Trigonometric Identities Question 65
Question: If $ \alpha =22{}^\circ 30’, $ then $ (1+\cos \alpha )(1+\cos 3\alpha ) $ $ (1+\cos 5\alpha )(1+\cos 7\alpha ) $ equals
[AMU 1999]
Options:
A) 1/8
B) 1/4
C) $ \frac{1+\sqrt{2}}{2\sqrt{2}} $
D) $ \frac{\sqrt{2}-1}{\sqrt{2}+1} $
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Answer:
Correct Answer: A
Solution:
We know, $ \sin 22\frac{1^{o}}{2}=\frac{1}{2}\sqrt{2-\sqrt{2}} $ and $ \cos 22\frac{1^{o}}{2}=\frac{1}{2}\sqrt{2+\sqrt{2}} $
$ \therefore ( 1+\cos 22\frac{1^{o}}{2} ),( 1+\cos 67\frac{1^{o}}{2} ),( 1+\cos 112\frac{1^{o}}{2} ) $ $ ( 1+\cos 157\frac{1^{o}}{2} ) $ $ =( 1+\frac{1}{2}\sqrt{2+\sqrt{2}} ),( 1+\frac{1}{2}\sqrt{2-\sqrt{2}} ),( 1-\frac{1}{2}\sqrt{2-\sqrt{2}} ), $ $ ( 1-\frac{1}{2}\sqrt{2+\sqrt{2}} ) $ $ =[ 1-\frac{1}{4}(2+\sqrt{2}) ],[ 1-\frac{1}{4}(2-\sqrt{2}) ] $ $ =\frac{(4-2-\sqrt{2})(4-2+\sqrt{2})}{16} $ $ =\frac{(2-\sqrt{2})(2+\sqrt{2})}{16}=\frac{4-2}{16}=\frac{1}{8} $ .
BETA
