Question

If $0<x<\frac{\pi }{4}$ and $cosx+sinx=\frac{5}{4}$, then the value of $16(cosx-sinx{)}^2$ is

Answer 1

$\cos x+\sin x=\frac{5}{4}\Rightarrow {(\cos x+\sin x)}^2=\frac{25}{16}$
$\Rightarrow {\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x=\frac{25}{16}$
Now ${(\cos x-\sin x)}^2={(\cos x+\sin x)}^2-4\sin x\cos x$
$=\frac{25}{16}-2\left(\frac{9}{16}\right)=\frac{7}{16}$
$\therefore 16{(\cos x-\sin x)}^2=7$