Question

If $\theta -\phi =\dfrac{\pi }{2}$ , then show that $\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\\ \end{matrix} \right]=0$ .

Answer 1

First, we rewrite the expression $\theta -\phi =\dfrac{\pi }{2}$ as $\phi =\theta -\dfrac{\pi }{2}$ . After that, using the two trigonometric identities $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ , we develop two expressions, which are $\sin \phi =-\cos \theta$ and $\cos \phi =\sin \theta$ . After that, we rewrite the matrix as,
$\Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\\ -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\\ \end{matrix} \right]$
We now multiply the two matrices to get $\left[ \begin{matrix} {{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta & -{{\cos }^{3}}\theta \sin \theta +\sin \theta {{\cos }^{3}}\theta \\\ \cos \theta {{\sin }^{3}}\theta -{{\sin }^{3}}\theta \cos \theta & -{{\cos }^{2}}\theta {{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\\ \end{matrix} \right]$ . Ruling out the like terms, we can arrive at the conclusion.

Complete step by step answer:
The condition that we are given is,
$\theta -\phi =\dfrac{\pi }{2}$
By taking the $\phi$ to the RHS, and $\dfrac{\pi }{2}$ to the LHS, we can write it as,
$\phi =\theta -\dfrac{\pi }{2}$
Now, we know the trigonometric identities, which are,
$\begin{aligned} & \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B \\\ & \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B \\\ \end{aligned}$
Using these trigonometric identities, we can show that,
$\begin{aligned} & \Rightarrow \sin \phi =\sin \left( \theta -\dfrac{\pi }{2} \right)=\sin \theta \cos \left( -\dfrac{\pi }{2} \right)-\cos \theta \sin \left( -\dfrac{\pi }{2} \right) \\\ & \Rightarrow \sin \phi =-\cos \theta \\\ \end{aligned}$
And that,
$\begin{aligned} & \Rightarrow \cos \phi =\cos \left( \theta -\dfrac{\pi }{2} \right)=\cos \theta \cos \left( -\dfrac{\pi }{2} \right)-\sin \theta \sin \left( -\dfrac{\pi }{2} \right) \\\ & \Rightarrow \cos \phi =\sin \theta \\\ \end{aligned}$
The matrix that we need to simplify and solve is,
$\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\\ \end{matrix} \right]$
Using the above expressions that we have derived of $\sin \phi$ and $\cos \phi$ , we can rewrite the matrix as,
$\Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\\ -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\\ \end{matrix} \right]$
Now, we know the multiplication of two matrices of order $2\times 2$ go as,
$\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\left[ \begin{matrix} e & f \\\ g & h \\\ \end{matrix} \right]=\left[ \begin{matrix} ae+bg & af+bh \\\ ce+dg & cf+dh \\\ \end{matrix} \right]$
So, multiplying the above matrices, we get,

& \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\\ -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} {{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\left( \cos \theta \sin \theta \right)}^{2}} & -{{\cos }^{2}}\theta \sin \theta \cos \theta +\cos \theta \sin \theta {{\cos }^{2}}\theta \\\ \cos \theta \sin \theta {{\sin }^{2}}\theta -{{\sin }^{2}}\theta \sin \theta \cos \theta & -{{\left( \cos \theta \sin \theta \right)}^{2}}{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\\ \end{matrix} \right] \\\ \end{aligned}$$ Simplifying the above matrix elements, we get, $$\begin{aligned} & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\\ -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} {{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta & -{{\cos }^{3}}\theta \sin \theta +\sin \theta {{\cos }^{3}}\theta \\\ \cos \theta {{\sin }^{3}}\theta -{{\sin }^{3}}\theta \cos \theta & -{{\cos }^{2}}\theta {{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\\ \end{matrix} \right] \\\ \end{aligned}$$ Ruling out the like terms in the element spaces, we get, $$\begin{aligned} & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\\ -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} 0 & 0 \\\ 0 & 0 \\\ \end{matrix} \right] \\\ \end{aligned}$$ Now, we know that the matrix $$\left[ \begin{matrix} 0 & 0 \\\ 0 & 0 \\\ \end{matrix} \right]$$ is called a null matrix or a zero matrix, and can also be represented by the digit $0$ as a short form. Thus, we have proved that the matrix $\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\\ \end{matrix} \right]=0$ if $\theta -\phi =\dfrac{\pi }{2}$ . **Note:** We can also solve the problem in a similar way, but now we replace $\theta $ with $2\theta $ . To do this, we rewrite the matrix as $\begin{aligned} & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\\ \end{matrix} \right] \\\ & =\dfrac{1}{4}\left[ \begin{matrix} 2{{\cos }^{2}}\theta & 2\cos \theta \sin \theta \\\ 2\cos \theta \sin \theta & 2{{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} 2{{\cos }^{2}}\phi & 2\cos \phi \sin \phi \\\ 2\cos \phi \sin \phi & 2{{\sin }^{2}}\phi \\\ \end{matrix} \right] \\\ \end{aligned}$ Now, we know $\begin{aligned} & 2{{\cos }^{2}}\theta =1+\cos 2\theta \\\ & 2{{\sin }^{2}}\theta =1-\cos 2\theta \\\ & 2\sin \theta \cos \theta =\sin 2\theta \\\ \end{aligned}$ Using these, we get, $\begin{aligned} & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right]\left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\\ \end{matrix} \right] \\\ & =\dfrac{1}{4}\left[ \begin{matrix} 1+\cos 2\theta & \sin 2\theta \\\ \sin 2\theta & 1-\cos 2\theta \\\ \end{matrix} \right]\left[ \begin{matrix} 1+\cos 2\phi & \sin 2\phi \\\ \sin 2\phi & 1-\cos 2\phi \\\ \end{matrix} \right] \\\ \end{aligned}$ Solving this in a similar way and incorporating $\phi =\theta -\dfrac{\pi }{2}$ , we get the same required answer.