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全部话题 - 话题: bcosx
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s*****r
发帖数: 11545
1
来自主题: Military版 - 几何题求助
Three basic rules:
c^2=a^2+b^2-2abcosC
cosC=(a^2+b^2-c^2)/2ab
sinC=c/2R
Solution:
(FD)^2
= (CF)^2+(a/2)^2-2(CF)(a/2)cos(C-X)
=(bcosX)^2+(a/2)^2-2(bcosX)(a/2)(cosCcosX+sinCsinX)
=(bcosX)^2+(a/2)^2-2(bcosX)(a/2)((a^2+b^2-c^2)/2abcosX+(c/2R)sinX))
=a^2/4+(b^2+c^2-a^2)(cosX)^2/2-abcsinX/2R
(ED)^2
=(BE)^2+(a/2)^2-2(BE)(a/2)cos(B-X)
=(ccosX)^2+(a/2)^2-2(ccosX)(a/2)(cosBcosX+sinBsinX)
=(ccosX)^2+(a/2)^2-2(ccosX)(a/2)((a^2+c^2-b^2)/2accosX+(b/2R)sinX))
=a^2/4+(b^2+c^2-a^2)(cosX)^2/2-abcsinX/2R
So, FD=ED
n***p
发帖数: 7668
2
OK, 闲着也是闲着,赚个包子吧。
a sinx +b sin(x+m)
= a sinx + bsinx cosm + bcosx sinm
= (a+bcosm)sinx + bsinm cosx
= c sin(x+n)
where
c =sqrt( (a+bcosm)^2 + (bsinm)^2 )
=sqrt( a^2 +b^2 +2abcosm)
And we need to take n so that
cos n = (a+bcosm) /c
sin n = bsinm /c.
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