Question

If $\sin x+\cos x=\sqrt{y^2+\frac{1}{y^2}}$ for $x\in [0,\pi ]$ has a solution, then which of the following is/are correct?

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

Answer 1

Answer: (A), (B), (C)

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

R.H.S.=
$y^2+\frac{1}{y^2}\ge 2\sqrt{y^2\times \frac{1}{y^2}}\Rightarrow y^2+\frac{1}{y^2}\ge 2\Rightarrow \sqrt{y^2+\frac{1}{y^2}}\ge \sqrt{2}$

$L.H.S.=\sin x+\cos x=\sqrt{2}\sin (\frac{\pi }{4}+x)\Rightarrow \sin x+\cos x\le \sqrt{2}$

So, the equality holds only when
$y^2+\frac{1}{y^2}=2\Rightarrow y=\pm 1$ and
$\sin x+\cos x=\sqrt{2}$
$\Rightarrow \sin (x+\frac{\pi }{4})=1\Rightarrow x+\frac{\pi }{4}=\frac{\pi }{2},\frac{5\pi }{2}\Rightarrow x=\frac{\pi }{4},\frac{9\pi }{4},\cdots \therefore x=\frac{\pi }{4}$