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Find the value of the determinant$\left| \begin{matrix}
   \cos 15{}^\circ & \sin 15{}^\circ \\
   \sin 75{}^\circ & \cos 75{}^\circ \\
\end{matrix} \right|$

seo-qna
Last updated date: 13th Jul 2024
Total views: 449.7k
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Answer
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Hint: Expansion of determinant $\left| \begin{matrix}
   {{x}_{1}} & {{x}_{2}} \\
   {{y}_{1}} & {{y}_{2}} \\
\end{matrix} \right|$ is ${{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}}.$ Use trigonometry identity to solve further.

Complete step-by-step answer:
We have the given determinant as
Let us suppose value of this determinant is M
M=$\left| \begin{matrix}
   \cos 15{}^\circ & \sin 15{}^\circ \\
   \sin 75{}^\circ & \cos 75{}^\circ \\
\end{matrix} \right|$………………..(1)
As, we know the rules of opening a determinant as
If we have any general determinant as
$\Delta =~\left| \begin{matrix}
   {{x}_{1}} & {{x}_{2}} \\
   {{y}_{1}} & {{y}_{2}} \\
\end{matrix} \right|$ Then we can expand it by following way:
\[\begin{align}
  & \Delta =\left( {{x}_{1}}\times {{y}_{2}} \right)-\left( {{x}_{2}}{{y}_{1}} \right) \\
 & or \\
 & \Delta ={{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}....................\left( 2 \right) \\
\end{align}\]
Now using the above expansion as expressed in equation (2), we can expand determinant given in equation (1) as;
$M=\left| \begin{matrix}
   \cos 15{}^\circ & \sin 15{}^\circ \\
   \sin 75{}^\circ & \cos 75{}^\circ \\
\end{matrix} \right|$
Where
$\begin{align}
  & {{x}_{1}}=\cos 15{}^\circ \\
 & {{x}_{2}}=\sin 15{}^\circ \\
 & {{y}_{1}}=\sin 75{}^\circ \\
 & {{y}_{2}}=\cos 75{}^\circ \\
\end{align}$
Therefore, we can write M as
$M=\cos 15{}^\circ \cos 75{}^\circ -\sin 15{}^\circ \sin 75{}^\circ .......\left( 3 \right)$
Now, we have a trigonometric identity of cos (A+B) as cosA cosB – sinA sinB or vice –versa will also be true.
Hence, equation (3) can be written as
$\begin{align}
  & M=\cos 15{}^\circ \cos 75{}^\circ -\sin 15{}^\circ \sin 75{}^\circ \\
 & M=\cos (15+75) \\
 & M=\cos 90{}^\circ \\
\end{align}$
We know value of $\cos 90{}^\circ $ as 0, hence we can get value of M as
M=0
Hence from equation (1), we get
$\left| \begin{matrix}
   \cos 15{}^\circ & \sin 15{}^\circ \\
   \sin 75{}^\circ & \cos 75{}^\circ \\
\end{matrix} \right|=0$
Note: One can calculate cos15, sin15, sin75 and cos75, then expand the given determinant.
We can calculate values as;
$\begin{align}
  & \cos 15{}^\circ =\cos \left( 45-30 \right)=\cos 45{}^\circ \cos 30{}^\circ +\sin 45{}^\circ \sin 30{}^\circ \\
 & \sin 15{}^\circ =\sin \left( 45-30 \right)=\sin 45{}^\circ \cos 30{}^\circ -\sin 30{}^\circ \cos 45{}^\circ \\
 & \sin 75{}^\circ =\sin \left( 90-15 \right)=\cos 15{}^\circ \\
 & \cos 75{}^\circ =\cos \left( 90-15 \right)=\sin 15{}^\circ \\
\end{align}$
But the above process will be much longer than given in the solution.
One can go wrong while expanding the determinant.
$\left| \begin{matrix}
   {{x}_{1}} & {{x}_{2}} \\
   {{y}_{1}} & {{y}_{2}} \\
\end{matrix} \right|={{x}_{2}}{{y}_{1}}-{{x}_{1}}{{y}_{2}}$
Which is wrong .Correct expression would be ${{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}}$.