Question

If $A,B,C$ are angle of a triangle $ABC$, then the value of the determinant $\left|\begin{array}{ccc}\sin \frac{A}{2}& \sin \frac{B}{2}& \sin \frac{C}{2}\\ \sin (A+B+C)& \sin \frac{B}{2}& \cos \frac{A}{2}\\ \cos \frac{(A+B+C)}{2}& \tan (A+B+C)& \sin \frac{C}{2}\end{array}\right|$ is less than or equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{8}$

Since A,B,C are the angles of a triangle so by angle sum property of a triangle, $A+B+C=\pi$

Putting in this value in the given determinant, we get-

$\Delta =\left|\begin{array}{ccc}\sin \frac{A}{2}& \sin \frac{B}{2}& \sin \frac{C}{2}\\ \sin \left(\pi \right)& \sin \frac{B}{2}& \cos \frac{A}{2}\\ \cos \frac{\pi }{2}& \tan \left(\pi \right)& \sin \frac{C}{2}\end{array}\right|$

$=\left|\begin{array}{ccc}\sin \frac{A}{2}& \sin \frac{B}{2}& \sin \frac{C}{2}\\ 0& \sin \frac{B}{2}& \cos \frac{A}{2}\\ 0& 0& \sin \frac{C}{2}\end{array}\right|$

Expanding along $R_3$ we get-

$=\sin \frac{C}{2}(\sin \frac{A}{2}\times \sin \frac{B}{2}-0)$

$=\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}$

Let $\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\le k$

Then, $k$ is the maximum value of $\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}$

Since, $A+B+C=\pi$

$\Rightarrow \frac{A+B+C}{2}=\frac{\pi }{2}\Rightarrow \frac{A+B}{2}+\frac{C}{2}=\frac{\pi }{2}$

$\Rightarrow \frac{C}{2}=\frac{\pi }{2}-\frac{A+B}{2}$

Thus, $\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}$

$=\sin \frac{A}{2}.\sin \frac{B}{2}.\sin (\frac{\pi }{2}-\frac{A+B}{2})$

$=\sin \frac{A}{2}.\sin \frac{B}{2}.\cos \left(\frac{A+B}{2}\right)$

Let $f(A,B)=\sin \frac{A}{2}.\sin \frac{B}{2}.\cos \left(\frac{A+B}{2}\right)$

Now, $\frac{\partial f}{\partial A}=\frac{1}{2}\sin \frac{B}{2}\cos (A+\frac{B}{2})$

and, $\frac{\partial f}{\partial B}=\frac{1}{2}\sin \frac{A}{2}\cos (B+\frac{A}{2})$

At the point of maxima of $f$, $\frac{\partial f}{\partial A}=\frac{\partial f}{\partial B}=0$

$\Rightarrow \frac{1}{2}\sin \frac{B}{2}\cos (A+\frac{B}{2})=\frac{1}{2}\sin \frac{A}{2}\cos (B+\frac{A}{2})=0$

Considering A,B to be non zero, we have-

$\cos (A+\frac{B}{2})=\cos (B+\frac{A}{2})=0$

$\Rightarrow A+\frac{B}{2}=B+\frac{A}{2}=\frac{\pi }{2}$

Solving we get, $A=B=\frac{\pi }{3}$ and thus, $C=\pi -A-B=\frac{\pi }{3}$

Now, $\frac{{\partial }^{2}f}{\partial A^2}=\frac{\partial }{\partial A}\left(\frac{\partial f}{\partial A}\right)$

$\frac{{\partial }^{2}f}{\partial A^2}=\frac{\partial }{\partial A}\left(\frac{1}{2}\sin \frac{B}{2}\cos \right(A+\frac{B}{2}\left)\right)$

$\Rightarrow \frac{{\partial }^{2}f}{\partial A^2}=-\frac{1}{2}\sin \frac{B}{2}\sin (A+\frac{B}{2})$

At $A=B=\frac{\pi }{3}$,

$\frac{{\partial }^{2}f}{\partial A^2}=-\frac{1}{2}\sin \frac{\pi }{6}\sin (\frac{\pi }{3}+\frac{\pi }{6})=-\frac{1}{4}$

Again,

$\frac{{\partial }^{2}f}{\partial B^2}=\frac{\partial }{\partial B}\left(\frac{\partial f}{\partial B}\right)$

$\frac{{\partial }^{2}f}{\partial B^2}=\frac{\partial }{\partial B}\left(\frac{1}{2}\sin \frac{A}{2}\cos \right(B+\frac{A}{2}\left)\right)$

$\Rightarrow \frac{{\partial }^{2}f}{\partial B^2}=-\frac{1}{2}\sin \frac{A}{2}\sin (B+\frac{A}{2})$

At $A=B=\frac{\pi }{3}$,

$\frac{{\partial }^{2}f}{\partial B^2}=-\frac{1}{2}\sin \frac{\pi }{6}\sin (\frac{\pi }{3}+\frac{\pi }{6})=-\frac{1}{4}$

And,

$\frac{{\partial }^{2}f}{\partial A.\partial B}=\frac{\partial }{\partial A}\left(\frac{\partial f}{\partial B}\right)$

$\frac{{\partial }^{2}f}{\partial A.\partial B}=\frac{\partial }{\partial A}\left(\frac{1}{2}\sin \frac{A}{2}\cos \right(B+\frac{A}{2}\left)\right)$

$\frac{{\partial }^{2}f}{\partial A.\partial B}=\frac{1}{4}\cos (A+B)$

Thus, at $A=B=\frac{\pi }{3}$,

$\frac{{\partial }^{2}f}{\partial A.\partial B}=\frac{1}{4}\cos (\frac{\pi }{3}+\frac{\pi }{3})=\frac{1}{4}\times \frac{1}{3}=\frac{1}{12}$

Now, $D=\frac{{\partial }^{2}f}{\partial A^2}.\frac{{\partial }^{2}f}{\partial B^2}-{\left(\frac{{\partial }^{2}f}{\partial A.\partial B}\right)}^2=(-\frac{1}{4})(-\frac{1}{4})-(\frac{1}{12}{)}^2=\frac{1}{16}-\frac{1}{144}=\frac{1}{18}>0$

And, $\frac{{\partial }^{2}f}{\partial A^2}=-\frac{1}{4}<0$

Hence, $A=B=C=\frac{\pi }{3}$ is a point of maxima.

Thus, the maximum value of $\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}=\sin \frac{\pi }{6}.\sin \frac{\pi }{6}.\sin \frac{\pi }{6}={\left(\frac{1}{2}\right)}^2=\frac{1}{8}$

Thus, $\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\le \frac{1}{8}$

Hence, the correct answer is option C.