JCECE Engineering JCECE Engineering Solved Paper-2009

  • question_answer
    \[{{\cos }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{3\pi }{8}+{{\cos }^{4}}\frac{5\pi }{8}+{{\cos }^{4}}\frac{7\pi }{8}\]is equal to

    A) \[\frac{3}{2}\]                                   

    B) \[-\frac{2}{3}\]

    C) \[-1\]                                    

    D) \[1\]

    Correct Answer: A

    Solution :

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


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