Question

If sinx+siny=3(cosy-cosx) then value of sin3x/sin3y is

Answer 1

sin3x/sin3y=-1. I hope this helps you

Answer 2

Answer:

The value of $\frac{\sin 3x}{\sin 3y}=-1$

Step-by-step explanation:

Given : Expression $\sin x+\sin y=3(\cos y-\cos x)$

To find : The value of $\frac{\sin 3x}{\sin 3y}$

Solution :

Expression $\sin x+\sin y=3(\cos y-\cos x)$

$\sin x+\sin y=3\cos y-3\cos x$

$\sin x+3\cos x=3\cos y-\sin y$

Substitute, $1=r\cos\theta$ and  $3=r\sin\theta$

$r\cos\theta\sin x+r\sin\theta\cos x=r\sin\theta\cos y-r\cos\theta\sin y$

$r(\cos\theta\sin x+\sin\theta\cos x)=r(\sin\theta\cos y-\cos\theta\sin y)$

$\sin(x+\theta)=\sin(\theta-y)$

$x+\theta=\theta-y$

$x=-y$

Now, We substitute the value of x in $\frac{\sin 3x}{\sin 3y}$

$=\frac{\sin 3(-y)}{\sin 3y}$

$=\frac{\sin (-3y)}{\sin 3y}$

We know, $\sin(-\theta)=-\sin\theta$

$=\frac{-\sin (3y)}{\sin 3y}$

$=-1$

Therefore, The value of $\frac{\sin 3x}{\sin 3y}=-1$