Trigonometric RatiosmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

sin10·sin50·sin70 and sin(10Kπ/3) = sin75° | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
If sin⁡(π18)sin⁡(5π18)sin⁡(7π18)=K\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⁡(10Kπ3)\sin\left(\dfrac{10K\pi}{3}\right) is
A3+122\dfrac{\sqrt3+1}{2\sqrt2}correct
B3−12\dfrac{\sqrt3-1}{\sqrt2}
C32\dfrac{\sqrt3}{2}
D12\dfrac12
Solution
Step 1: Convert to degrees: π18=10∘, 5π18=50∘, 7π18=70∘\dfrac{\pi}{18}=10^\circ,\ \dfrac{5\pi}{18}=50^\circ,\ \dfrac{7\pi}{18}=70^\circ. So
K=sin⁡10∘sin⁡50∘sin⁡70∘.K=\sin10^\circ\sin50^\circ\sin70^\circ.
 Step 2: Use the identity sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)=14sin⁡3θ\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\dfrac14\sin3\theta with θ=10∘\theta=10^\circ (since 50∘=60∘−10∘50^\circ=60^\circ-10^\circ, 70∘=60∘+10∘70^\circ=60^\circ+10^\circ):
K=14sin⁡30∘=14⋅12=18.K=\frac14\sin30^\circ=\frac14\cdot\frac12=\frac18.
 Step 3: Then
sin⁡(10Kπ3)=sin⁡(10π3⋅18)=sin⁡(10π24)=sin⁡(5π12)=sin⁡75∘.\sin\left(\frac{10K\pi}{3}\right)=\sin\left(\frac{10\pi}{3}\cdot\frac18\right)=\sin\left(\frac{10\pi}{24}\right)=\sin\left(\frac{5\pi}{12}\right)=\sin75^\circ.
 Step 4:
sin⁡75∘=3+122.\sin75^\circ=\frac{\sqrt3+1}{2\sqrt2}.
 Correct answer: (1)
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