Question

What will be the value of \[\vartriangle \,=\,\left| \begin{align}
  & a\,\,\,\,\,\,b\,\,\,\,\,\,\,c \\
 & x\,\,\,\,\,\,y\,\,\,\,\,\,z \\
 & bc\,\,\,ca\,\,\,ab \\
\end{align} \right|\]. If we are given that \[a=sin\theta ,\,b=sin\left( \theta +\dfrac{2\pi }{3} \right),~c=sin\left( \theta +\dfrac{4\pi }{3} \right),~x=cos\theta ,~y=cos\left( \theta +\dfrac{2\pi }{3} \right),~\]\[z=cos\left( \theta +\dfrac{4\pi }{3} \right)\] then
(a) $\dfrac{1}{8}$
(b) $\dfrac{3\sqrt{3}}{4}$
(c) $\dfrac{3\sqrt{3}}{8}$
(d) $\dfrac{\sqrt{3}}{4}$

Answer 1

Hint: We will try to simplify the determinant using trigonometric properties and properties of the determinant. To simplify we need to know that \[2~sinA\text{ }sinB~=cos\left( A-B \right)-cos\left( A+B \right)~\] and after simplifying the determinant we will apply operations \[{{C}_{2}}\,\to {{C}_{2}}\,-{{C}_{1}}\,\,and\,\,{{C}_{3}}\,\to {{C}_{3}}\,-{{C}_{1}}\] on the determinant and then expand the determinant accordingly to find it’s value.

Complete step by step answer:
We need to put the given values of the variables in the given determinant, we are given that,
\[\begin{align}
  & a=sin\theta ,\, \\
 & b=sin\left( \theta +\dfrac{2\pi }{3} \right), \\
 & ~c=sin\left( \theta +\dfrac{4\pi }{3} \right),~ \\
 & x=cos\theta ,~ \\
 & y=cos\left( \theta +\dfrac{2\pi }{3} \right),\,and \\
 & z=cos\left( \theta +\dfrac{4\pi }{3} \right) \\
\end{align}\]
Now putting the above values in given determinant, we get
\[\vartriangle \,=\,\left| \begin{matrix}
   a & b & c \\
   x & y & z \\
   bc & ca & ab \\
\end{matrix} \right|\]
\[=\left| \begin{matrix}
   sin\theta & sin\left( \theta +\dfrac{2\pi }{3} \right) & sin\left( \theta +\dfrac{4\pi }{3} \right) \\
   cos\theta & cos\left( \theta +\dfrac{2\pi }{3} \right) & cos\left( \theta +\dfrac{4\pi }{3} \right) \\
   sin\left( \theta +\dfrac{2\pi }{3} \right)sin\left( \theta +\dfrac{4\pi }{3} \right) & sin\left( \theta +\dfrac{4\pi }{3} \right)sin\theta & sin\theta sin\left( \theta +\dfrac{2\pi }{3} \right) \\
\end{matrix} \right|\],
Now, using the formula \[2~sinA\text{ }sinB~=cos\left( A-B \right)-cos\left( A+B \right)~\] in the third row of the above determinant, we get
\[=\left| \begin{matrix}
   sin\theta & sin\left( \theta +\dfrac{2\pi }{3} \right) & sin\left( \theta +\dfrac{4\pi }{3} \right) \\
   cos\theta & cos\left( \theta +\dfrac{2\pi }{3} \right) & cos\left( \theta +\dfrac{4\pi }{3} \right) \\
   \dfrac{1}{2}\left( cos\left( -\dfrac{2\pi }{3} \right)-cos(2\theta +2\pi ) \right)\, & \dfrac{1}{2}\left( cos\left( \dfrac{4\pi }{3} \right)-cos\left( 2\theta +\dfrac{4\pi }{3} \right) \right) & \,\dfrac{1}{2}\left( cos\left( -\dfrac{2\pi }{3} \right)-cos\left( 2\theta +\dfrac{2\pi }{3} \right) \right) \\
\end{matrix} \right|\]
Taking $\dfrac{1}{2}$ common from the third row, we get
\[=\dfrac{1}{2}\left| \begin{matrix}
   sin\theta & sin\left( \theta +\dfrac{2\pi }{3} \right) & sin\left( \theta +\dfrac{4\pi }{3} \right) \\
   cos\theta & cos\left( \theta +\dfrac{2\pi }{3} \right) & cos\left( \theta +\dfrac{4\pi }{3} \right) \\
   \left( cos\left( -\dfrac{2\pi }{3} \right)-cos(2\theta +2\pi ) \right)\, & \left( cos\left( \dfrac{4\pi }{3} \right)-cos\left( 2\theta +\dfrac{4\pi }{3} \right) \right) & \,\left( cos\left( -\dfrac{2\pi }{3} \right)-cos\left( 2\theta +\dfrac{2\pi }{3} \right) \right) \\
\end{matrix} \right|\]
We can also write above matrix as,
\[=\dfrac{1}{2}\left( \left| \begin{matrix}
   sin\theta & sin\left( \theta +\dfrac{2\pi }{3} \right) & sin\left( \theta +\dfrac{4\pi }{3} \right) \\
   cos\theta & cos\left( \theta +\dfrac{2\pi }{3} \right) & cos\left( \theta +\dfrac{4\pi }{3} \right) \\
   \,\cos \left( -\dfrac{2\pi }{3} \right) & \cos \left( \dfrac{4\pi }{3} \right) & \cos \left( -\dfrac{2\pi }{3} \right) \\
\end{matrix} \right|+\left| \begin{matrix}
   0 & 0 & 0 \\
   0 & 0 & 0 \\
   -cos(2\theta +2\pi ) & -cos\left( 2\theta +\dfrac{4\pi }{3} \right) & -cos\left( 2\theta +\dfrac{2\pi }{3} \right) \\
\end{matrix} \right| \right)\]
But the second matrix i.e. on the RHS is 0 as all 3 elements of row 1 are 0, so we get
\[=\dfrac{1}{2}\left| \begin{matrix}
   sin\theta & sin\left( \theta +\dfrac{2\pi }{3} \right) & sin\left( \theta +\dfrac{4\pi }{3} \right) \\
   cos\theta & cos\left( \theta +\dfrac{2\pi }{3} \right) & cos\left( \theta +\dfrac{4\pi }{3} \right) \\
   \,\cos \left( -\dfrac{2\pi }{3} \right) & \cos \left( \dfrac{4\pi }{3} \right) & \cos \left( -\dfrac{2\pi }{3} \right) \\
\end{matrix} \right|\]
Now putting the values \[\cos \left( -\dfrac{2\pi }{3} \right)=-\dfrac{1}{2}\,and\,\cos \left( \dfrac{4\pi }{3} \right)=-\dfrac{1}{2}\], we get
\[=\dfrac{1}{2}\left| \begin{matrix}
   sin\theta & sin\left( \theta +\dfrac{2\pi }{3} \right) & sin\left( \theta +\dfrac{4\pi }{3} \right) \\
   cos\theta & cos\left( \theta +\dfrac{2\pi }{3} \right) & cos\left( \theta +\dfrac{4\pi }{3} \right) \\
   -\dfrac{1}{2} & -\dfrac{1}{2} & -\dfrac{1}{2} \\
\end{matrix} \right|\],
Now taking $-\dfrac{1}{2}$ common from third row we get,
\[=-\dfrac{1}{4}\left| \begin{matrix}
   sin\theta & sin\left( \theta +\dfrac{2\pi }{3} \right) & sin\left( \theta +\dfrac{4\pi }{3} \right) \\
   cos\theta & cos\left( \theta +\dfrac{2\pi }{3} \right) & cos\left( \theta +\dfrac{4\pi }{3} \right) \\
   1 & 1 & 1 \\
\end{matrix} \right|\]
Now, applying the operations, \[{{C}_{2}}\,\to {{C}_{2}}\,-{{C}_{1}}\,\,and\,\,{{C}_{3}}\,\to {{C}_{3}}\,-{{C}_{1}}\] we get
\[=-\dfrac{1}{4}\left| \begin{matrix}
   sin\theta & sin\left( \theta +\dfrac{2\pi }{3} \right)-\sin \theta & sin\left( \theta +\dfrac{4\pi }{3} \right)-\sin \theta \\
   cos\theta & cos\left( \theta +\dfrac{2\pi }{3} \right)-\cos \theta & cos\left( \theta +\dfrac{4\pi }{3} \right)-\cos \theta \\
   1 & 0 & 0 \\
\end{matrix} \right|\],
Expanding the determinant along row 3, we get
\[=-\dfrac{1}{4}\times \left( \left( sin\left( \theta +\dfrac{2\pi }{3} \right)-\sin \theta \right)\times \left( cos\left( \theta +\dfrac{4\pi }{3} \right)-\cos \theta \right)-\left( cos\left( \theta +\dfrac{2\pi }{3} \right)-\cos \theta \right)\times \left( sin\left( \theta +\dfrac{4\pi }{3} \right)-\sin \theta \right) \right)\]
\[=-\dfrac{1}{4}\times \left( \begin{align}
  & \left( sin\left( \theta +\dfrac{2\pi }{3} \right)\times cos\left( \theta +\dfrac{4\pi }{3} \right)-cos\left( \theta +\dfrac{2\pi }{3} \right)\times sin\left( \theta +\dfrac{4\pi }{3} \right) \right)+ \\
 & \left( \sin \theta .cos\left( \theta +\dfrac{2\pi }{3} \right)-sin\left( \theta +\dfrac{2\pi }{3} \right).\cos \theta \right)+ \\
 & \left( sin\left( \theta +\dfrac{4\pi }{3} \right).\cos \theta -cos\left( \theta +\dfrac{4\pi }{3} \right).\sin \theta \right) \\
\end{align} \right)\]
Now applying identity $\sin (A-B)=\sin A\cos B-\cos A\sin B$, we get
$=-\dfrac{1}{4}\times \left( \sin \left( -\dfrac{2\pi }{3} \right)+\sin \left( -\dfrac{2\pi }{3} \right)+\sin \left( \dfrac{4\pi }{3} \right) \right)$
Putting \[\sin \left( -\dfrac{2\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}\,and\,\sin \left( \dfrac{4\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}\], we get
$=-\dfrac{1}{4}\times \left( -\dfrac{3\sqrt{3}}{2} \right)=\dfrac{3\sqrt{3}}{8}$
Hence we got our answer as $\dfrac{3\sqrt{3}}{8}$, so the value of the $\vartriangle $ given in the question is $\dfrac{3\sqrt{3}}{8}$.

So, the correct answer is “Option C”.

Note: While solving this question you need to be patient while solving these kinds of problems as this involves very large calculations and even if we go for a one step wrong we will not be able to get our answer. So be careful while doing calculations and also you need to practice trigonometric identities so that you can easily understand when and where to use them to simplify the expression.