Question

Using the relation 2 (1 - cos x) < x^2 , x ≠ 0 or otherwise, prove that sin (tan x) ≥ x, ∀ x ∈ [0, pi/4]

Answer 1

Mark it brainliest answer please
It is enough to show sin(x)≥arctan(x)sin⁡(x)≥arctan⁡(x)over x∈[0,1]x∈[0,1], or cos(x)≥11+x2cos⁡(x)≥11+x2, then integrate both sides. Given the hint, that is straightforward:

∀x∈[0,1],(1+x2)cos(x)≥(1+x2)(1−x22)≥1(1)(1)∀x∈[0,1],(1+x2)cos⁡(x)≥(1+x2)(1−x22)≥1

hence for any z∈[0,1]z∈[0,1] we have cos(z)≥11+z2cos⁡(z)≥11+z2 and

∀u∈[0,1],sin(u)=∫u0cos(z)dz≥∫u0dzz2+1=arctan(u)(2)(2)∀u∈[0,1],sin⁡(u)=∫0ucos⁡(z)dz≥∫0udzz2+1=arctan⁡(u)

so by setting u=tan(x)u=tan⁡(x) we have:

∀x∈[0,π4],sin(tan(x))≥x