Let n be a fixed positive integer such that sin(2nπ ) +cos(2nπ ) =2n , then

We have, sin(2nπ ) +cos(2nπ ) =2sin(4π +2nπ ) ⇒2n =2sin(4π +2nπ ) So for n>1, 22 n =sin(4π +2nπ ) >sin4π =2 1Thus n>4Since sin(4π +2nπ ) 2,we get 22 n <1⇒n<8∴4<n<8

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Let $n$ be a fixed positive integer such that $sin (\frac{π}{2 n}) + cos (\frac{π}{2 n}) = \frac{n}{2}$ , then

n = 4

n = 5

n = 6

none of these

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Liam Discussed

Let

n

be a fixed positive integer such that

sin (\frac{π}{2 n}) + cos (\frac{π}{2 n}) = \frac{n}{2}

, then

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

We have,

sin (\frac{π}{2 n}) + cos (\frac{π}{2 n}) = 2 sin (\frac{π}{4} + \frac{π}{2 n})

\Rightarrow \frac{n}{2} = 2 sin (\frac{π}{4} + \frac{π}{2 n})

So for

n > 1, \frac{n}{2 2} = sin (\frac{π}{4} + \frac{π}{2 n}) > sin \frac{π}{4} = \frac{1}{2}

Thus

n > 4

Since

sin (\frac{π}{4} + \frac{π}{2 n}) < 1

for all

n > 2

we get

\frac{n}{2 2} < 1 \Rightarrow n < 8

∴ 4 < n < 8

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Question Text	Let $n$ be a fixed positive integer such that $sin (\frac{π}{2 n}) + cos (\frac{π}{2 n}) = \frac{n}{2}$ , then
Answer Type	Text solution:1
Upvotes	150

Let n be a fixed positive integer such that sin(2nπ ​ ) +cos(2nπ ​ ) =2n​ ​, then

Text solutionVerified

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Let $n$ be a fixed positive integer such that $sin (\frac{π}{2 n}) + cos (\frac{π}{2 n}) = \frac{n}{2}$ , then