Prove-that-sin-x-x-tan-x-for-x-lt-pi-2-

Question Number 56681 by Joel578 last updated on 21/Mar/19

Prove that

\sin ∣ x ∣ ⩽ ∣ x ∣ ⩽ \tan ∣ x ∣ for ∣ x ∣ < \frac{π}{2}

Commented by Abdo msup. last updated on 22/Mar/19

l e t ∣ x ∣= t l e t p r o v e t h a t s i n t ⩽ t ⩽ t a n t f o r 0 ⩽ t < \frac{π}{2}

l e t W (x) = t - s i n t \Rightarrow W^{,} (t) = 1 - c o s t ⩾ 0

t 0 \frac{π}{2}

W^{^{'}} +

W 0 i n c r 1 \Rightarrow W (t) ⩾ 0 \Rightarrow s i n t ⩽ t

l e t φ (t) = t a n t - t \Rightarrow φ^{^{'}} (t) = 1 + {t a n}^{2} t - 1 = {t a n}^{2} t ⩾ 0 \Rightarrow

t 0 \frac{π}{2}

φ^{^{'}} +

φ 0 + \infty \Rightarrow φ (t) ⩾ 0 \Rightarrow t ⩽ t a n t

s o t h e r e s u l t i s p r o v e d .

Answered by ajfour last updated on 21/Mar/19

let ∣ x ∣= x

\cos x ⩽ 1 ⩽ 1 + \tan^{2} x

\frac{d (\sin x)}{dx} ⩽ \frac{d (x)}{dx} ⩽ \frac{d (\tan x)}{dx}

and at x = 0, \sin x = x = \tan x;

hence \sin x ⩽ x ⩽ \tan x .

Answered by mr W last updated on 21/Mar/19

Commented by mr W last updated on 21/Mar/19

O A = 1

∠ A O B = x

\overset{⌢}{A B} = x

A C = \sin x

A D = \tan x

A C < \overset{⌢}{A B} < A D

\Rightarrow \sin x < x < \tan x

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