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已知函数f(x)=2√3sinxcosx+2(cosx)^2 -1 若f(a)=6/5 a∈[π/4,π/2],求cos

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已知函数f(x)=2√3sinxcosx+2(cosx)^2 -1 若f(a)=6/5 a∈[π/4,π/2],求cos2a的值
已知函数f(x)=2√3sinxcosx+2(cosx)^2 -1 若f(a)=6/5 a∈[π/4,π/2],求cos
f(x)=2√3sinxcosx+2(cosx)^2 -1
=√3*sin2x+cos2x
=2sin(2x+π/6)
f(a)=2sin(2a+π/6)=6/5
所以sin(2a+π/6)=3/5
因为 a∈[π/4,π/2],所以 2a+π/6∈[2π/3,7π/6] 所以cos(2a+π/6)=-4/5
cos2a=cos(2a+π/6-π/6)=cos(2a+π/6)*√3/2+sin(2a+π/6)*1/2
=-4/5*√3/2+3/5*1/2=-2√3/5+3/10
所以cos2a=-2√3/5+3/10