Trigonometric Identities Question 55
Question: $ \sin 36{}^\circ \sin 72{}^\circ \sin 108{}^\circ \sin 144{}^\circ = $
[IIT 1965]
Options:
A) 1/4
B) 1/16
C) 3/4
D) 5/16
Show Answer
Answer:
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},{ (2{{\sin }^{2}}36^{o})(2{{\sin }^{2}}72^{o}) } $ $ =\frac{1}{4}{ (1-\cos 72^{o})(1-\cos 144^{o}) } $ $ =\frac{1}{4}{ (1-\sin 18^{o})(1+\cos 36^{o}) } $ $ =\frac{1}{4}[ ( 1-\frac{\sqrt{5}-1}{4} )( 1+\frac{\sqrt{5}+1}{4} ) ]=\frac{20}{16}\times \frac{1}{4}=\frac{5}{16} $ .
BETA
