Re: [其他] 三角函數 內心
前一陣子講的
亦可再參考"標準奧林匹克數競教程"
當中習題3.1的2nd題
證明1/a = 1/b + 1/c,b+c=bc/a,1/(b+c)=a/bc
CD=ab/(b+c),CE=ab/(c+a)
ab{[1/(b+c)]+[1/(c+a)]}=12[2cos^2(B/2)-1]
(a^2/c)+ab/(c+a)=24cos^2(B/2)-12
(a^2/c)+(a^2/b)=24cos^2(B/2)-12...代b^2=a(c+a),參考黃家禮所編著的"幾明".
a+12=24cos^2(B/2),a+12=24cos^2A=12(1+cos2A)
a=12cos2A,B=2A,C=2B=4A,A=pi-6A,A=pi/7
CF=[2ab/(a+b)]cos(B/2)...陳一理所編著的"三角"皆有證明
=[2ab/(a+b)]cosA=[2ab/(a+b)]*[(b^2+c^2-a^2)/2bc]
=(a/c)[a(c+a)+c^2-a^2]/(a+b)
=a(c+a)/(a+b)=b^2/(a+b)
=b^3/[b(a+b)]=b^3/c^2...同理C=2B時,c^2=b(a+b).
=b(b/c)^2=(12sin2Acos2A/sinA)(sin2A/sin4A)^2
=6(sin4A/sinA)(sin2A/sin4A)^2
=6[sin^2(2A)]/(sinAsin4A)
=6sin2A/(2sinAcos2A)
=6(cosA/cos2A)...大概就是6
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※ 編輯: wayne2011 (61.58.103.35), 09/25/2017 15:32:04
※ 編輯: wayne2011 (61.58.103.35), 09/25/2017 15:32:47
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