Let `n` be a positive integer such that `sinpi/(2n)+cospi/(2n)=(sqrt(n))/2dot` Then `6lt=nlt=8` (b) `4 ltnle8` c) `4lenle8` d) `4ltnlt8`

Explanation

`sin(pi/(2n))+cos(pi/(2n)) = sqrtn/2`

Squaring both sides,

`=>sin^2(pi/(2n)) +cos^2(pi/(2n)) +2sin(pi/(2n))cos(pi/(2n)) = n/4`

`=>1+sin(pi/n) = n/4`

`=>sin(pi/n) = n/4-1`

We know, `-1 le sintheta le 1`

As, `n` is positive,

`:. 0 lt sin(pi/n) lt 1`

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