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)=sin2x+cos2x=√2sin(2x+π/4)
∴f(θ+π/8)=√2sin[2(θ+π/8)+π/4]
=√2sin(2θ+π/4+π/4)
=√2sin(2θ+π/2)
然后可使用诱导公式
sin(θ+π/2)=cosθ
∴f(θ+π/8)=√2sin(2θ+π/2)
=√2cos2θ=√2/3
解得tan2θ=2√2

f(x) = sin2x + cos2x = sqrt(2)sin(2x+pi/4)
f(θ+π/8) = sqrt(2) sin(2θ + pi/4+pi/4) = sqrt(2)cos(2θ) = sqrt(2)/3
cos(2θ) = 1/3
sin(2θ ) = sqrt(8)/3
tan(2θ) = sqrt(8)

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