11th Class Mathematics Trigonometric Identities Question Bank Critical Thinking

  • question_answer \[\sin {{20}^{o}}\,\sin {{40}^{o}}\,\sin {{60}^{o}}\,\sin {{80}^{o}}=\] [MNR 1976, 81]

    A) \[-3/16\]

    B) \[5/16\]

    C) \[3/16\]

    D) \[-5/16\]

    Correct Answer: C

    Solution :

    \[\sin 20{}^\circ \sin {{40}^{o}}\sin 60{}^\circ \sin 80{}^\circ \]\[=\frac{1}{2}\sin 20{}^\circ \sin 60{}^\circ \,(2\sin {{40}^{o}}\sin 80{}^\circ )\] \[=\frac{1}{2}\sin 20{}^\circ \sin 60{}^\circ (\cos 40{}^\circ -\cos 120{}^\circ )\] \[=\frac{1}{2}.\frac{\sqrt{3}}{2}\sin 20{}^\circ \left( 1-2{{\sin }^{2}}20{}^\circ +\frac{1}{2} \right)\] \[=\frac{\sqrt{3}}{4}\sin 20{}^\circ \left( \frac{3}{2}-2{{\sin }^{2}}20{}^\circ  \right)\] \[=\frac{\sqrt{3}}{8}(3\sin 20{}^\circ -4{{\sin }^{3}}20{}^\circ )\]\[=\frac{\sqrt{3}}{8}\sin 60{}^\circ =\frac{\sqrt{3}}{8}.\frac{\sqrt{3}}{2}=\frac{3}{16}\].


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