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
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
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
Was Aristarchus of Samos wrong because he failed to meet the burden of proof?
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