(1)
f(x)
=b.c
= (m,sin2x).(cos2x,n)
= mcos2x+ nsin2x
f(0) = m = 1
f(π/4) = n = 1
(2)
f(x)= cos2x+sin2x
= √2(sin(2x+π/4))
最小正周期 = π
min f(x) = f(0) = 1
(3)
f(α/2)=1/5
cosα+sinα = 1/5
(5cosα)^2 = (1- 5sinα)^2
25(sinα)^2 -5sinα-12 =0
(5sinα+3)(5sinα-4)=0
sinα =4/5 or sinα = -3/5 ( rejected )
ie sinα=4/5