11Use this fact sina+cosa=√(1+sin2a)
c).
Question. Let n b a +ve integer such that sinπ2n+cosπ2n=√n2.
then
(a) 6≤n≤8 (b) 4≤n≤8 (c) 4<n≤8 (d) 4<n<8
Please show ur steps........am doubtful about a step.
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8 Answers
1but doing that also how are u getting range of n????
unable to understand
1i m getting b................plz tell the right ans............
1i donno.............whether i m rite bt plz see & correct me if i m wrong
applying √(1+sin2a) =√n/2 ,,,,
=> 1+ sin 2a = n/4
=>sin2a =(n-4)/4
=> n-4 can be zero at minimum ..................so n≥4
also max(n-4) = 4 => n max =8
so, n≤8
1Answer is c only....................expert help needed for detailed solution
11My dear jhunjhunwala....(don't mind).
Squaring I've 1+sin2a=n/4
So n≤8.
What is the min value of sin(Î 21-n) ? it is 0 - but it never attains that value.
So n>4.