Question

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

(a) $\left[-\sqrt{2},\sqrt{2}\right]$
(b) $\left[-1,1\right]$
(c) $\left[-\sqrt{2},1\right]$
(d) $\left[-1,-\sqrt{2}\right]$

Answer 1

$$\begin{gathered} ∆=\left|\begin{array}{ccc}\cos x& -\sin x& 1\\ \sin x& \cos x& 1\\ \cos \left(x+y\right)& -\sin \left(x+y\right)& 0\end{array}\right| \\ =\left|\begin{array}{ccc}\cos x& -\sin x& 1\\ \sin x& \cos x& 1\\ 0& 0& \sin y-\cos y\end{array}\right|\left[\mathrm{Applying}{R}_3\to R_3-\cos y{R}_1+\sin y{R}_2\right] \\ =\left(\sin y-\cos y\right)\left({\cos }^{2}x+{\sin }^{2}x\right) \\ =\sin y-\cos y \\ =\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin y-\frac{1}{\sqrt{2}}\cos y\right) \\ =\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{4}\sin y-\sin \frac{\mathrm{\pi }}{4}\cos y\right) \\ =\sqrt{2}\sin \left(y-\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{Therefore},-\sqrt{2}\le ∆\le \sqrt{2}. \end{gathered}$$

Hence, the correct option is (a).