已知函数F(x)=2sin(2x+π/6)-1,当cos(β-α)=4/5,cos(β+α)=-4/5(0<α<β≤π/2)时,求F(β)的值?

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已知函数F(x)=2sin(2x+π/6)-1,当cos(β-α)=4/5,cos(β+α)=-4/5(0<α<β≤π/2)时,求F(β)的值?

已知函数F(x)=2sin(2x+π/6)-1,当cos(β-α)=4/5,cos(β+α)=-4/5(0<α<β≤π/2)时,求F(β)的值?
已知函数F(x)=2sin(2x+π/6)-1,当cos(β-α)=4/5,cos(β+α)=-4/5(0<α<β≤π/2)时,求F(β)的值?

已知函数F(x)=2sin(2x+π/6)-1,当cos(β-α)=4/5,cos(β+α)=-4/5(0<α<β≤π/2)时,求F(β)的值?
根据cos(β-α)=4/5,cos(β+α)=-4/5(0<α<β≤π/2),得β=π/2;
F(x)=2sin(2x+π/6)-1,所以F(β)=2sin(2β+π/6)-1
F(π/2)=2sin(2π/2+π/6)-1=2sin(π+π/6)-1=-2

F(β)=2sin(2β+π/6)-1,2β=β+α+β-α