Mathematics · 2022
JEE Main · 25 June 2022, Shift 2 · Q15
The value of 2 sin (12^°)- sin (72^°) is:
The value of $\displaystyle 2 \sin \left(12^{\circ}\right)-\sin \left(72^{\circ}\right)$ is :
Official answer
(4)
$\displaystyle \frac{\sqrt{3}(1-\sqrt{5})}{4}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.