:

*

*On a unit circle, the hypotenuse will always be 1

*On a unit circle, the distance around the circle is the same as the angle in radians

*The change angle x, as the distance on the unit circle approaches zero, closer and closer approximates a LINEAR change

So, we will use $a$ for the adjacent, $p$ for the opposite, and $x$ for the angle in radians.
The definition of sine is $\frac{\text{opposite}}{\text{hypotenuse}}$ and the definition of cosine is $\frac{\text{adjacent}}{\text{hypotenuse}}$.  Since the hypotenuse is $1$, this means that $\sin(x) = p$ and $\cos(x) = a$.  The derivative of $\sin(x)$ with respect to $x$, therefore, is $\frac{dp}{dx}$.  If we can show that $\frac{dp}{dx} = a$, then we win.
Okay, the pythagorean theorem (plus #1 above) gives us:
$$ a^2 + p^2 = 1 \\ p^2 = 1 - a^2 $$
Differentiating yields:
$$2p\,dp = -2a\,da \\ dp = -\frac{a}{p}da \\ da = -\frac{p}{a}dp$$
Now, because of #2 and #3, we can use the distance formula to say:
$$dx = \sqrt{dp^2 + da^2} \\ 
dx = \sqrt{dp^2 + (-1)^2\frac{p^2}{a^2} dp^2} \\
dx = \sqrt{dp^2 + \frac{1 - a^2}{a^2} dp^2} \\
dx = \sqrt{dp^2 + \frac{dp^2}{a^2} - dp^2} \\
dx = \sqrt{\frac{dp^2}{a^2}} \\
dx = \frac{dp}{a}$$
So, this means that we can transform $\frac{dp}{dx}$ like this:
$$ \frac{dp}{dx} = \frac{dp}{\frac{dp}{a}} = \frac{dp}{1}\frac{a}{dp} = a$$
QED
