JEE Main 2026 — Mathematics Trigonometry

Given $K = \sin\left(\dfrac{\pi}{18}\right) \sin\left(\dfrac{5\pi}{18}\right) \sin\left(\dfrac{7\pi}{18}\right)$K=sin(18π​)sin(185π​)sin(187π​)

Using the standard trigonometric identity $\sin(\theta) \sin\left(\dfrac{\pi}{3} - \theta\right) \sin\left(\dfrac{\pi}{3} + \theta\right) = \dfrac{1}{4} \sin(3\theta)$sin(θ)sin(3π​−θ)sin(3π​+θ)=41​sin(3θ) with $\theta = \dfrac{\pi}{18}$θ=18π​:

$$\dfrac{\pi}{3} - \theta = \dfrac{6\pi}{18} - \dfrac{\pi}{18} = \dfrac{5\pi}{18}$$3π​−θ=186π​−18π​=185π​

$$\dfrac{\pi}{3} + \theta = \dfrac{6\pi}{18} + \dfrac{\pi}{18} = \dfrac{7\pi}{18}$$3π​+θ=186π​+18π​=187π​

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=41​sin(3×18π​)=41​sin(6π​)

$$K = \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}$$K=41​×21​=81​

We need to find the value of $\sin\left(\dfrac{10K\pi}{3}\right)$sin(310Kπ​). Substituting $K = \dfrac{1}{8}$K=81​:

$$\sin\left(\dfrac{10 \times \dfrac{1}{8} \times \pi}{3}\right) = \sin\left(\dfrac{10\pi}{24}\right) = \sin\left(\dfrac{5\pi}{12}\right)$$sin​310×81​×π​​=sin(2410π​)=sin(125π​)

Expanding $\dfrac{5\pi}{12}$125π​ as $\dfrac{\pi}{4} + \dfrac{\pi}{6}$4π​+6π​:

$$\sin\left(\dfrac{5\pi}{12}\right) = \sin\left(\dfrac{\pi}{4} + \dfrac{\pi}{6}\right)$$sin(125π​)=sin(4π​+6π​)

$$\sin\left(\dfrac{5\pi}{12}\right) = \sin\left(\dfrac{\pi}{4}\right)\cos\left(\dfrac{\pi}{6}\right) + \cos\left(\dfrac{\pi}{4}\right)\sin\left(\dfrac{\pi}{6}\right)$$sin(125π​)=sin(4π​)cos(6π​)+cos(4π​)sin(6π​)

$$\sin\left(\dfrac{5\pi}{12}\right) = \left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{\sqrt{3}}{2}\right) + \left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{1}{2}\right) = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}$$sin(125π​)=(2​1​)(23​​)+(2​1​)(21​)=22​3​+1​

Answer: $\dfrac{\sqrt{3} + 1}{2\sqrt{2}}$22​3​+1​