已知函数f(x)=2sinxcosx+cos2x(x属于R) 若θ为锐角,且f(θ+π/8)=3分之根号2,求tan2θ的值

问题描述:

已知函数f(x)=2sinxcosx+cos2x(x属于R) 若θ为锐角,且f(θ+π/8)=3分之根号2,求tan2θ的值
f(x)=sin2x+cos2x=√2sin(2x+π/4)
f(θ+π/8)=√2sin(2θ+π/4+π/4) ←为什么要化为这个?
=√2cos2θ
=√2/3
不是应该是f(θ+π/8)=√2sin(2θ+π/4+π/8)=√2/3

f(x)=2sinxcosx+cos2x
=sin2x+cos2x
=√2sin(2x+π/4)
∵f(θ+π/8)=√2/3
∴f(θ+π/8)=√2sin[2^(θ+π/8)+π/4]
=√2sin(2θ+π/2)
=√2cos2θ=√2/3
∴cos2θ=1/3
又 θ为锐角
所以 0