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
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Hint: For answering this question we will expand this determinant and simplify it by using the below formula
$ \begin{align}
& \cos 2x=1-2{{\sin }^{2}}x \\
& {{\cos }^{2}}x=1-{{\sin }^{2}}x \\
\end{align} $
From these two we can get
$ \begin{align}
& \cos 4x=2\left( 1-2{{\sin }^{2}}x \right)-1 \\
& {{\cos }^{4}}x={{\left( 1-{{\sin }^{2}}x \right)}^{2}} \\
\end{align} $
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{align}
& \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{align} $
By using $ {{\cos }^{2}}x=1-{{\sin }^{2}}x $ identity, we will get
$ \begin{align}
& \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{align} $
Adding up similar terms, we will get
$ \begin{align}
& \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{align} $
By using $ {{\cos }^{4}}x={{\left( 1-{{\sin }^{2}}x \right)}^{2}} $ , we will get
$ \begin{align}
& \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{align} $
By using $ \cos 4x=2\left( 1-2{{\sin }^{2}}x \right)-1 $ , we will get
$ \begin{align}
& \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{align} $
By simplifying $ 2\left( 1-2{{\sin }^{2}}x \right)-1=1-4{{\sin }^{2}}x $ , we will get
$ \begin{align}
& \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{align} $
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{align}
& \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{align} $
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{align}
& \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{align} $
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.
$ \begin{align}
& \cos 2x=1-2{{\sin }^{2}}x \\
& {{\cos }^{2}}x=1-{{\sin }^{2}}x \\
\end{align} $
From these two we can get
$ \begin{align}
& \cos 4x=2\left( 1-2{{\sin }^{2}}x \right)-1 \\
& {{\cos }^{4}}x={{\left( 1-{{\sin }^{2}}x \right)}^{2}} \\
\end{align} $
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{align}
& \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{align} $
By using $ {{\cos }^{2}}x=1-{{\sin }^{2}}x $ identity, we will get
$ \begin{align}
& \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{align} $
Adding up similar terms, we will get
$ \begin{align}
& \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{align} $
By using $ {{\cos }^{4}}x={{\left( 1-{{\sin }^{2}}x \right)}^{2}} $ , we will get
$ \begin{align}
& \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{align} $
By using $ \cos 4x=2\left( 1-2{{\sin }^{2}}x \right)-1 $ , we will get
$ \begin{align}
& \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{align} $
By simplifying $ 2\left( 1-2{{\sin }^{2}}x \right)-1=1-4{{\sin }^{2}}x $ , we will get
$ \begin{align}
& \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{align} $
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{align}
& \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{align} $
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{align}
& \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{align} $
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.
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