It can be concluded from the figure that A= 1, T/4=π/3-π/ 12=π/4.
∴T=π，
ω= 2π/T = 2π/π= 2；
∴f(x)=sin(2x+φ).
And f(x)=sin2(x+φ/2), φ/2- initial phase angle.
From the picture: φ/2+π/ 12=π/4.
φ/2=π/6.
f(x)=sin2(x+π/6)
(1) ∴ f(x) = sin (2x+π/3)-that is, the expression of the function f (x);
(2) If f (α+π/ 12) = 1/3. (α ∈ (0, π/2), find the value of tanα.
sin[2(α+π/ 12)+π/3]= 1/3。
sin(2α+π/6+π/3)= 1/3。
sin(2α+π/2)= 1/3。
cos2α= 1/3。
2cos^2- 1= 1/3.
2cos^2=4/3.
cos^2α=2/3.
1/sec^2=2/3.
∴sec^2α=3/2.
tan^2α=sec^2- 1.
=3/2- 1.
tan^2α= 1/2.
∴ tanα=√2/2。 [0