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Y=Cos 2 (45+x)+(sinx-cosx) 2 find the max and min value of y with all the steps please

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2 years ago

```							Dear Avni y=cos^2(45+x)+(sin x-cos x)^2.We kn ow that:-2.cos ^2A-1=cos2Aor cos^2.A=(1+cos2A)/2or 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) = -3cos2xFor maxima or minima put dy/dx=0-3cos2x =0cos2x =0 => 2x =90x=45°d2y/dx^2= 6 sin2xd2y/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
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2 years ago
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