Q54 / 125 · BITSAT 2020 · Mathematics
The value of $\sin \left[\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}\right], \mathrm{n} \in \mathrm{I}$ is
- A0
- Correct answerB$\frac{1}{\sqrt{2}}$
- C$-\frac{1}{\sqrt{2}}$
- DNone of these
Solution
$\sin \left[\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}\right]=(-1)^{\mathrm{n}} \sin \left[(-1)^{\mathrm{n}} \frac{\pi}{4}\right]$
$\left[\because \sin (\mathrm{n} \pi+\theta)=(-1)^{\mathrm{n}} \sin \theta\right]$
$\begin{aligned} &=(-1)^{\mathrm{n}}(-1)^{\mathrm{n}} \sin \frac{\pi}{4} \\ \therefore \sin &\left.\left[(-1)^{\mathrm{n}} \theta\right]=(-1)^{\mathrm{n}} \sin \theta\right] \\ &=(-1)^{2 \mathrm{n}} \sin \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \end{aligned}$
Hence, the correct answer is option: 'B'.