作业帮已知函数f(x)=[2√3cos(x/2)+2sin(x/2)]cos(x/2)（1）求f(17π/12)（2）在⊿ABC中,角A,B,C所对的边分别为a,b,c,若f(C)=√3+1,且b^2=ac,求sinA的值
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![已知函数f(x)=[2√3cos(x/2)+2sin(x/2)]cos(x/2)（1）求f(17π/12)（2）在⊿ABC中,角A,B,C所对的边分别为a,b,c,若f(C)=√3+1,且b^2=ac,求sinA的值](/uploads/image/z/5529735-63-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%5B2%E2%88%9A3cos%28x%2F2%29%2B2sin%28x%2F2%29%5Dcos%28x%2F2%29%EF%BC%881%EF%BC%89%E6%B1%82f%2817%CF%80%2F12%29%EF%BC%882%EF%BC%89%E5%9C%A8%E2%8A%BFABC%E4%B8%AD%2C%E8%A7%92A%2CB%2CC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2C%E8%8B%A5f%28C%29%3D%E2%88%9A3%2B1%2C%E4%B8%94b%5E2%3Dac%2C%E6%B1%82sinA%E7%9A%84%E5%80%BC)
已知函数f(x)=[2√3cos(x/2)+2sin(x/2)]cos(x/2)（1）求f(17π/12)（2）在⊿ABC中,角A,B,C所对的边分别为a,b,c,若f(C)=√3+1,且b^2=ac,求sinA的值
已知函数f(x)=[2√3cos(x/2)+2sin(x/2)]cos(x/2)
（1）求f(17π/12)
（2）在⊿ABC中,角A,B,C所对的边分别为a,b,c,若f(C)=√3+1,且b^2=ac,求sinA的值
已知函数f(x)=[2√3cos(x/2)+2sin(x/2)]cos(x/2)（1）求f(17π/12)（2）在⊿ABC中,角A,B,C所对的边分别为a,b,c,若f(C)=√3+1,且b^2=ac,求sinA的值
f(x)=[2√3cos(x/2)+2sin(x/2)]cos(x/2)
=2√3[cos(x/2)]^2+sinx
=√3{2[cos(x/2)]^2-1}+√3+sinx
=√3cosx+sinx+√3
=2(sinx*1/2+√3/2*cosx)+√3
=2sin(x+π/3)+√3
(1)求f(17π/12)=2sin(7π/4)+√3=√3-√2
(2)f(C)=√3+1得2sin(C+π/3)+√3=√3+1
故C=π/2,sinC=1,A+B=π/2,则sinB=cosA
且b^2=ac由正弦定理得
sinB^2=sinAsinC=sinA
代入的cosA^2=sinA
1-sinA^2=sinA
sinA^2+sinA-1=0
sinA=(√5-1)/2
1。先把[2√3cos(x/2)+2sin(x/2)]cos(x/2)转化为4[sin(π/3)cos(x/2)+cos(π/3)sin(x/2)]cos(x/2)
下面就利用两角和公式和积化和差公式就能算了。（提示）1/24+17/24=3/4 1/24-17/24=-2/3
2。将上式化为f(x)=√3(1+cosx)+sinx=2sin(x+π/3)+√3,知sin(x+π/...
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1。先把[2√3cos(x/2)+2sin(x/2)]cos(x/2)转化为4[sin(π/3)cos(x/2)+cos(π/3)sin(x/2)]cos(x/2)
下面就利用两角和公式和积化和差公式就能算了。（提示）1/24+17/24=3/4 1/24-17/24=-2/3
2。将上式化为f(x)=√3(1+cosx)+sinx=2sin(x+π/3)+√3,知sin(x+π/3)=1/2,x=π/3
可见ABC为直角三角形，有a^2+b^2=c^2,把b^2=ac带入，得到
(a/c)^2+(a/c)-1=0,故a/c=(√5-1)/2,由于a/c=sinA(直角三角形),故sinA=(√5-1)/2
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