Question

If the determinant $\left|\begin{array}{ccc}1+{\sin }^2\theta & {\sin }^2\theta & {\sin }^2\theta \\ {\cos }^2\theta & 1+{\cos }^2\theta & {\cos }^2\theta \\ 4\sin 4\theta & 4\sin 4\theta & 1+4\sin 4\theta \end{array}\right|=0$,
then $\sin 4\theta$ equals to
$(\text{where}\theta \in [0,2\pi \left]\right)$

• A. $\frac{1}{2}$
• B. $1$
• C. $-\frac{1}{2}$
• D. $-1$

Answer 1

The correct option is C $-\frac{1}{2}$

Let$A=\left|\begin{array}{ccc}1+{\sin }^2\theta & {\sin }^2\theta & {\sin }^2\theta \\ {\cos }^2\theta & 1+{\cos }^2\theta & {\cos }^2\theta \\ 4\sin 4\theta & 4\sin 4\theta & 1+4\sin 4\theta \end{array}\right|=0$
Applying, $R_1\to R_1+R_2$
$\left|\begin{array}{ccc}2& 2& 1\\ {\cos }^2\theta & 1+{\cos }^2\theta & {\cos }^2\theta \\ 4\sin 4\theta & 4\sin 4\theta & 1+4\sin 4\theta \end{array}\right|=0$
Applying, $C_1\to C_1-C_3,C_2\to C_2-C_3$
$\left|\begin{array}{ccc}1& 1& 1\\ 0& 1& {\cos }^2\theta \\ -1& -1& 1+4\sin 4\theta \end{array}\right|=0$
$\Rightarrow 2+4\sin 4\theta =0\Rightarrow \sin 4\theta =-\frac{1}{2}$