Question

If $x,y\in R$, then the determinant $\left|\begin{array}{ccc}\cos x& -\sin x& 1\\ \sin x& \cos x& 1\\ \cos (x+y)& -\sin (x+y)& 0\end{array}\right|$ lie lies in the interval

• A. $[-\sqrt{2},\sqrt{2}]$
• B. $[-1,1]$
• C. $[-\sqrt{2},1]$
• D. $[-1,-\sqrt{2}]$

Answer 1

Answer: (A)

The given determinant is
$\left|\begin{array}{ccc}\cos x& -\sin x& 1\\ \sin x& \cos x& 1\\ \cos (x+y)& -\sin (x+y)& 0\end{array}\right|$
Applying $R_3\to R_3-\cos y{R}_1+\sin y{R}_2$, we get
$\Delta =\left|\begin{array}{ccc}\cos x& -\sin x& 1\\ \sin x& \cos x& 1\\ 0& 0& \sin y-\cos y\end{array}\right|$
By expanding along $R_3$, we have
$\Delta =(\sin y-\cos y)\left({\cos }^{2}x+{\sin }^{2}x\right)$
$=(\sin y-\cos y)=\sqrt{2}\left[\frac{1}{\sqrt{2}}\sin y-\frac{1}{\sqrt{2}}\cos y\right]$
$=\sqrt{2}\left[\cos \frac{\pi }{4}\sin y-\sin \frac{\pi }{4}\cos y\right]=\sqrt{2}\sin \left(y-\frac{\pi }{4}\right)$
Hence, $-\sqrt{2}\le \Delta \le \sqrt{2}$