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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

BITSAT 2020 · Mathematics · Trigonometric Equations

The value of [n π+(-1)^ n (π)/(4)], n ∈ I is

The value of sin⁡[nπ+(−1)nπ4],n∈I\sin \left[\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}\right], \mathrm{n} \in \mathrm{I} is

Options

  1. A0
  2. B12\frac{1}{\sqrt{2}}
  3. C−12-\frac{1}{\sqrt{2}}
  4. DNone of these

Correct answer

B. 12\frac{1}{\sqrt{2}}

Step-by-step solution

sin⁡[nπ+(−1)nπ4]=(−1)nsin⁡[(−1)nπ4][∵sin⁡(nπ+θ)=(−1)nsin⁡θ]\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]=(−1)n(−1)nsin⁡π4∴sin⁡[(−1)nθ]=(−1)nsin⁡θ]=(−1)2nsin⁡π4=sin⁡π4=12\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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