Trigonometric Functions — MATHS STD 11 Science — Question

2. $0$

Solution:

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, $\tan\Big(\frac{\text{x}-\text{y}}{2}\Big)=\frac{1}{\sqrt3}=\tan\frac{\pi}{6}$

$\Rightarrow\ \text{x}=-\text{y}\text{ Or }, \frac{\text{x}-\text{y}}{2}=\frac{\pi}{6}$

$\Rightarrow\ \text{x}=-\text{y}\text{ Or }, \text{x}-\text{y}=\frac{\pi}{3}$

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, $\sin3\text{x}+\sin3\text{y}=\sin(\pi+3\text{y})+\sin3\text{y}$

$=\ -\sin3\text{y}+\sin3\text{y}$

$=\ 0$