在△ABC中,a,b,c分別為內(nèi)角A,B,C的對(duì)邊,且2asinA=(2b+c)sinB+(2c+b)sinC.
(Ⅰ)求A的大�。�
(Ⅱ)求sinB+sinC的最大值.
【答案】
分析:(Ⅰ)根據(jù)正弦定理,設(shè)
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,把sinA,sinB,sinC代入2asinA=(2b+c)sinB+(2c+b)sinC求出a
2=b
2+c
2+bc
再與余弦定理聯(lián)立方程,可求出cosA的值,進(jìn)而求出A的值.
(Ⅱ)根據(jù)(Ⅰ)中A的值,可知c=60°-B,化簡(jiǎn)得sin(60°+B)根據(jù)三角函數(shù)的性質(zhì),得出最大值.
解答:解:(Ⅰ)設(shè)
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則a=2RsinA,b=2RsinB,c=2RsinC
∵2asinA=(2b+c)sinB+(2c+b)sinC
方程兩邊同乘以2R
∴2a
2=(2b+c)b+(2c+b)c
整理得a
2=b
2+c
2+bc
∵由余弦定理得a
2=b
2+c
2-2bccosA
故cosA=-
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,A=120°
(Ⅱ)由(Ⅰ)得:sinB+sinC
=sinB+sin(60°-B)
=
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cosB+
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sinB
=sin(60°+B)
故當(dāng)B=30°時(shí),sinB+sinC取得最大值1.
點(diǎn)評(píng):本題主要考查了余弦函數(shù)的應(yīng)用.其主要用來(lái)解決三角形中邊、角問(wèn)題,故應(yīng)熟練掌握.