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Mathematics · Trigonometry Ratios and Identities

JEE Main 2026 — 2 April, Morning Shift — Question 29

If sin⁡(π18)sin⁡(5π18)sin⁡(7π18)=K\sin \left(\frac{\pi}{18}\right)\sin \left(\frac{5\pi}{18}\right)\sin \left(\frac{7\pi}{18}\right) = \mathrm{K} , then the value of sin⁡(10Kπ3)\sin \left(\frac{10\mathrm{K}\pi}{3}\right) is :

  1. Option A:

    3+122\frac{\sqrt{3} + 1}{2\sqrt{2}}

    Correct
  2. Option B:

    3−12\frac{\sqrt{3} - 1}{\sqrt{2}}

  3. Option C:

    32\frac{\sqrt{3}}{2}

  4. Option D:

    12\frac{1}{2}

Answer: A

Step-by-step solution

K=sin⁡10∘sin⁡50∘sin⁡70∘\mathrm{K}=\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ} =14sin⁡30∘=18=\frac{1}{4} \sin 30^{\circ}=\frac{1}{8} sin⁡10Kπ3=sin⁡(10×18⋅π3)=sin⁡5π12=3+122\sin 10 K \frac{\pi}{3}=\sin \left(10 \times \frac{1}{8} \cdot \frac{\pi}{3}\right)=\sin \frac{5 \pi}{12}=\frac{\sqrt{3}+1}{2 \sqrt{2}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Morning Shift
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Continued Sum or Product of Series of Trigonometric Ratios

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