1applying am gm for x2 and 1 (since given that x≥0) => x2+1 ≥ 2x but since sin-1 is there , the only possibility is equality .:. rhs = 2pi sin-1(1) = 1
and on rationalisation lhs = in .:. minimum n = 4
The least positive integer n for which (1+i1-i)n=2πsin-1(1+x22x) x≥0
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3 Answers
262hint : just rationalise lhs
rhs : only possible values =±1
ans : n=2
11Plz check if this solution is rite??
[(i+1)2i2-12]n=2Ï€sin-1(1+x22x)
==> in=2Ï€sin-1(1+x22x)
let x=tanθ
==>sin-1(1+x22x)=sin-1(sec2θ2tanθ)
=sin-11sin2θ
=sin-1cosec2θ
since the only possible values of cosec θ for which it can satisfy sin-1θ are ±1 and x≥0 hence the only possible value of sin θ is π2 hence in=1
==>n=4