Trigonometric Identities Question 264
Question: Let n be a fixed positive integer such that $ \sin ( \frac{\pi }{2n} )+\cos ( \frac{\pi }{2n} )=\frac{\sqrt{n}}{2}, $ then:
Options:
A) $ n=4 $
B) $ n=5 $
C) $ n=6 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \sin \frac{\pi }{2n}+\cos \frac{\pi }{2n}=\sqrt{2}\sin ( \frac{\pi }{4}+\frac{\pi }{2n} ) $
$ \Rightarrow \frac{\sqrt{n}}{2}=\sqrt{2}\sin ( \frac{\pi }{4}+\frac{\pi }{2n} ) $ So, for $ n>1, $ $ \frac{\sqrt{n}}{2\sqrt{2}}=\sin ( \frac{\pi }{4}+\frac{\pi }{2n} )>\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}} $ Thus, $ n>4 $ Since, $ \sin ( \frac{\pi }{4}+\frac{\pi }{2n} )<1 $ for all $ n>2, $ we get $ \frac{\sqrt{n}}{2\sqrt{2}}<1 $ or $ n<8, $ So that $ 4<n<8 $ . By actual verification we find that only $ n=6 $ satisfies the given relation.
BETA
