A) 1/4
B) 1/16
C) 3/4
D) 5/16
Correct Answer: D
Solution :
[d]: \[\sin 36{}^\circ \sin 72{}^\circ \sin 108{}^\circ \sin 144{}^\circ \] \[={{\sin }^{2}}{{36}^{o}}{{\sin }^{2}}{{72}^{o}}=\frac{1}{4}\{(2si{{n}^{2}}{{36}^{o}})(2si{{n}^{2}}{{72}^{o}})\}\] \[=\frac{1}{4}\{(1-cos{{72}^{o}})(1-cos{{114}^{o}})\}\] \[=\frac{1}{4}\{(1-sin{{18}^{o}})(1+cos{{36}^{o}})\}\] \[=\frac{1}{4}\left[ \left( 1-\frac{\sqrt{5}-1}{4} \right)\left( 1+\frac{\sqrt{5}+1}{4} \right) \right]=\frac{20}{16}\times \frac{1}{4}=\frac{5}{16}\]
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