$\dfrac{\sin (2x+y)}{\sin (2x)} =\dfrac{\sin (x+2y)}{\sin (2y)}$,where $0
Can I show that $x=y $ or find two numbers $x,y$ such that $x\not=y$?
$\dfrac{\sin (2x+y)}{\sin (2x)} =\dfrac{\sin (x+2y)}{\sin (2y)}$,where $0
Can I show that $x=y $ or find two numbers $x,y$ such that $x\not=y$?
Edited: I think I got it...there was typo in the first try.
After cross-multiplication, we get $[2\sin y\sin(2x+y)]\cos y-[2\sin x\sin(x+2y)]\cos x=0$
$\Rightarrow[\cos(2x)-\cos2(x+y)]\cos y-[\cos(2y)-\cos2(x+y)]\cos x=0$
$\Rightarrow\cos2(x+y)[\cos x-\cos y]+[(2\cos^2x-1)\cos y-(2\cos^2y-1)\cos x]=0$
$\Rightarrow[\cos x-\cos y][\cos2(x+y)+2\cos x\cos y+1]=0$
$\Rightarrow[\cos x-\cos y][\cos^2(x+y)+\cos x\cos y]=0$
Note that in the specified range, the second factor is strictly positive. Hence we must have $\cos x=\cos y\Rightarrow x=y$
Start with $ \frac{\sin (2x+y)}{\sin (2x)} =\frac{\sin (x+2y)}{\sin (2y)}\tag{1} $ Regrouping $(1)$, we get $ \frac{\sin\left(\frac{3}{2}(x+y)+\frac{1}{2}(x-y)\right)}{\sin\left((x+y)+(x-y)\right)}=\frac{\sin\left(\frac{3}{2}(x+y)-\frac{1}{2}(x-y)\right)}{\sin\left((x+y)-(x-y)\right)}\tag{2} $ Expanding $(2)$, yields $ \begin{align} &\frac{\sin\frac{3}{2}\!\!(x+y)\;\cos\frac{1}{2}\!\!(x-y)+\cos\frac{3}{2}\!\!(x+y)\;\sin\frac{1}{2}\!\!(x-y)}{\sin(x+y)\;\cos(x-y)+\cos(x+y)\;\sin(x-y)}\\ &=\frac{\sin\frac{3}{2}\!\!(x+y)\;\cos\frac{1}{2}\!\!(x-y)-\cos\frac{3}{2}\!\!(x+y)\;\sin\frac{1}{2}\!\!(x-y)}{\sin(x+y)\;\cos(x-y)-\cos(x+y)\;\sin(x-y)}\tag{3} \end{align} $ Since $\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{b}{a}=\frac{d}{c}$, $(3)$ implies $ \frac{\tan\frac{1}{2}\!\!(x-y)}{\tan\frac{3}{2}\!\!(x+y)}=\frac{\tan(x-y)}{\tan(x+y)}\tag{4} $ Assume $x\not=y$. We can rearrange $(4)$ to get $ \frac{\tan\frac{1}{2}\!\!(x-y)}{\tan(x-y)}=\frac{\tan\frac{3}{2}\!\!(x+y)}{\tan(x+y)}\tag{5} $ Since $0