Find the derivative of the function $\sin (x+a)$ with respect to $x$ using the first principle,where $a$ is a fixed non-zero constant.

Let $f(x) = \sin (x+a)$.
By the first principle of derivatives:
$f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Substituting the function:
$f^{\prime}(x) = \lim_{h \to 0} \frac{\sin (x+h+a) - \sin (x+a)}{h}$
Using the trigonometric identity $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$:
$f^{\prime}(x) = \lim_{h \to 0} \frac{2 \cos \left(\frac{x+h+a+x+a}{2}\right) \sin \left(\frac{x+h+a-x-a}{2}\right)}{h}$
$f^{\prime}(x) = \lim_{h \to 0} \frac{2 \cos \left(\frac{2x+2a+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$
$f^{\prime}(x) = \lim_{h \to 0} \cos \left(x+a+\frac{h}{2}\right) \cdot \frac{\sin (h/2)}{h/2}$
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$f^{\prime}(x) = \cos (x+a) \cdot 1 = \cos (x+a)$.