Questions about trigonometric functions (F.4)

1) (sin^2)x +3(cos^2)x-2
= 1-(cos^2)x +3(cos^2)x-2
= 2(cos^2)x - 1
The maximum value of (cos^2)x = 1
So The maximum value of 2(cos^2)x - 1 = 2 x 1 - 1 = 1

2) Let y=x/2.
2sin(x/2) +(cos^2)(x/2)
= 2 sin y + (cos^2) y
= 2 sin y + 1 - (sin^2) y
= -(sin y - 1)^2 + 2
The minimum value of (sin y - 1)^2 = 0 ( when sin y = 1)
So The maximum value of -(sin y - 1)^2 + 2 = -0+2 = 2

1)Find the maximum value of sin2x +3cos2x-2.

sin2x +3cos2x-2=sin2x+3(1-sin2x)-2
=sin2x+3-3sin2x-2
=-2sin2x+1
the maximum value =-2(0)+1
=1
2)Find the maximum value of 2sin(x/2) +cos2(x/2).
2sin(x/2) +cos2(x/2)=2sin(x/2)+(1-sin2(x/2)]
=-sin2(x/2)+2sin(x/2)+1
=-[sin2(x/2)-2sin(x/2)+1]+2
=-[sin(x/2)-1]2+2
the maximum value=-(0)2+2
=2