Answer
Verified
479.4k+ views
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 ${{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}}$.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Change the following sentences into negative and interrogative class 10 english CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Application to your principal for the character ce class 8 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE