# If x, y ∈ R, then the determinant

Question:

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

(a) $[-\sqrt{2}, \sqrt{2}]$

(b) $[-1,1]$

(c) $[-\sqrt{2}, 1]$

(d) $[-1,-\sqrt{2}]$

Solution:

$\Delta=\left|\begin{array}{ccc}\cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x+y) & -\sin (x+y) & 0\end{array}\right|$

$=\left|\begin{array}{ccc}\cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ 0 & 0 & \sin y-\cos y\end{array}\right|$    [Applying $R_{3} \rightarrow R_{3}-\cos y R_{1}+\sin y R_{2}$ ]

$=(\sin y-\cos y)\left(\cos ^{2} x+\sin ^{2} x\right)$

$=\sin y-\cos y$

$=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin y-\frac{1}{\sqrt{2}} \cos y\right)$

$=\sqrt{2}\left(\cos \frac{\pi}{4} \sin y-\sin \frac{\pi}{4} \cos y\right)$

$=\sqrt{2} \sin \left(y-\frac{\pi}{4}\right)$

Therefore, $-\sqrt{2} \leq \Delta \leq \sqrt{2}$.

Hence, the correct option is (a).