在△ABC中.
(1)已知sinA=cosBcosC,求證:tanC+tanB=1;
(2)求證:a2-2ab cos(60°+C)=b2-2bc cos(60°+A).
【答案】
分析:(1)根據A=B+C把sinA轉換成sin(A+B),進而利用兩角和公式化簡整理,等式兩邊同時除以cosBcosC,即可證明原式.
(2)先利用兩角和公式對要證的結論化簡整理可得a
2-abcosC+ab
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sinC=c
2-bccosA+bc
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sinA 再利用余弦定理分別把cosC,cosA代入整理asinC=csinA,根據正弦定理可知在三角形中此等式恒成立,進而使原式得證.
解答:解:(1)因為在三角形ABC中,sinA=cosBcosC
∴sin(B+C)=cosBcosC
即sinBcosC+cosBsinC=cosBcosC
等式兩邊同時除以cosBcosC,得
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+
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=1
即tanB+tanC=1,原式得證.
(2)證明:要使a
2-2ab cos(60°+C)=b
2-2bc cos(60°+A).
需a
2-2ab(
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cosC-
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sinC)=c
2-2bc(
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cosA-
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sinA)
需a
2-abcosC+ab
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sinC=c
2-bccosA+bc
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sinA
需a
2-
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(a
2+b
2-c
2)+ab
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sinC=c
2-
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(b
2+c
2-a
2)+bc
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sinA
需a
2-b
2+c
2+2ab
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sinC=c
2-b
2+a
2+2bc
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sinA
需asinC=csinA
在三角形ABC中,根據正弦定理可知
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即asinC=csinA恒成立,
所以等式得證
點評:本題主要考查了三角函數恒等式的證明,涉及了正弦定理,余弦定理,同角三角函數基本關系的應用等.考查了學生綜合分析問題和演繹推理的能力.