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The value of $\sin \left[\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}\right], \mathrm{n} \in \mathrm{I}$ is
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$\frac{1}{\sqrt{2}}$
$\sin \left[\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}\right]=(-1)^{\mathrm{n}} \sin \left[(-1)^{\mathrm{n}} \frac{\pi}{4}\right]$
$\left[\because \sin (\mathrm{n} \pi+\theta)=(-1)^{\mathrm{n}} \sin \theta\right]$
$\begin{aligned} &=(-1)^{\mathrm{n}}(-1)^{\mathrm{n}} \sin \frac{\pi}{4} \\ \therefore \sin &\left.\left[(-1)^{\mathrm{n}} \theta\right]=(-1)^{\mathrm{n}} \sin \theta\right] \\ &=(-1)^{2 \mathrm{n}} \sin \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \end{aligned}$
$\left[\because \sin (\mathrm{n} \pi+\theta)=(-1)^{\mathrm{n}} \sin \theta\right]$
$\begin{aligned} &=(-1)^{\mathrm{n}}(-1)^{\mathrm{n}} \sin \frac{\pi}{4} \\ \therefore \sin &\left.\left[(-1)^{\mathrm{n}} \theta\right]=(-1)^{\mathrm{n}} \sin \theta\right] \\ &=(-1)^{2 \mathrm{n}} \sin \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \end{aligned}$
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