已知函数f(x)=cos^2x/2-sin^2x/2+sinx.求:(2)当x属于(0,π/4)且f(x0)=4根号2/5时,
问题描述:
已知函数f(x)=cos^2x/2-sin^2x/2+sinx.求:(2)当x属于(0,π/4)且f(x0)=4根号2/5时,
f(x0+π/6)的值.
答
原式=cosx+sinx
=根号2*sin(x+π/4)
f(x0)=4根号2/5,即
sin(x0+π/4)=4/5
∵x∈(0,π/4),
∴(x0+π/4)∈(π/4,π/2),
sin(x0+π/4)=4/5
cos(x0+π/4)=3/5
f(x0+π/6)
=根号2*sin(x0+π/4+π/6)
=根号2*[(4/5*0.5根号3)+3/5*0.5]
=(4*根号6+3*根号2)/10
≈1.404