Trigonometric-Identities Question 377
Question: $ \sin 12{}^\circ \sin 24{}^\circ \sin 48{}^\circ \sin 84{}^\circ = $
[EAMCET 1989]
Options:
A) $ \cos 20{}^\circ \cos 40{}^\circ \cos 60{}^\circ \cos 80{}^\circ $
B) $ \sin 20{}^\circ \sin 40{}^\circ \sin 60{}^\circ \sin 80{}^\circ $
C) $ \frac{3}{15} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \sin 12^{o}\sin 24^{o}\sin 48^{o}\sin 84^{o} $ $ =\frac{1}{4},(2\sin 12^{o},\sin 48^{o})(2\sin 24^{o}\sin 84^{o}) $ $ =\frac{1}{2}(\cos 36^{o}-\cos 60^{o})(\cos 60^{o}-\cos 108^{o}) $ $ =\frac{1}{4},( \cos 36^{o}-\frac{1}{2} )( \frac{1}{2}+\sin 18^{o} ) $ $ =\frac{1}{4}{ \frac{1}{4}(\sqrt{5}+1)-\frac{1}{2} },{ \frac{1}{2}+\frac{1}{4}(\sqrt{5}-1) }=\frac{1}{16} $ and $ \cos 20^{o},\cos 40^{o}\cos 60\cos 80^{o} $ $ =\frac{1}{2}[\cos ,(60^{o}-20^{o}),\cos 20^{o},\cos ,(60^{o}+20^{o})] $ $ =\frac{1}{2},[ \frac{1}{4}\cos 3(20^{o}) ]=\frac{1}{8}\cos 60^{o}=\frac{1}{2}\times \frac{1}{8}=\frac{1}{16} $ .
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