If ((π)/(18)) ((5π)/(18)) ((7π)/(18)) = K , then the value of ((10Kπ)/(3)) is :

Correct answer: A. 3 + 1 2 2 Given K = ((π)/(18)) ((5π)/(18)) ((7π)/(18)) Using the standard trigonometric identity (θ) ((π)/(3) - θ) ((π)/(3) + θ) = (1)/(4) (3θ) with θ = (π)/(18) : (π)/(3) - θ = (6π)/(18) - (π)/(18) = (5π)/(18) (π)/(3) + θ = (6π)/(18) + (π)/(18) = (7π)/(18) Applying the identity, we get: K = (1)/(4) (3 × (π)/(18)) = (1)/(4) ((π)/(6)) K = (1)/(4) × (1)/(2) = (1)/(8) We need to find the value of ((10Kπ)/(3)) . Substituting K = (1)/(8) : ((10 × (1)/(8) × π)/(3)) = ((10π)/(24)) = ((5π)/(12)) Expanding (5π)/(12) as (π)/(4) + (π)/(6) : ((5π)/(12)) = ((π)/(4) + (π)/(6)) ((5π)/(12)) = ((π)/(4)) ((π)/(6)) + ((π)/(4)) ((π)/(6)) ((5π)/(12)) = ( 1 2 )( 3 2 ) + ( 1 2 )((1)/(2)) = 3 + 1 2 2 Answer: 3 + 1 2 2

JEE Main 20262 AprilMorning Shift

Exam: JEE Main
Year: 2026
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2026 · 2 April · Morning Shift · Mathematics · Trigonometric Ratios & Identities

If ((π)/(18)) ((5π)/(18)) ((7π)/(18)) = K , then the value of...

\sin\left(\dfrac{\pi}{18}\right) \sin\left(\dfrac{5\pi}{18}\right) \sin\left(\dfrac{7\pi}{18}\right) = K

, then the value of

\sin\left(\dfrac{10K\pi}{3}\right)

is :

Options

A $\dfrac{\sqrt{3} + 1}{2\sqrt{2}}$
B $\dfrac{\sqrt{3} - 1}{\sqrt{2}}$
C $\dfrac{\sqrt{3}}{2}$
D $\dfrac{1}{2}$

Correct answer

A. $\dfrac{\sqrt{3} + 1}{2\sqrt{2}}$

Step-by-step solution

Given

K = \sin\left(\dfrac{\pi}{18}\right) \sin\left(\dfrac{5\pi}{18}\right) \sin\left(\dfrac{7\pi}{18}\right)

Using the standard trigonometric identity

\sin(\theta) \sin\left(\dfrac{\pi}{3} - \theta\right) \sin\left(\dfrac{\pi}{3} + \theta\right) = \dfrac{1}{4} \sin(3\theta)

with

\theta = \dfrac{\pi}{18}

\dfrac{\pi}{3} - \theta = \dfrac{6\pi}{18} - \dfrac{\pi}{18} = \dfrac{5\pi}{18}

\dfrac{\pi}{3} + \theta = \dfrac{6\pi}{18} + \dfrac{\pi}{18} = \dfrac{7\pi}{18}

Applying the identity, we get:

K = \dfrac{1}{4} \sin\left(3 \times \dfrac{\pi}{18}\right) = \dfrac{1}{4} \sin\left(\dfrac{\pi}{6}\right)

K = \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}

We need to find the value of

\sin\left(\dfrac{10K\pi}{3}\right)

. Substituting

K = \dfrac{1}{8}

\sin\left(\dfrac{10 \times \dfrac{1}{8} \times \pi}{3}\right) = \sin\left(\dfrac{10\pi}{24}\right) = \sin\left(\dfrac{5\pi}{12}\right)

Expanding

\dfrac{5\pi}{12}

\dfrac{\pi}{4} + \dfrac{\pi}{6}

\sin\left(\dfrac{5\pi}{12}\right) = \sin\left(\dfrac{\pi}{4} + \dfrac{\pi}{6}\right)

\sin\left(\dfrac{5\pi}{12}\right) = \sin\left(\dfrac{\pi}{4}\right)\cos\left(\dfrac{\pi}{6}\right) + \cos\left(\dfrac{\pi}{4}\right)\sin\left(\dfrac{\pi}{6}\right)

\sin\left(\dfrac{5\pi}{12}\right) = \left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{\sqrt{3}}{2}\right) + \left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{1}{2}\right) = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}

Answer:

\dfrac{\sqrt{3} + 1}{2\sqrt{2}}

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