Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
$x=0.1$ 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
3  
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 Speyer Nov 8 '11 at 12:07
    
([curved surface]i.imgur.com/UN8qA.jpg[/IMG]) –  pedja Nov 8 '11 at 12:40

2 Answers 2

up vote 6 down vote accepted

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

share|improve this answer
    
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, $x=y$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.