Question

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

Answer 1

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:

& \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{aligned}$$ 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 $\begin{aligned} & {{x}_{1}}=\cos 15{}^\circ \\\ & {{x}_{2}}=\sin 15{}^\circ \\\ & {{y}_{1}}=\sin 75{}^\circ \\\ & {{y}_{2}}=\cos 75{}^\circ \\\ \end{aligned}$ 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 $\begin{aligned} & M=\cos 15{}^\circ \cos 75{}^\circ -\sin 15{}^\circ \sin 75{}^\circ \\\ & M=\cos (15+75) \\\ & M=\cos 90{}^\circ \\\ \end{aligned}$ 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; $\begin{aligned} & \cos 15{}^\circ =\cos \left( 45-30 \right)=\cos 45{}^\circ \cos 30{}^\circ +\sin 45{}^\circ \sin 30{}^\circ \\\ & \sin 15{}^\circ =\sin \left( 45-30 \right)=\sin 45{}^\circ \cos 30{}^\circ -\sin 30{}^\circ \cos 45{}^\circ \\\ & \sin 75{}^\circ =\sin \left( 90-15 \right)=\cos 15{}^\circ \\\ & \cos 75{}^\circ =\cos \left( 90-15 \right)=\sin 15{}^\circ \\\ \end{aligned}$ 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 ${{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}}$.