Question

If $A+B+C=\pi$, then $\sin 2A+\sin 2B+\sin 2C$ is equal to

• A. $4\sin A\sin B\sin C$
• B. $4\cos A\cos B\cos C$
• C. $2\cos A\cos B\cos C$
• D. $2\sin A\sin B\sin C$

Answer 1

The correct option is A

$4\sin A\sin B\sin C$

Find the value of $\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{A}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{B}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{C}\mathbf{:}$

Given data:$\begin{array}{rcl}A+B+C& =& \pi \Rightarrow 2A+2B+2C=2\pi \end{array}$

$\begin{array}{rcl}sin2A+sin2B+sin2C& =& 2\sin \frac{(2A+2B)}{2}cos\frac{(2A–2B)}{2}+sin[2\pi –2(A+B\left)\right]\\ & =& 2sin(A+B)cos(A–B)–sin2(A+B)\\ & =& 2sin(A+B)cos(A–B)–2sin(A+B)cos(A+B)(\because sin2x=2sinxcosx)\\ & =& 2sin(A+B)\left[cos\right(A–B)–cos(A+B\left)\right]\\ & =& 2sin(\pi –C)\left(2sinAsinB\right)\\ & =& 4sinCsinAsinB\\ & =& 4sinAsinBsinC\end{array}$

Hence option $\mathbf{\left(}\mathit{A}\mathbf{\right)}$ is the correct option.