11th Class Mathematics Trigonometric Identities Question Bank Critical Thinking

  • question_answer \[\left( 1+\cos \frac{\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8} \right)\,\left( 1+\cos \frac{5\pi }{8} \right)\,\left( 1+\cos \frac{7\pi }{8} \right)=\]  [IIT 1984; WB JEE 1992]

    A) \[\frac{1}{2}\]

    B) \[\frac{1}{4}\]

    C) \[\frac{1}{8}\]

    D) \[\frac{1}{16}\]

    Correct Answer: C

    Solution :

    \[\left( 1+\cos \frac{\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8} \right)\,\left( 1+\cos \frac{5\pi }{8} \right)\,\left( 1+\cos \frac{7\pi }{8} \right)\] \[=\left( 1+\cos \frac{\pi }{8}+\cos \frac{7\pi }{8}+\cos \frac{\pi }{8}\cos \frac{7\pi }{8} \right)\]\[\left( 1+\cos \frac{5\pi }{8}+\cos \frac{3\pi }{8}+\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} \right)\] \[=\left( 1+\cos \frac{\pi }{8}-\cos \frac{\pi }{8}+\cos \frac{\pi }{8}\cos \frac{7\pi }{8} \right)\]\[\left( 1+\cos \frac{5\pi }{8}-\cos \frac{5\pi }{8}+\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} \right)\] \[=\left( 1+\cos \frac{\pi }{8}\cos \frac{7\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} \right)\] \[=\frac{1}{4}\,\,\left( 2+2\cos \frac{\pi }{8}\cos \frac{7\pi }{8} \right)\,\,\left( 2+2\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} \right)\] \[=\frac{1}{4}\left( 2+\cos \frac{3\pi }{4}+\cos \pi  \right)\left( 2+\cos \frac{\pi }{4}+\cos \pi  \right)\] \[=\frac{1}{4}\,\left( 1+\cos \frac{3\pi }{4} \right)\,\left( 1+\cos \frac{\pi }{4} \right)=\frac{1}{4}\left( 1-\cos \frac{\pi }{4} \right)\,\left( 1+\cos \frac{\pi }{4} \right)\] \[=\frac{1}{4}\left( 1-{{\cos }^{2}}\frac{\pi }{4} \right)=\frac{1}{4}\left( 1-\frac{1}{2} \right)=\frac{1}{8}\]. Aliter:\[\left( 1+\cos \frac{\pi }{8} \right)\,\left( 1+\cos \frac{7\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8} \right)\,\left( 1+\cos \frac{5\pi }{8} \right)\] \[=\left( 1+\cos \frac{\pi }{8} \right)\,\left( 1-\cos \frac{\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8} \right)\,\left( 1-\cos \frac{3\pi }{8} \right)\] \[=\left( 1-{{\cos }^{2}}\frac{\pi }{8} \right)\text{ }\left( 1-{{\cos }^{2}}\frac{3\pi }{8} \right)={{\sin }^{2}}\frac{\pi }{8}{{\sin }^{2}}\frac{3\pi }{8}\] \[=\frac{1}{4}{{\left( 2\sin \frac{\pi }{8}.\sin \frac{3\pi }{8} \right)}^{2}}\]\[=\frac{1}{4}{{\left( \cos \frac{\pi }{4}-\cos \frac{\pi }{2} \right)}^{2}}=\frac{1}{8}\].


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