Question

If $\sin x+\cos x=\sqrt{y+\frac{1}{y}},x\in [0,\pi ]$ then

• A. $x=\frac{\pi }{4}$
• B. $y=0$
• C. $y=1$
• D. $x=\frac{3\pi }{4}$

Answer 1

Answer: (A), (C)

The correct options are
A $x=\frac{\pi }{4}$
C $y=1$

$\sin x+\cos x=\sqrt{y+\frac{1}{y}}$

$\frac{2}{\sqrt{2}}(\sin x+\cos x)=\sqrt{(\sqrt{y}{)}^2+(\frac{1}{\sqrt{y}}{)}^2}$

$\sqrt{2}(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)=\sqrt{(\sqrt{y}-\frac{1}{\sqrt{y}}{)}^2+2\sqrt{y}\times \frac{1}{\sqrt{y}}}$

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

$\sqrt{2}\sin (x+\frac{\pi }{4})=\sqrt{(\sqrt{y}-\frac{1}{\sqrt{y}}{)}^2+2}$

Now, LHS =$\sqrt{2}\sin (x+\frac{\pi }{4})$

$-1\le \sin (x+\frac{\pi }{4})\le 1$

$-\sqrt{2}\le \sin (x+\frac{\pi }{4})\le \sqrt{2}$

RHS = $\sqrt{(\sqrt{y}-\frac{1}{\sqrt{y}}{)}^2+2}$

$0\le (\sqrt{y}-\frac{1}{\sqrt{y}}{)}^2<\infty$

$2\le (\sqrt{y}-\frac{1}{\sqrt{y}}{)}^2+2<\infty$

$\sqrt{2}\le \sqrt{(\sqrt{y}-\frac{1}{\sqrt{y}}{)}^2+2}<\infty$

LHS = RHS = $\sqrt{2}$

$\sqrt{2}\sin (x+\frac{\pi }{4})=\sqrt{2}$

$\sin (x+\frac{\pi }{4})=\sin \frac{\pi }{2}$

$x+\frac{\pi }{4}=\frac{\pi }{2}$

$x=\frac{\pi }{4}$

RHS = $\sqrt{(\sqrt{y}-\frac{1}{\sqrt{y}}{)}^2+2}=\sqrt{2}$

$\sqrt{y}-\frac{1}{\sqrt{y}}=0$

$y=1$

Hence, option A and C are correct.