A) 1/4
B) 1/16
C) 3/4
D) 5/16
Correct Answer: D
Solution :
\[\sin \,\,{{36}^{o}}\,\,\sin \,\,{{72}^{o}}\,\sin \,\,{{108}^{o}}\,\,\sin \,\,{{144}^{o}}\] \[={{\sin }^{2}}{{36}^{o}}\,\,{{\sin }^{2}}\,{{72}^{o}}=\frac{1}{4}\,\left\{ (2\,\,{{\sin }^{2}}{{36}^{o}})\,\,(2\,\,{{\sin }^{2}}\,\,{{72}^{o}}) \right\}\] \[=\frac{1}{4}\left\{ (1-\cos \,\,{{72}^{o}})\,\,(1-\cos \,\,{{144}^{o}}) \right\}\] \[=\frac{1}{4}\left\{ (1-\sin \,\,{{18}^{o}})\,\,(1+\cos \,\,{{36}^{o}}) \right\}\] \[=\frac{1}{4}\left[ \left( 1-\frac{\sqrt{5}-1}{4} \right)\,\,\left( 1+\frac{\sqrt{5}+1}{4} \right) \right]=\frac{20}{16}\times \frac{1}{4}=\frac{5}{16}\].
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