JEE Main 2022 · MathsMultiple choiceSingle correctEasyCalculation

JEE Main 26 June 2022, Shift 2, Maths Q18

Question 18 of 90 in this shift, Maths question 18 of 30, Section A.

16 \sin(20^\circ) \sin(40^\circ) \sin(80^\circ)

is equal to :

(1) $\sqrt{3}$
(2) $2\sqrt{3}$ Official answer
(3) $3$
(4) $4\sqrt{3}$

Official answer

Option 2

NTA final key.

Chapter: Trigonometric Functions
Topic: Trigonometric Identities
Idea tested: Sine Product Identity with 60 Degrees

Same idea in other shifts

Asked 3× in all

2 Apr 2026, Shift 1 · Q11If $\sin\left(\frac{\pi}{18}\right)\sin\left(\frac{5\pi}{18}\right)\sin\left(\frac{7\pi}{18}\right)=K$ , then the value of…MediumSingle correct
26 Jun 2022, Shift 1 · Q26If $\sin^2(10^\circ)\,\sin(20^\circ)\,\sin(40^\circ)\,\sin(50^\circ)\,\sin(70^\circ) = \alpha - \frac{1}{16}\sin(10^\circ)$ , then…HardNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.