JEE Main 2019TrigonometryTrigonometric Ratio and IdentitesEasyQuestion16sin(20∘)sin(40∘)sin(80∘)16\sin (20^\circ )\sin (40^\circ )\sin (80^\circ )16sin(20∘)sin(40∘)sin(80∘) is equal to :OptionsA3\sqrt 3 3B23\sqrt 3 3C3D43\sqrt 3 3Check AnswerHide SolutionSolution16sin20∘ . sin40∘ . sin80∘16\sin 20^\circ \,.\,\sin 40^\circ \,.\,\sin 80^\circ 16sin20∘.sin40∘.sin80∘ =4sin60∘= 4\sin 60^\circ=4sin60∘ {∵\because∵ 4sinθ . sin(60∘−θ) . sin(60∘+θ)=sin3θ4\sin \theta \,.\,\sin (60^\circ - \theta )\,.\,\sin (60^\circ + \theta ) = \sin 3\theta 4sinθ.sin(60∘−θ).sin(60∘+θ)=sin3θ} =23= 2\sqrt 3=23