Question

If $y=co{s}^2({45}^o+x)(sinx-cosx{)}^2$ then the maximum value of y is:

Answer 1

$y={\cos }^2(45°+x){(\sin x-\cos x)}^2$

Expanding $\cos (45°+x)$

$y={[\cos 45°\cos x-\sin 45°\times \sin x]}^2{(\sin x-\cos x)}^2$

$y={\left[\frac{1}{\sqrt{2}}(\sin x-\cos x)\right]}^2{(\sin x-\cos x)}^2$

$y=\frac{1}{2}{(\sin x-\cos x)}^4\to 1$

Let $z=\sin x-\cos x$

For find critical points

$\frac{dz}{dx}=0$

$\cos x+\sin x=0$

$\tan x=-1$

$x=135°$

Putting this in z

$z=\sin \left(135°\right)-\cos \left(135°\right)$

$z=\frac{1}{\sqrt{2}}-(-\frac{1}{\sqrt{2}})$

$z=\sqrt{2}$

Putting this value of z in equation $1$

$y=\frac{1}{2}\times {\left(\sqrt{2}\right)}^4$

$y=\frac{1}{2}\times 4$

$y=2$ ( maximum value )