Re: [中學] 三角函數疑問已刪文
※ 引述《HANDREAD0819 (小欣)》之銘言:
: 想請問板上高手
: a(b^2+c^2)cosA + b(c^2+a^2)cosB + c(a^2+b^2)cosC=????
: 這題帶餘弦公式沒看到什們特別的地方可以消
: 想破頭一陣子,想請板上高手指點一下,感激不盡><
這題如果要像
D大所說用餘弦
爆開來算的話
其實也都可以
(起碼就此題來看,去年八月初那題不會太難
想到用"射影"來解,即使今年二月下旬
回文所po"奧數競教"當中解答也是提供
"餘弦"作法,否則此題也頗難想到用"射影"來解題.)
原式
=a(b^2+c^2)[(b^2+c^2-a^2)/2bc]+b(c^2+a^2)[(c^2+a^2-b^2)/2ca]
+c(a^2+b^2)[(a^2+b^2-c^2)/2ab]
=[a^2(b^2+c^2)(b^2+c^2-a^2)+b^2(c^2+a^2)(c^2+a^2-b^2)+
c^2(a^2+b^2)(a^2+b^2-c^2)]/(2abc)
=[a^2(b^2+c^2)^2-a^4(b^2+c^2)+b^2(c^2+a^2)^2-b^4(c^2+a^2)
+c^2(a^2+b^2)^2-c^4(a^2+b^2)]/(2abc)
=[6a^2b^2c^2+a^2(b^4+c^4)+b^2(c^4+a^4)+c^2(a^4+b^4)
-a^4(b^2+c^2)-b^4(c^2+a^2)-c^4(a^2+b^2)]/(2abc)
觀察分子倒數前三項
也是最難看出可消去的地方
亦可整理成-[a^2(b^4+c^4)+b^2(c^4+a^4)+c^2(a^4+b^4)]
於是乎
正負相抵後
可得
6(abc)^2/(2abc)=3abc
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