Question

Without expanding as far as possible, evaluate
$\left|\begin{array}{ccc}{\sin }^{2}A& \sin A\cos A& {\cos }^{2}A\\ {\sin }^{2}B& \sin B\cos B& {\cos }^{2}B\\ {\sin }^{2}C& \sin C\cos C& {\cos }^{2}C\end{array}\right|$
given that $A+B+C=\pi$

Answer 1

Apply $C_1+C_3$ and then multiply $C_2$ and $C_3$ by $2$ each and divide $\Delta$ by $4$ and write $2\sin A\cos$ $A=\sin 2A$ and $2{\cos }^2$ $A=1+\cos 2A$
$\therefore \Delta =\frac{1}{4}\left|\begin{array}{ccc}1& \sin 2A& 1+\cos 2A\\ 1& \sin 2B& 1+\cos 2B\\ 1& \sin 2C& 1+\cos 2C\end{array}\right|$
Split into two determinants one of which will vanish
$\Delta =\frac{1}{4}\left|\begin{array}{ccc}1& \sin 2A& \cos 2A\\ 1& \sin 2B& \cos 2B\\ 1& \sin 2C& \cos 2C\end{array}\right|$.
Expanding we get
$\Delta =\frac{1}{4}[\sin 2(B-C)+\sin 2(C-A)+\sin 2(A-B)]$
$=\frac{1}{4}[\sin 2X+\sin 2Y+\sin 2Z]$
Where $X+Y+Z=0$
$\therefore \sin (X+Y)=-\sin Z,\cos (X+Y)=\cos Z$
$=\frac{1}{4}.-4\sin X\sin Y\sin Z$ as in $Q.1\left(i\right)$ identities of Trigonometry
$=-\sin (A-B)\sin (B-C)\sin (C-A)$.