已知函数f(x)=[2√3cos(x/2)+2sin(x/2)]cos(x/2)
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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
(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
=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
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