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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
人气:112 ℃ 时间:2019-09-29 04:31:14
解答
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...
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