JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer 65) The value of\[\cos {{36}^{o}}\cos {{42}^{o}}\cos {{78}^{o}}\]is\[\left[ Given:\sin 18\frac{\sqrt{5}-1}{4}\,\,and\,\,\cos 36=\frac{\sqrt{5}+1}{4} \right]\]

    A) \[1/4\]                          

    B) \[1/8\]

    C) \[1/16\]                        

    D) \[{{[(\sqrt{5}-1)/4]}^{2}}\]

    Correct Answer: B

    Solution :

    \[\cos {{36}^{o}}\cos {{42}^{o}}\cos {{78}^{o}}\] \[=\cos {{36}^{o}}\cos ({{60}^{o}}-{{18}^{o}})\cos ({{60}^{o}}+{{18}^{o}})\] \[=\frac{\sqrt{5}+1}{4}\left( {{\cos }^{2}}{{60}^{o}}-{{\sin }^{2}}{{18}^{o}} \right)\] \[=\left( \frac{\sqrt{5}+1}{4} \right)\left[ \frac{1}{4}-{{\left( \frac{\sqrt{5}-1}{4} \right)}^{2}} \right]\] \[=\left( \frac{\sqrt{5}+1}{4} \right)\frac{1}{4}-\left( \frac{\sqrt{5}+1}{4} \right)\left( \frac{5+1-2\sqrt{5}}{16} \right)\] \[=\left( \frac{\sqrt{5}+1}{16} \right)-\frac{\left( \sqrt{5}+1 \right){{\left( \sqrt{5}-1 \right)}^{2}}}{64}=\frac{\sqrt{5}+1}{16}\left[ 1-\frac{{{\left( \sqrt{5}-1 \right)}^{2}}}{4} \right]\]\[=\frac{\sqrt{5}+1}{16}\left[ \frac{4-6+2\sqrt{5}}{4} \right]=\frac{1}{8}\]

adversite



LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

You need to login to perform this action.
You will be redirected in 3 sec spinner

Free
Videos