Question

Find the derivation of $\sin x$ from first principle.

Answer 1

Let $f\left(x\right)=sinx$

then,

$f(x+y)=sin(x+y)$

Therefore,

$\begin{array}{c}\frac{d}{dx}f\left(x\right)={lim}_{y\to 0}\frac{\sin \left(x+y\right)-\sin x}{y}\\ \Rightarrow \frac{d}{dx}f\left(x\right)={lim}_{y\to 0}\frac{2\cos \left(\frac{x+y+x}{2}\right)-\sin \left(\frac{x+y-x}{2}\right)}{y}\\ \\ \sin x-\sin y=2\cos \left(\frac{x+y}{2}\right)-\sin \left(\frac{x-y}{2}\right)\\ \Rightarrow \frac{d}{dx}f\left(x\right)={lim}_{y\to 0}\frac{2\cos \left(\frac{2x+y}{2}\right)-\sin \left(\frac{y}{2}\right)}{y}\\ \Rightarrow \frac{d}{dx}f\left(x\right)={lim}_{y\to 0}\frac{2\cos \left(x+\frac{y}{2}\right)-\sin \left(\frac{y}{2}\right)}{2\left(\frac{y}{2}\right)}\\ \Rightarrow \frac{d}{dx}f\left(x\right)={lim}_{y\to 0}\cos \left(x+\frac{h}{2}\right)\times {lim}_{y\to 0}\frac{\sin \frac{y}{2}}{\frac{y}{2}}\\ \Rightarrow \frac{d}{dx}f\left(x\right)={lim}_{y\to 0}\cos x\times 1\left[{lim}_{y\to 0}\frac{\sin \frac{y}{2}}{\frac{y}{2}}=1\right]\\ \Rightarrow \frac{d}{dx}\left(\sin x\right)=\cos x\end{array}$

Which is the required answer.