Question

Prove that
$\Delta =\left| \begin{matrix}
   \sin A & \sin B & \sin C \\
   \cos A & \cos B & \cos C \\
   {{\cos }^{3}}A & {{\cos }^{3}}B & {{\cos }^{3}}C \\
\end{matrix} \right|=\sin \left( A-B \right)\sin \left( B-c \right)\sin \left( C-A \right)\cos \left( A+B+C \right)$
Also determine when $\Delta =0.$

Answer 1

Hint: Take $\cos A,\cos B,\cos C$ common from the three columns ${{C}_{1}},{{C}_{2}},{{C}_{3}}$ respectively. Subtract two third column with its corresponding elements. Use the trigonometric identities, wherever required. Some of them are
$\begin{align}
  & 2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right) \\
 & \sin x-\sin y=2\sin \left( \dfrac{x-y}{2} \right)\cos \left( \dfrac{x+y}{2} \right) \\
\end{align}$

Complete step-by-step answer:
We have to prove
$\Delta =\left| \begin{matrix}
   \sin A & \sin B & \sin C \\
   \cos A & \cos B & \cos C \\
   {{\cos }^{3}}A & {{\cos }^{3}}B & {{\cos }^{3}}C \\
\end{matrix} \right|=\sin \left( A-B \right)\sin \left( B-c \right)\sin \left( C-A \right)\cos \left( A+B+C \right)...............\left( i \right)$
Hence, we need to determine the relation in A, B, C if $\Delta =0$ . So let us solve the left hand side of the equation (i) and try to simplify it and hence prove it to be equal to RHS of the equation (i). So, we have
\[\Delta =\left| \begin{matrix}
   \sin A & \sin B & \sin C \\
   \cos A & \cos B & \cos C \\
   {{\cos }^{3}}A & {{\cos }^{3}}B & {{\cos }^{3}}C \\
\end{matrix} \right|\]
Now let us apply properties of determinant with the above $\Delta $ . So apply
$\begin{align}
  & {{C}_{1}}\to \dfrac{{{C}_{1}}}{\cos A} \\
 & {{C}_{2}}\to \dfrac{{{C}_{2}}}{\cos B} \\
 & {{C}_{3}}\to \dfrac{{{C}_{3}}}{\cos C} \\
\end{align}$
As, we are dividing the columns of determinant by $\cos A,\cos B,\cos C$ respectively. So, we need to multiply the determinant by $\cos A,\cos B,\cos C$ as well. So, we get
$\Delta =\cos A\cos B\cos C\left| \begin{matrix}
   \dfrac{\sin A}{\cos A} & \dfrac{\sin B}{\cos B} & \dfrac{\sin C}{\cos C} \\
   1 & 1 & 1 \\
   {{\cos }^{2}}A & {{\cos }^{2}}B & {{\cos }^{2}}C \\
\end{matrix} \right|$
Now, apply the property,
$\begin{align}
  & {{C}_{1}}\to {{C}_{1}}-{{C}_{3}} \\
 & {{C}_{2}}\to {{C}_{2}}-{{C}_{3}} \\
\end{align}$
Hence, we get value of $\Delta $ as
$\begin{align}
  & \Delta =\cos A\cos B\cos C\left| \begin{matrix}
   \dfrac{\sin A}{\cos A}-\dfrac{\sin C}{\cos C} & \dfrac{\sin B}{\cos B}-\dfrac{\sin C}{\cos C} & \dfrac{\sin C}{\cos C} \\
   1-1 & 1-1 & 1 \\
   {{\cos }^{2}}A-{{\cos }^{2}}C & {{\cos }^{2}}B-{{\cos }^{2}}C & {{\cos }^{2}}C \\
\end{matrix} \right| \\
 & \Delta =\cos A\cos B\cos C\left| \begin{matrix}
   \dfrac{\operatorname{sinAcosC}-cosAsinC}{\cos A\cos C} & \dfrac{\sin B\cos C-\sin C\cos B}{\cos B\cos C} & \dfrac{\sin C}{\cos C} \\
   0 & 0 & 1 \\
   {{\cos }^{2}}A-{{\cos }^{2}}C & {{\cos }^{2}}B-{{\cos }^{2}}C & {{\cos }^{2}}C \\
\end{matrix} \right| \\
\end{align}$
We know
$\begin{align}
  & \sin \left( x-y \right)=\sin x\cos y-\cos x\sin y \\
 & \sin \left( x+y \right)\sin \left( x-y \right)={{\sin }^{2}}x-{{\sin }^{2}}y={{\cos }^{2}}y-{{\cos }^{2}}x \\
\end{align}$
Hence, we can get the value of $\Delta $ with the help of identities as
$\Delta =\cos A\operatorname{cosB}\cos C\left| \begin{matrix}
   \dfrac{\sin \left( A-C \right)}{\cos A\cos C} & \dfrac{\sin \left( B-c \right)}{\cos B\cos C} & \dfrac{\sin C}{\cos C} \\
   0 & 0 & 1 \\
   \sin \left( C+A \right)\sin \left( C-A \right) & \sin \left( C+B \right) & {{\cos }^{2}}C \\
\end{matrix} \right|$
Now, we know
$\sin \left( -\theta \right)=-\sin \theta $
So, we get value of $\Delta $ as
$\Delta =\cos A\operatorname{cosB}\cos C\left| \begin{matrix}
   \dfrac{\sin \left( A-C \right)}{\cos A\cos C} & \dfrac{\sin \left( B-c \right)}{\cos B\cos C} & \dfrac{\sin C}{\cos C} \\
   0 & 0 & 1 \\
   -\sin \left( A+C \right)\sin \left( A-C \right) & -\sin \left( B+C \right)\sin \left( B-C \right) & {{\cos }^{2}}C \\
\end{matrix} \right|$
Taking $\sin \left( A-C \right)$ common from column 1, and $\sin \left( B-C \right)$ from column 2. We get
$\Delta =\cos A\operatorname{cosB}\cos C\sin \left( A-C \right)\sin \left( B-C \right)\left| \begin{matrix}
   \dfrac{1}{\cos A\cos C} & \dfrac{1}{\cos B\cos C} & \dfrac{\sin C}{\cos C} \\
   0 & 0 & 1 \\
   -\sin \left( A+C \right) & -\sin \left( B+C \right) & {{\cos }^{2}}C \\
\end{matrix} \right|$
Now, expand the above determinant along row 2 i.e. ${{R}_{2}}$ we get
$\begin{align}
  & \Delta =\cos A\operatorname{cosB}\cos C\sin \left( A-C \right)\sin \left( B-C \right)\times \left( -1 \right)\left[ \dfrac{\sin \left( B+C \right)}{\cos A\cos C}-\left( \dfrac{-\sin \left( A+C \right)}{\cos B\cos C} \right) \right] \\
 & \Delta =-\cos A\operatorname{cosB}\cos C\sin \left( A-C \right)\sin \left( B-C \right)\left[ \dfrac{\sin \left( A+C \right)}{\operatorname{cosB}\cos C}-\dfrac{\sin \left( B+C \right)}{\cos A\operatorname{cosC}} \right] \\
 & \Delta =-\cos A\operatorname{cosB}\cos C\sin \left( A-C \right)\sin \left( B-C \right)\left[ \dfrac{\sin \left( A+C \right)\cos A-\cos B\sin \left( B+C \right)}{\cos A\cos B\cos C} \right] \\
\end{align}$
Now, cancelling the term $\cos A\cos B\cos C$ from numerator and denominator, we get
$\Delta =-\sin \left( A-C \right)\sin \left( B-C \right)\left[ \sin \left( A+C \right)\cos A-\cos B\sin \left( B+C \right) \right]$
Now, multiply and divide the whole expression by 2. We get
$\Delta =\dfrac{-\sin \left( A-C \right)\sin \left( B-C \right)}{2}\left[ 2\sin \left( A+C \right)\cos A-2\sin \left( B+C \right)\operatorname{cosB} \right]$
We know the trigonometric identity of $2\sin x\cos y$ can be given as
$2\sin x\cos y=\sin \left( x+y \right)+\sin \left( x-y \right)$
Hence, we can simplify $\Delta $ with the help of the above expression
$\begin{align}
  & \Delta =\dfrac{-\sin \left( A-C \right)\sin \left( B-C \right)}{2}\left[ \left( \sin \left( A+C+A \right) \right)+\sin \left( A+C-A \right)-\left( \sin \left( B+C+B \right) \right)+\sin \left( B+C-B \right) \right] \\
 & \Delta =\dfrac{-\sin \left( A-C \right)\sin \left( B-C \right)}{2}\left[ \sin \left( 2A+C \right)+\sin C-\sin \left( 2B+C \right)-\sin C \right] \\
 & \Delta =\dfrac{-\sin \left( A-C \right)\sin \left( B-C \right)}{2}\left[ \sin \left( 2A+C \right)-\sin \left( 2B+C \right) \right] \\
\end{align}$
Now, we know the trigonometric identity of $\left( \sin x-\sin y \right)$ can be given as
$\begin{align}
  & \sin x-\sin y=2\sin \left( \dfrac{x-y}{2} \right)\cos \left( \dfrac{x+y}{2} \right) \\
 & \Delta =\dfrac{-\sin \left( A-C \right)\sin \left( B-C \right)}{2}\left[ 2\sin \left( \dfrac{2A+C-2B-C}{2} \right)\times \cos \left( \dfrac{2A+C+2B+C}{2} \right) \right] \\
 & \Delta =-\sin \left( A-C \right)\sin \left( B-C \right)\sin \left( \dfrac{2A-2B}{2} \right)\cos \dfrac{2A+2B+2C}{2} \\
 & \Delta =-\sin \left( A-C \right)\sin \left( B-C \right)\sin \left( A-B \right)\cos \left( A+B+C \right) \\
 & \Delta =-\sin \left( A-B \right)\sin \left( B-C \right)\sin \left( C-A \right)\cos \left( A+B+C \right) \\
\end{align}$
Where we use
$\sin \left( -\theta \right)=-\sin \theta $
Hence, it is proved that
$\Delta =\left| \begin{matrix}
   \sin A & \sin B & \sin C \\
   \cos A & \cos B & \cos C \\
   {{\cos }^{3}}A & {{\cos }^{3}}B & {{\cos }^{3}}C \\
\end{matrix} \right|=\sin \left( A-B \right)\sin \left( B-c \right)\sin \left( C-A \right)\cos \left( A+B+C \right)$
Now, if $\Delta =0$ , then we get
$\sin \left( A-B \right)\sin \left( B-C \right)\sin \left( C-A \right)\cos \left( A+B+C \right)=0$
As, four numbers are in multiplication and the result is 0. So, either of the four numbers should be 0. Hence, we get
$\begin{align}
  & \sin \left( A-B \right)=0 \\
 & \Rightarrow \sin \left( B-C \right)=0 \\
 & \Rightarrow \sin \left( C-A \right)=0 \\
 & \Rightarrow \cos \left( A+B+C \right)=0 \\
\end{align}$
We know
$\sin {{0}^{\circ }}=0,\cos \dfrac{\pi }{2}=0$
Hence, we get
$\begin{align}
  & A-B=0 \\
 & \Rightarrow B-C=0 \\
 & \Rightarrow C-A=0 \\
 & \Rightarrow A+B+C+\dfrac{\pi }{2} \\
 & A=B \\
 & \Rightarrow B=C \\
 & \Rightarrow C=A \\
 & \Rightarrow A+B+C=\dfrac{\pi }{2} \\
\end{align}$
Hence, the determinant will be 0 if angles of A, B, C are equal or their summation of A, B, C is ${{90}^{\circ }},\dfrac{\pi }{2}$

Note: Another approach for the question would be that we can apply the property in given determinant as
${{C}_{1}}\to {{C}_{1}}-{{C}_{3}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$
So, use the proper trigonometric and algebraic identities, whenever required. But it will be longer process that provided in the solution. Clear with trigonometric identities and take care which property need to use where. So, be focused while solving these kinds of questions.