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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
If $\sin\text{x}+\sin\text{y}=\sqrt3(\cos\text{y}-\cos\text{x}),$ than $\sin3\text{x}+\sin3\text{y}=$
  • A
    $2\sin3\text{x}$
  • ✓
    $0$
  • C
    $1$
  • D
    None of these
✓

Answer

Correct option: B.
$0$
We have,
$\sin\text{x}+\sin\text{y}=\sqrt3(\cos\text{y}-\cos\text{x})$
$\Rightarrow\ 2\sin\Big(\frac{\text{x+y}}{2}\Big)\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)=2\sqrt3\sin\Big(\frac{\text{x+y}}{2}\Big)\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)$
$\Rightarrow\ 2\sin\Big(\frac{\text{x+y}}{2}\Big)\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-2\sqrt3\sin\Big(\frac{\text{x+y}}{2}\Big)\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)=0$
$\Rightarrow\ 2\sin\Big(\frac{\text{x+y}}{2}\Big)\Big[\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-\sqrt3\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)\Big]=0$
$\Rightarrow\ \sin\Big(\frac{\text{x+y}}{2}\Big)\Big[\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-\sqrt3\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)\Big]=0$
$\Rightarrow\ \sin\frac{\text{x+y}}{2}=0$ Or, $\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-\sqrt3\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)=0$
$\Rightarrow\ \frac{\text{x+y}}{2}=0$ Or, $\tan\Big(\frac{\text{x}-\text{y}}{2}\Big)=\frac{1}{\sqrt3}=\tan\frac{\pi}{6}$
$\Rightarrow\ \text{x}=-\text{y}$ Or $, \frac{\text{x}-\text{y}}{2}=\frac{\pi}{6}$
$\Rightarrow\ \text{x}=-\text{y}$ Or $, \text{x}-\text{y}=\frac{\pi}{3}$
Case $1:$
When $\text{x}=-\text{y}$
In this case,
$\sin3\text{x}+\sin3\text{y}=\sin(-3\text{y})+\sin3\text{y}=-\sin3\text{y}+\sin3\text{y}=0$
Case $2:$
When $\text{x}-\text{y}=\frac{\pi}{3}$
Or, $3\text{x}=\pi+3\text{y}$
So, $\sin3\text{x}+\sin3\text{y}=\sin(\pi+3\text{y})+\sin3\text{y}$
$=\ -\sin3\text{y}+\sin3\text{y}$
$=\ 0$

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