Question
In $\triangle A B C$ prove that
$\sin 2 A+\sin 2 B-\sin 2 C=4 \cos A \cos B \sin C $
$\sin 2 A+\sin 2 B-\sin 2 C=4 \cos A \cos B \sin C $
✓
Answer
$
\begin{aligned}
& \text { L.H.S. }=\sin 2 A+\sin 2 B-\sin 2 C \\
&= 2 \sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)-\sin 2 \mathrm{C} \\
&= 2 \sin (A+B) \cos (A-B)-2 \sin \mathrm{C} \cos \mathrm{C} \\
&= 2 \sin (\pi-C) \cos (A-B)-2 \sin C \cos [\pi-(A+B)] \\
&= 2 \sin C \cos (A-B)+2 \sin C \cos (A+B) \\
&= 2 \sin \mathrm{C}[\cos (A-B)+\cos (A+B) \\
&= 2 \sin \mathrm{C} \cdot 2 \cos \left(\frac{A-B+A+B}{2}\right) \cos \left(\frac{A-B-A-B}{2}\right) \\
&= 4 \sin \mathrm{C} \cos A \cos B \\
&= 4 \cos A \cos B \sin C \\
&= \text { R.H.S. }
\end{aligned}
$
\begin{aligned}
& \text { L.H.S. }=\sin 2 A+\sin 2 B-\sin 2 C \\
&= 2 \sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)-\sin 2 \mathrm{C} \\
&= 2 \sin (A+B) \cos (A-B)-2 \sin \mathrm{C} \cos \mathrm{C} \\
&= 2 \sin (\pi-C) \cos (A-B)-2 \sin C \cos [\pi-(A+B)] \\
&= 2 \sin C \cos (A-B)+2 \sin C \cos (A+B) \\
&= 2 \sin \mathrm{C}[\cos (A-B)+\cos (A+B) \\
&= 2 \sin \mathrm{C} \cdot 2 \cos \left(\frac{A-B+A+B}{2}\right) \cos \left(\frac{A-B-A-B}{2}\right) \\
&= 4 \sin \mathrm{C} \cos A \cos B \\
&= 4 \cos A \cos B \sin C \\
&= \text { R.H.S. }
\end{aligned}
$
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