Answer
See the proof below.
Work Step by Step
Consider the function $f\left( x \right)=\sin x$ ,
Find the derivative of the function $f\left( x \right)=\sin x$ using the formula ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ ,
${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \left( x+h \right)-\sin x}{h}$
Now, expand $\sin \left( x+h \right)$ by the formula $\sin \left( a+b \right)=\sin a\cos b+\sin b\cos a$ ,
$\sin \left( x+h \right)=\sin x\cosh +\sinh \cos x$
Now, substitute the value of $\sin \left( x+h \right)$ in ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \left( x+h \right)-\sin x}{h}$ and solve further,
$\begin{align}
& {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin x\cosh +\sinh \cos x-\sin x}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin x\cosh -\sin x+\sinh \cos x}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin x\left( \cosh -1 \right)+\sinh \cos x}{h}
\end{align}$
Now, take the limit inside and solve further,
$\begin{align}
& {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin x\left( \cosh -1 \right)}{h}+\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh \cos x}{h} \\
& =\sin x\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( \cosh -1 \right)}{h}+\cos x\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh }{h}
\end{align}$
Substitute the values $\text{ }\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh }{h}=1\text{ and }\underset{h\to 0}{\mathop{\lim }}\,\frac{\cosh -1}{h}=0$ in the above expression and solve further.
$\begin{align}
& {f}'\left( x \right)=\sin x\left( 0 \right)+\cos x\left( 1 \right) \\
& =\cos x
\end{align}$
Thus, $f'\left( x \right)=\cos x$
