Trigonometric Identities Question 7
Question: The value of $ \sin 47^{o}+\sin 61{}^\circ -\sin 11{}^\circ -\sin 25{}^\circ = $
[MP PET 2001; EAMCET 2003]
Options:
A) $ \sin 36{}^\circ $
B) $ \cos 36{}^\circ $
C) $ \sin 7{}^\circ $
D) $ \cos 7{}^\circ $
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Answer:
Correct Answer: D
Solution:
$ \sin 47^{o}+\sin 61^{o}-(\sin 11^{o}+\sin 25^{o}) $ $ =\frac{\sin 20^{o}\sin 40^{o}\sin 80^{o}}{\cos 20^{o}\cos 40^{o}\cos 80^{o}} $ $ =2\cos 7^{o},(\sin 54^{o}-\sin 18^{o}) $ $ =2\cos 7^{o}.2\cos 36^{o}.\sin 18^{o} $ $ =4.,\cos 7^{o}.,\frac{\sqrt{5}+1}{4}.\frac{\sqrt{5}-1}{4}=\cos {7^{o.}} $ .
BETA
