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JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 15

The value of 2sin⁡(12°)−sin⁡(72°)2\sin(12°)-\sin(72°) is :

Choose one option
Answer and worked solution

Official answer: D

Rewrite. 2sin⁡12°−sin⁡72°=sin⁡12°+(sin⁡12°−sin⁡72°)=sin⁡12°−2cos⁡42°sin⁡30°=sin⁡12°−sin⁡48°2\sin12°-\sin72°=\sin12°+(\sin12°-\sin72°)=\sin12°-2\cos42°\sin30°=\sin12°-\sin48°.

Again. sin⁡12°−sin⁡48°=−2cos⁡30°sin⁡18°=−3⋅5−14=3(1−5)4\sin12°-\sin48°=-2\cos30°\sin18°=-\sqrt3\cdot\dfrac{\sqrt5-1}4=\dfrac{\sqrt3\left(1-\sqrt5\right)}4.

(D)

Topics:Trigonometric ratios and identities
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