Question

已知函数f(x)=sin2x+con2x,最小正周期是π,若f(α/2+π/8)=3√2/5,求

Answer 1

f(x)=sin2x+cos2x.

f(x) √2sin(2x+π/4).

f(α/2+π/8)=√2sin[2(α/2+/8+)π/4]=3√2/5.

sin(α+π/4+π/4)=3/5.

sin(α+π/2)=3/5.

cosα=3/5.

cos2α＝2ccos^2α-1.

=2*(3/5)^2-1.

∴cos2α=-7/25.