Hello,

I have been trying to find out if 'Aristarchus' Theorem' can be solved geometrically.

It states that if 0 < α < β < pi/2 then Sin(α)/α > Sin(β)/β

Can you folks think of a possible way of proving this geometrically ?

Many thanks

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- October 31st 2012, 04:42 PMsab456proof of Aristarchus' Theorem geomterically
Hello,

I have been trying to find out if 'Aristarchus' Theorem' can be solved geometrically.

It states that if 0 < α < β < pi/2 then Sin(α)/α > Sin(β)/β

Can you folks think of a possible way of proving this geometrically ?

Many thanks - October 31st 2012, 09:21 PMSalahuddin559Re: proof of Aristarchus' Theorem geomterically
How about a unit circle? The length on perimeter is equals to angle theta, and the side approximating it in the right triangle will be sin(theta).

Salahuddin

Maths online - November 1st 2012, 02:52 AMelisaevedentRe: proof of Aristarchus' Theorem geomterically
Was Aristarchus of Samos wrong because he failed to meet the burden of proof?

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