A geometry problem by A Former Brilliant Member

Geometry Level 3

Which is larger?

(A) sin ⁡ x x \dfrac{\sin x}{x}

(B) x tan ⁡ x \dfrac{x}{\tan x}

Details

x ∈ ( 0 , π 2 ) x\in {(0 , \dfrac{π}{2})}

B A

A Former Brilliant Member by A Former Brilliant Member

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1 solution

G.M. of sin ⁡ x \sin x and tan ⁡ x \tan x is greater than their H.M. i.e.

sin ⁡ x × tan ⁡ x > ( 2 cosec ⁡ x + cot ⁡ x ) 2 \sin x\times \tan x>\displaystyle (\dfrac{2}{\cosec x+\cot x})^2

= 4 ( cosec ⁡ x − cot ⁡ x ) 2 4(\cosec x-\cot x)^2

= 4 × ( 1 − cos ⁡ x sin ⁡ x ) 2 4\times \displaystyle (\dfrac{1-\cos x}{\sin x})^2

= 4 × ( tan ⁡ ( x 2 ) ) 2 > 4 × ( x 2 ) 2 4\times \displaystyle (\tan (\dfrac{x}{2}))^2>4\times (\dfrac{x}{2})^2

= x 2 x^2 (Since tan ⁡ ( x 2 ) > x 2 \tan (\dfrac{x}{2})>\dfrac{x}{2} ).

Therefore sin ⁡ x x > x tan ⁡ x \dfrac{\sin x}{x}>\dfrac{x}{\tan x}

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