$\sin(n\pi+(-1)^n\large\frac{\pi}{4})$
$\Rightarrow (-1)^n\sin[(-1)^n\large\frac{\pi}{4}]$
$\sin(n\pi+\theta)=(-1)^n\sin \theta$
$\Rightarrow (-1)^n(-1)^n\sin \large\frac{\pi}{4}$
$\sin(-\theta)=-\sin \theta$
$\Rightarrow (-1)^{2n}\sin \large\frac{\pi}{4}$
$\Rightarrow \sin \large\frac{\pi}{4}$
$2n$ is even
$\Rightarrow \large\frac{1}{\sqrt 2}$
Hence (A) is the correct answer.