Trigonometric Identities Question 218
Question: If $ A=(\cos 12{}^\circ -\cos 36{}^\circ )(\sin 96{}^\circ +\sin 24{}^\circ ) $ and $ B=(\sin 60{}^\circ -\sin 12{}^\circ )(\cos 48{}^\circ -\cos 72{}^\circ ), $ then what is $ \frac{A}{B} $ equal to?
Options:
A) -1
B) $ 0 $
C) $ 1 $
D) $ 2 $
Show Answer
Answer:
Correct Answer: C
Solution:
Given $ A=(\cos 12{}^\circ -\cos 36{}^\circ )(\sin 96{}^\circ +sin24{}^\circ ) $ $ B=(\sin 60{}^\circ -\sin 12{}^\circ )(\cos 48{}^\circ -\cos 72{}^\circ ) $ $ \frac{A}{B}=\frac{[-2\sin 24{}^\circ \sin 12{}^\circ ][2\sin 60{}^\circ \cos 36{}^\circ ]}{[2\cos 36{}^\circ \sin 24{}^\circ ][-2\sin 60{}^\circ \sin 12{}^\circ ]} $
$ \Rightarrow \frac{A}{B}=1 $
BETA
