Question

数学题16题求解答

Answer 1

（1）f(x)=(1/2)sin2x-(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x+2cos²x-1
=sin2x+cos2x=√2sin(2x+π/4) 即f(x)最小正周期为π
（2）因为α∈[π/4，π/2]，所以2α+π/4∈[3π/4，,5π/4]，因为f(α)=3√2/5，即cos(2α+π/4)=-4/5
因为cos2α=cos[(2α+π/4)-π/4]=cos(2α+π/4)·cos(π/4)+sin(2α+π/4)·sin(π/4)
=(√2/2)(-4/5+3/5)=√2/10。

Answer 2

cos^2 (x) = (1 + cos (2x)) /2
sin (2 x - pi / 3) = - sin(pi /3 - 2x) = - cos (pi / 2 - pi / 3 + 2x) = - cos (2x + pi / 6)
We know f(x) = - cos (2x + pi / 6) + cos (2 x - pi / 6) + cos (2x)
= - 2 sin (2x) * sin (- pi / 6) + cos (2x)
= sin(2x) + cos (2x)
= 1/sin (Pi/4) * sin (2x + Pi/4)
(1) The minimal period is Pi
(2) Sin(2x) + cos(2x) = 3 * Sqrt (2)/ 5 => tan (x) = bla bla => cos (2x) = f(tan (x) ) => Solution

Answer 3

f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos²x
=根号下3sin2x/2+cos2x/2-cos2x/2+根号下3sin2x/2+2cos²x-1+1
=根号下3sin2x+cos2x+1
=2(根号下3sin2x/2+cos2x/2)+1
=2sin(2x+π/6)+1
f(π/12)=2sin(π/3)+1=根号下3+1

Answer 4

原式=sin(2x-∏/3)+cos[∏/2-(∏/3-2x)]+2*(1+cos2x)/2-1
=sin(2x-∏/3)-sin(2x-∏/3)+1+cos2x-1
=cos2x
T=∏

Answer 5

f(x)=(1/2)sin2x-(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x+2cos²x-1
=sin2x+cos2x=√2sin(2x+π/4)。所以t为pi