Showing whether two numbers are equal or not

$\dfrac{\sin (2x+y)}{\sin (2x)} =\dfrac{\sin (x+2y)}{\sin (2y)}$,where $0<x,y\le\dfrac{\pi}{4}$ .

Can I show that $x=y$ or find two numbers $x,y$ such that $x\not=y$?

• $x=0.1$ and $y=0.2$ are two numbers with $x\not= y$. – Phira Nov 8 '11 at 11:14
• @Phira by $x=0.1$ and $y=0.2$ do not solve the equality, the lhs is $1.96$ and rhs is $1.23$ – Sasha Nov 8 '11 at 11:16
• Interesting structure here. – Ragib Zaman Nov 8 '11 at 11:49
• In particular, looking at Ragib's graph, there are no solutions with $0 \leq x,y \leq \pi/4$. – David E Speyer Nov 8 '11 at 12:07
• ([curved surface]i.imgur.com/UN8qA.jpg[/IMG]) – Peđa Terzić Nov 8 '11 at 12:40

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 $\cos x=\cos y\Rightarrow x=y$

• That's nice.Thanks – Leitingok Nov 8 '11 at 13:51

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 $0<x,y\le\frac{\pi}{4}$, the left hand side of $(5)$ is positive, and the denominator of the right hand side is also positive; therefore, the numerator of the right hand side must be positive; that is, $0<\frac{3}{2}\!\!(x+y)\le\frac{\pi}{2}$. Thus, the left hand side is less than $1$ and the right hand side is greater than $1$. Contradiction; therefore, $x=y$.