Y=Cos²(45+x)+(sinx-cosx)^{2 find the max and min value of y}

^{with all the steps please}

y=cos^2(45+x)+(sin x-cos x)^2.

We kn ow that:-

2.cos ^2A-1=cos2A

or cos^2.A=(1+cos2A)/2

or cos^2.(45+x) =[1+cos(90+2x)]^2=(1-sin2x)/2.

Now y=(1-sin2x)/2+sin^2x+cos^2x-2.sin x.cos x.

y=(1-sin2x)/2+(1-sin2x)=3/2.(1-sin2x)

y= 3/2 -3/2.sin2x.

dy/dx= (-3/2).cos2x.(2) = -3cos2x

For maxima or minima put dy/dx=0

-3cos2x =0

cos2x =0 => 2x =90

x=45°

d2y/dx^2= 6 sin2x

d2y/dx^2 at x=45° is +ve , therefore there exist minima at x= 45°.

Minimum value y=3/2(1-sin90°) =3(1–1) =0

while maximum value = 3