Question

If the determinant $\left| \begin{matrix} \cos 2x & {{\sin }^{2}}x & \cos 4x \\\ {{\sin }^{2}}x & \cos 2x & {{\cos }^{2}}x \\\ \cos 4x & {{\cos }^{2}}x & \cos 2x \\\ \end{matrix} \right|$ is expanded in the powers of $\sin x$ then the negative of the constant term in the expansion is___________.
(A) 1
(B) -1
(C) -2
(D) 0

Answer 1

For answering this question we will expand this determinant and simplify it by using the below formula
$\begin{aligned} & \cos 2x=1-2{{\sin }^{2}}x \\\ & {{\cos }^{2}}x=1-{{\sin }^{2}}x \\\ \end{aligned}$
From these two we can get
$\begin{aligned} & \cos 4x=2\left( 1-2{{\sin }^{2}}x \right)-1 \\\ & {{\cos }^{4}}x={{\left( 1-{{\sin }^{2}}x \right)}^{2}} \\\ \end{aligned}$
The minor of ${{a}_{ij}}$ is represented by ${{M}_{ij}}$ and for example for any $3\times 3$ matrix the minor of ${{a}_{21}}$ is represented by ${{M}_{21}}$ and is given by $\left| \begin{matrix} {{a}_{12}} & {{a}_{13}} \\\ {{a}_{32}} & {{a}_{23}} \\\ \end{matrix} \right|$ . The cofactor of ${{a}_{ij}}$ is represented by ${{C}_{ij}}$ is given by ${{C}_{ij}}={{\left( -1 \right)}^{i+j}}{{M}_{ij}}$ .The determinant of any $3\times 3$ matrix is given by $={{a}_{11}}{{C}_{11}}+{{a}_{12}}{{C}_{12}}+{{a}_{13}}{{C}_{13}}.$

Complete step-by-step answer:
Let us get started by expanding the determinant
$\left| \begin{matrix} \cos 2x & {{\sin }^{2}}x & \cos 4x \\\ {{\sin }^{2}}x & \cos 2x & {{\cos }^{2}}x \\\ \cos 4x & {{\cos }^{2}}x & \cos 2x \\\ \end{matrix} \right|=\cos 2x\left( {{\cos }^{2}}2x-{{\cos }^{4}}x \right)-{{\sin }^{2}}x\left( \sin {{x}^{2}}\cos 2x-{{\cos }^{2}}x\cos 4x \right)+\cos 4x\left( {{\sin }^{2}}x{{\cos }^{2}}x-\cos 2x\cos 4x \right)$ $\Rightarrow {{\cos }^{3}}2x-\cos 2x{{\cos }^{4}}x-{{\sin }^{4}}x\cos 2x+{{\sin }^{2}}x{{\cos }^{2}}x\cos 4x+\cos 4x{{\sin }^{2}}x{{\cos }^{2}}x-{{\cos }^{2}}4x\cos 2x$
By using $\cos 2x=1-2{{\sin }^{2}}x$ identity, we will get
$\begin{aligned} & \Rightarrow {{\left( 1-2{{\sin }^{2}}x \right)}^{3}}-\left( 1-2{{\sin }^{2}}x \right){{\cos }^{4}}x-{{\sin }^{4}}x\left( 1-2{{\sin }^{2}}x \right)+{{\sin }^{2}}x{{\cos }^{2}}x\cos 4x+\cos 4x{{\sin }^{2}}x{{\cos }^{2}}x- \\\ & {{\cos }^{2}}4x\left( 1-2{{\sin }^{2}}x \right) \\\ \end{aligned}$
By using ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ identity, we will get
$\begin{aligned} & \Rightarrow {{\left( 1-2{{\sin }^{2}}x \right)}^{3}}-\left( 1-2{{\sin }^{2}}x \right){{\cos }^{4}}x-{{\sin }^{4}}x\left( 1-2{{\sin }^{2}}x \right)+{{\sin }^{2}}x\left( 1-{{\sin }^{2}}x \right)\cos 4x+ \\\ & \cos 4x{{\sin }^{2}}x\left( 1-{{\sin }^{2}}x \right)-{{\cos }^{2}}4x\left( 1-2{{\sin }^{2}}x \right) \\\ \end{aligned}$
Adding up similar terms, we will get
$\begin{aligned} & \Rightarrow {{\left( 1-2{{\sin }^{2}}x \right)}^{3}}-\left( 1-2{{\sin }^{2}}x \right){{\cos }^{4}}x-{{\sin }^{4}}x\left( 1-2{{\sin }^{2}}x \right)+2{{\sin }^{2}}x\left( 1-{{\sin }^{2}}x \right)\cos 4x \\\ & -{{\cos }^{2}}4x\left( 1-2{{\sin }^{2}}x \right) \\\ \end{aligned}$
By using ${{\cos }^{4}}x={{\left( 1-{{\sin }^{2}}x \right)}^{2}}$ , we will get
$\begin{aligned} & \Rightarrow {{\left( 1-2{{\sin }^{2}}x \right)}^{3}}-\left( 1-2{{\sin }^{2}}x \right){{\left( 1-{{\sin }^{2}}x \right)}^{2}}-{{\sin }^{4}}x\left( 1-2{{\sin }^{2}}x \right)+2{{\sin }^{2}}x\left( 1-{{\sin }^{2}}x \right)\cos 4x \\\ & -{{\cos }^{2}}4x\left( 1-2{{\sin }^{2}}x \right) \\\ \end{aligned}$
By using $\cos 4x=2\left( 1-2{{\sin }^{2}}x \right)-1$ , we will get
$\begin{aligned} & \Rightarrow {{\left( 1-2{{\sin }^{2}}x \right)}^{3}}-\left( 1-2{{\sin }^{2}}x \right){{\left( 1-{{\sin }^{2}}x \right)}^{2}}-{{\sin }^{4}}x\left( 1-2{{\sin }^{2}}x \right)+2{{\sin }^{2}}x\left( 1-{{\sin }^{2}}x \right)\left( 2\left( 1-2{{\sin }^{2}}x \right)-1 \right) \\\ & -{{\left( 2\left( 1-2{{\sin }^{2}}x \right)-1 \right)}^{2}}\left( 1-2{{\sin }^{2}}x \right) \\\ \end{aligned}$
By simplifying $2\left( 1-2{{\sin }^{2}}x \right)-1=1-4{{\sin }^{2}}x$ , we will get
$\begin{aligned} & \Rightarrow {{\left( 1-2{{\sin }^{2}}x \right)}^{3}}-\left( 1-2{{\sin }^{2}}x \right){{\left( 1-{{\sin }^{2}}x \right)}^{2}}-{{\sin }^{4}}x\left( 1-2{{\sin }^{2}}x \right)+2{{\sin }^{2}}x\left( 1-{{\sin }^{2}}x \right)\left( 1-4{{\sin }^{2}}x \right) \\\ & -{{\left( 1-4{{\sin }^{2}}x \right)}^{2}}\left( 1-2{{\sin }^{2}}x \right) \\\ \end{aligned}$
Here if we observe that now we have all terms as the powers of $\sin x$ now we need to find the value of the constant term in the expansion for that if we assume $x=0$ then all other terms containing the powers of $\sin x$ will become zero.
Let us do that
We know that $\sin 0=0$ so by substituting that we will get
$\begin{aligned} & \Rightarrow {{\left( 1-0 \right)}^{3}}-\left( 1-0 \right){{\left( 1-0 \right)}^{2}}-0\left( 1-0 \right)+2.0\left( 1-0 \right)\left( 1-0 \right)-{{\left( 1-0 \right)}^{2}}\left( 1-0 \right) \\\ & \Rightarrow 1-1-1 \\\ & \Rightarrow -1 \\\ \end{aligned}$
Here we need a negative of the constant term of the expansion of the determinant that is 1.

So, the correct answer is “Option A”.

Note: We can also answer this question in other way as it is given that the determinant is expanded in the powers of $\sin x$ we can assume that expansion to be
$\left| \begin{matrix} \cos 2x & {{\sin }^{2}}x & \cos 4x \\\ {{\sin }^{2}}x & \cos 2x & {{\cos }^{2}}x \\\ \cos 4x & {{\cos }^{2}}x & \cos 2x \\\ \end{matrix} \right|={{a}_{0}}+{{a}_{1}}\sin x+{{a}_{2}}{{\sin }^{2}}x+..........and\text{ so on}$
Here if we observe this now we have all terms as the powers of $\sin x$ Now we need to find the value of the constant term in the expansion for that if we assume $x=0$ then all other terms containing the powers of $\sin x$ will become zero. Let us apply the value of $x$ in the determinant.
$\left| \begin{matrix} \cos 0 & {{\sin }^{2}}0 & \cos 0 \\\ {{\sin }^{2}}0 & \cos 0 & {{\cos }^{2}}0 \\\ \cos 0 & {{\cos }^{2}}0 & \cos 0 \\\ \end{matrix} \right|={{a}_{0}}$
Since we know that $\sin 0=0$ and $\cos 0=1$ we will have
$\left| \begin{matrix} 1 & 0 & 1 \\\ 0 & 1 & 1 \\\ 1 & 1 & 1 \\\ \end{matrix} \right|={{a}_{0}}$
By expanding the above determinant we will have
$\begin{aligned} & \left| \begin{matrix} 1 & 0 & 1 \\\ 0 & 1 & 1 \\\ 1 & 1 & 1 \\\ \end{matrix} \right|={{a}_{0}} \\\ & \Rightarrow 1\left( 1-1 \right)-0\left( 0-1 \right)+1\left( 0-1 \right)={{a}_{0}} \\\ & \Rightarrow -1={{a}_{0}} \\\ \end{aligned}$
As we need the negative of the constant term that is negative of ${{a}_{0}}$ that is 1.
Hence we end up with a conclusion saying that option (A) is correct.