# Find the value of the determinant$\left| \begin{matrix}

\cos 15{}^\circ & \sin 15{}^\circ \\

\sin 75{}^\circ & \cos 75{}^\circ \\

\end{matrix} \right|$

Last updated date: 18th Mar 2023

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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 ${{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}}$.

{{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 ${{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}}$.

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