Question

If $0<x<\frac{\pi }{4}$ and $\cos x+\sin x=\frac{5}{4}$, the value of $16(\cos x-\sin x{)}^2$ is _______.

Answer 1

We have,

$\cos x+\sin x=\frac{5}{4}$ ……. (1) $0<x<\frac{\pi }{4}$

Squaring both side and we get,

${(\cos x+\sin x)}^2={\left(\frac{5}{4}\right)}^2$

${\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x=\frac{25}{16}$

$1+\sin 2x=\frac{25}{16}$

$\sin 2x=\frac{25}{16}-1$

$\sin 2x=\frac{25-16}{16}$

$\sin 2x=\frac{9}{16}$

Value of

$16{(\cos x-\sin x)}^2$

$=16({\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x)$

$=16(1-\sin 2x)$

$=16(1-\frac{9}{16})\therefore \sin 2x=\frac{9}{16}$

$=16\times \frac{(16-9)}{16}$

$=7$

Hence, this is the answer