Trigonometric Identities Question 243
Question: What is $ {{\sin }^{2}}66\frac{1{}^\circ }{2}-{{\sin }^{2}}23\frac{1{}^\circ }{2} $ equal to?
Options:
A) $ sin47{}^\circ $
B) $ \text{cos }47{}^\circ $
C) $ 2,\sin 47{}^\circ $
D) $ 2,\cos 47{}^\circ $
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\sin }^{2}}66\frac{1{}^\circ }{2}-{{\sin }^{2}}23\frac{1{}^\circ }{2} $ $ ={{[ \sin ( 90{}^\circ -23\frac{1{}^\circ }{2} ) ]}^{2}}-{{\sin }^{2}}23\frac{1{}^\circ }{2} $ $ ={{\cos }^{2}}23\frac{1{}^\circ }{2}-{{\sin }^{2}}23\frac{1{}^\circ }{2} $ $ =\cos 2( 23\frac{1{}^\circ }{2} )=\cos 47{}^\circ $ $ (\because ,\cos 2A=cos^{2}A-{{\sin }^{2}}A) $ $ =\cos [ 2\times ( \frac{47}{2} ) ]=\cos 47{}^\circ $
BETA
