Q191 / 270 · NDA 2016 (Phase 2) · Mathematics
$\sin \mathrm{A}+2 \sin 2 \mathrm{~A}+\sin 3 \mathrm{~A}$ is equal to which of the following?
1. $4 \sin 2 \mathrm{~A} \cos ^2\left(\frac{\mathrm{~A}}{2}\right)$
2. $2 \sin 2 \mathrm{~A}\left(\sin \frac{\mathrm{~A}}{2}+\cos \frac{\mathrm{A}}{2}\right)^2$
3. $8 \sin \mathrm{~A} \cos \mathrm{~A} \cos ^2\left(\frac{\mathrm{~A}}{2}\right)$
Select the correct answer using the code given below:
- A1 and 2 only
- B2 and 3 only
- Correct answerC1 and 3 only
- D1,2 and 3
Solution
Let $\mathrm{A}=30^{\circ}$
$\begin{aligned}
\Rightarrow & \sin \mathrm{A}+2 \sin 2 \mathrm{~A}+\sin 3 \mathrm{~A}=\sin 30^{\circ}+2 \sin 60^{\circ}+\sin 90^{\circ} \\
\quad & \frac{1}{2}+\frac{2 \sqrt{3}}{2}+1=\frac{2 \sqrt{3}+3}{2} \\
& \left(\because 2 \cos ^2 \mathrm{~A}=1+\cos 2 \mathrm{~A}\right)
\end{aligned}$
Now, $4 \sin 2 \mathrm{~A} \cos ^2\left(\frac{\mathrm{~A}}{2}\right)=2 \sin 2 \mathrm{~A}[1+\cos \mathrm{A}]$
$=2 \sin 60^{\circ}\left[1+\cos 30^{\circ}\right]=\frac{2 \sqrt{3}+3}{2}$
Also, $\sin 2 \mathrm{~A}=2 \sin \mathrm{~A} \cos \mathrm{~A} \& \sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}=1$
$\begin{aligned}
& 2 \sin 2 \mathrm{~A}\left[\sin \frac{\mathrm{~A}}{2}+\cos \frac{\mathrm{A}}{2}\right]^2 \\
& =2 \sin 2 \mathrm{~A}\left[\sin ^2 \frac{\mathrm{~A}}{2}+\cos ^2 \frac{\mathrm{~A}}{2}+2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}\right] \\
& =2 \sin 2 \mathrm{~A}[1+\sin \mathrm{A}]=2 \sin 60^{\circ}\left[1+\sin 30^{\circ}\right] \\
& =\frac{3 \sqrt{3}}{2}
\end{aligned}$
$\begin{aligned}
& \& 8 \sin A \cos A \cos ^2\left(\frac{\mathrm{~A}}{2}\right) \\
& =4 \sin \mathrm{~A} \cos \mathrm{~A}[1+\cos \mathrm{A}] \\
& =4 \sin 30^{\circ} \cos 30^{\circ}\left[1+\cos 30^{\circ}\right] \\
& =\frac{2 \sqrt{3}+3}{2}
\end{aligned}$
Hence, the correct answer is option: 'C'.