α is acute. Show that α < (sin α + tan α)/2.
Solution
Take A and B on a circle center O radius 1 with ∠AOB = α. Let the tangents at A and B meet at D, and let the tangent at A meet the line OB at C. Then the area of the sector OAB = α/2 < area OADB. Area OAC = ½ tan α and area OAB = ½ sin α. So it is sufficient to show that area OADB < (area OAB + area OAC)/2 or equivalently that area ABD < ½ area ABC. That in turn is equivalent to AD < ½ AC or AD < CD. But AD = BD (equal tangents), and BD < CD, because CD is the hypoteneuse.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03