Prove for all integer n>1?

Question

1/(2n) < sin(1/n) < 1/n

Answer 1

Since for every integer n≥1, 0 < 1/n < π/3 it suffices to show that x/2 < sin x < x for all x∈(0, π/3). Consider the function:

f(x) = sin x - x/2

Now, f(π/3) = sin (π/3) - π/6 = √3/2 - π/6 > 0. Suppose that there was some x∈(0, π/3) such that f(x) ≤ 0. Then there must be a point c∈(0, π/3) such that f(c) = 0 -- if f(x) = 0, then we simply take c=x, otherwise we have f(π/3) > 0 > f(x) and so by the intermediate value theorem there is a point c∈(x, π/3)⊆(0, π/3) such that f(c) = 0.

Now, since f(0) also equals 0, applying the mean value theorem to the interval (0, c) yields that there is a point ξ∈(0, c) ⊆ (0, π/3) such that f'(ξ) = 0. But f'(ξ) = cos ξ - 1/2, so this implies cos ξ = 1/2, which means ξ must be either π/3 + 2πk or -π/3 + 2πk for some integer k, and in any case would not be on the open interval (0, π/3) (f'(π/3) = 0, but (0, π/3) does not contain its endpoints)). This is a contradiction, so we must have that f(x) > 0 for all x∈(0, π/3), so in fact sin x > x/2 for all x∈(0, π/3)

To show that sin x < x on (0, π/3), we apply similar reasoning to the function g(x) = sin x - x -- we observe that g(π/3) = sin π/3 - π/3 < 0, so if there were a point x∈(0, π/3) such that g(x) ≥ 0, then by the intermediate value theorem, there would be a point c∈(0, π/3) such that g(c) = 0, so then applying the mean value theorem gives us a point ξ∈(0, c)⊆(0, π/3) such that g'(ξ) = 0. But g'(ξ) = cos ξ - 1, so this implies cos ξ = 1 and thus that ξ = π/2 + 2πk for some integer k, and thus ξ is not on the interval (0, π/3), which is a contradiction. Therefore g(x) < 0 for all x∈(0, π/3), so sin x < x for all x in the same interval.

Note that while we've established that x/2 < sin x < x only for x∈(0, π/3), this inequality actually holds for all x between 0 and about 1.895494267033981 (a little more than 3π/5). This was found through numerical computation -- I don't think the number has a closed-form expression.