Re: [解題] 高三 數學 三角函數跟微分
sina + sinb = 2sin[(a+b)/2]cos[(a-b)/2] ≦ 2sin[(a+b)/2]×1
(sina + sinb)/2 ≦ sin[(a+b)/2]
sin[(a+b+c+d)/4]
= sin{{[(a+b)/2]+[(c+d)/2]}/2}
≧{sin[(a+b)/2]+sin[(c+d)/2]}
≧{[(sina+sinb)/2] + [(sina+sinb)/2]}/2
= (sina+sinb+sinc+sind)/4
令 d = (a+b+c)/3
=> 3d = a+b+c
=> a+b+c+d = 4d
=> (a+b+c+d)/4 = d
sind≧(sina+sinb+sinc+sind)/4
=> 4sind≧sina+sinb+sinc+sind
=> 3sind≧sina+sinb+sinc
=> sind≧(sina+sinb+sinc)/3
=> sin[(a+b+c)/3]≧(sina+sinb+sinc)/3
而本題abc為三角形三內角
故a+b+c＝180度
(sina+sinb+sinc)/3 ≦ sin[(a+b+c)/3] = sin(180度/3) = sin 60度 = (根號3)/2
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06/09 00:03, , 1F
06/09 00:03, 1F
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