Question

The maximum value of ${\cos }^2({45}^{\circ }+x)+(\sin x-\cos x{)}^2$ is

Answer 1

Given expression ${\cos }^2({45}^{\circ }+x)+(\sin x-\cos x{)}^2$

$=\frac{1}{2}(1+\cos (90+2x\left)\right)+(\sin x-\cos x{)}^2$

$[\because cos2x=2co{s}^{2}x-1]$

$=\frac{1}{2}(1-\sin 2x)+({\sin }^{2}x-2sinx\cos x+co{s}^{2}x)$

$=\frac{1}{2}(1-\sin 2x)+1-2sinx\cos x$

$=\frac{1}{2}(1-\sin 2x)+1-sin2x$

$=\frac{3}{2}(1-\sin 2x)$
Now, since $-1\le \sin 2x\le 1$

$-1\le -\sin 2x\le 1$

$\Rightarrow 0\le 1-\sin 2x\le 2$

$\Rightarrow 0\le \frac{3}{2}(1-\sin 2x)\le 3$

Hence,maximum value is $3$