Odpowiedzi

2010-04-10T00:09:35+02:00
Γ=180-(α+β)=2*[90-(α+β)/2]
sinα+sinβ+sinγ=2sin(α+β)/2*cos((α-β)/2+2sin[90-(α+β)/2]*cos[90-(α+β)/2]=
2sin(α+β)/2*cos((α-β)/2+2cos(α+β)/2* sin(α+β)/2=
2sin(α+β)/2*[cos(α-β)/2+cos(α+β)/2]=
4sin(α+β)/2*cosα/2*cosβ/2
ale
(α+β)/2=90-γ/2
wiec
sinα+sinβ+sinγ=4cosα/2 * cosβ/2 * cosγ/2

cbdu

wykorzystano wzory:
sin2α=2sinα*cosα
cosα+cosβ=2cos(α+β)/2 * cos(α-β)/2

pozdrawiam

Hnas


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