JEE Main 2024TrigonometryTrigonometric Ratio and IdentitesEasyQuestionThe value of 2sin (12∘^\circ∘) −-− sin (72∘^\circ∘) is :OptionsA5(1−3)4{{\sqrt 5 (1 - \sqrt 3 )} \over 4}45(1−3)B1−58{{1 - \sqrt 5 } \over 8}81−5C3(1−5)2{{\sqrt 3 (1 - \sqrt 5 )} \over 2}23(1−5)D3(1−5)4{{\sqrt 3 (1 - \sqrt 5 )} \over 4}43(1−5)Check AnswerHide SolutionSolution2sin12∘−sin72∘2\sin 12^\circ - \sin 72^\circ 2sin12∘−sin72∘ =sin12∘+(−2cos42∘ . sin30∘) = \sin 12^\circ + ( - 2\cos 42^\circ \,.\,\sin 30^\circ )=sin12∘+(−2cos42∘.sin30∘) =sin12∘−cos42∘= \sin 12^\circ - \cos 42^\circ=sin12∘−cos42∘ =sin12∘−sin48∘= \sin 12^\circ - \sin 48^\circ=sin12∘−sin48∘ =2sin18∘ . cos30∘= 2\sin 18^\circ \,.\,\cos 30^\circ=2sin18∘.cos30∘ =−2(5−14) . 32 = - 2\left( {{{\sqrt 5 - 1} \over 4}} \right)\,.\,{{\sqrt 3 } \over 2}=−2(45−1).23 =3(1−5)4 = {{\sqrt 3 \left( {1 - \sqrt 5 } \right)} \over 4}=43(1−5)