• # question_answer 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) ${{[(\sqrt{5}-1)/4]}^{2}}$

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}$

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