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Please help! I don't know how to solve this question. I tried putting the whole thing equal to "k" and then calculating values of x,y and z in terms of k and putting there. But it messes up the question even more. Any help would be greatly appreciated.

Also, I have no clue what Σxy means but I assume it means xy + yz + zx = 0

Shubham
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2 Answers2

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Assuming that your guess is correct and we are trying to find $xy+yz+zx$. . .

Case 1: $x=0$. Since $\sin(\theta+2\pi/3)$ and $\sin(\theta+4\pi/3)$ cannot both be zero we have $y=0$ or $z=0$, and so $$xy+yz+zx=yz=0\ .$$

Case 2: $x\ne0$. Then $$\eqalign{\frac{xy+yz+zx}{x^2} &=\frac{y}{x}+\frac{z}{x}+\frac{y}{x}\frac{z}{x}\cr &=\frac{\sin\theta}{\sin(\theta+2\pi/3)} +\frac{\sin\theta}{\sin(\theta+4\pi/3)} +\frac{\sin^2\theta}{\sin(\theta+2\pi/3)\sin(\theta+4\pi/3)}\cr &=\frac{(\sin\theta)(\sin(\theta+4\pi/3)+\sin(\theta+2\pi/3))+\sin^2\theta}{\sin(\theta+2\pi/3)\sin(\theta+4\pi/3)}\cr &=\frac{(\sin\theta)(\sin(\theta+2\pi/3)+\sin(\theta-2\pi/3))+\sin^2\theta}{\sin(\theta+2\pi/3)\sin(\theta+4\pi/3)}\ .\cr}$$ Now use a standard trig formula to simplify $\sin(\theta+\alpha)+\sin(\theta-\alpha)$.

David
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  • I would seriously give you a cookie if I were there. A big cyber hug from me. But please can you elaborate on how this sin(θ+4π/3)+sin(θ+2π/3) got converted to sin(θ+2π/3)+sin(θ−2π/3) ? I couldn't catch this step. Instead I converted that to (2sin(θ+π)cos(2π/3)) which gives 2sin^2θ in the numerator. – Shubham Sep 18 '14 at 01:02
  • Just $\sin(x)=\sin(x-2\pi)$, so$$\sin(\theta+4\pi/3)=\sin((\theta+4\pi/3)-2\pi)=\sin(\theta-2\pi/3)\ .$$There are other methods which will also work fine. – David Sep 18 '14 at 01:06
  • Ahh that is my weakest area in this topic. Thanks for clarifying. :) – Shubham Sep 18 '14 at 01:08
  • Hey, can you answer this question too? I will be very grateful to you http://math.stackexchange.com/questions/935816/if-%CE%B1-%CE%B2-are-two-values-of-%CE%B8-satisfying-equation-cos%CE%B8-a-sin%CE%B8-b-1-c-then-prove – Shubham Sep 18 '14 at 01:28
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Observe that $\sin\left(3A\right)=\sin3A,$

$\sin3\left(\dfrac{2\pi}3+A\right)=\sin3A$ and $\sin3\left(\dfrac{4\pi}3+A\right)=\sin3A$

So, if $\sin3x=\sin3A,3\sin x-4\sin^3x=\sin3A\iff4\sin^3x-3\sin x+\sin3A=0$

The roots of the above cubic equation are $\sin A,\sin\left(\dfrac{2\pi}3+A\right),\sin\left(\dfrac{4\pi}3+A\right)$

Using Vieta's formula, $\implies\sum \sin A=\dfrac04$

$\implies\sum\sin A\sin\left(\dfrac{2\pi}3+A\right)=-\dfrac34$

$\implies\sin A\sin\left(\dfrac{2\pi}3+A\right)\sin\left(\dfrac{4\pi}3+A\right)=-\dfrac{\sin3A}4$

lab bhattacharjee
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  • This was an interesting way of doing it but I can't understand how that proves Σxy = 0 Sorry but can you please elaborate it further? This was really short and I plan on using it once I get a good hold of it. :) – Shubham Sep 18 '14 at 11:12
  • @Shubham, If $x\sin\theta=\cdots=k, \frac1x=\frac{\sin\theta}k$ So, $$\sum xy=xyz(\sum\frac1z)=0$$ – lab bhattacharjee Sep 18 '14 at 12:29
  • Sorry I still couldn't understand :(( Can you please write the complete solution and then maybe I will ask which steps aren't clear to me. It will be very helpful to me. – Shubham Sep 18 '14 at 12:36
  • @Shubham, We have $$xy+yz+zx=xyz\left(\frac1x+\frac1y+\frac1z\right)=\frac{xyz}k\left(\sin A+\sin\left(\dfrac{2\pi}3+A\right)+\sin\left(\dfrac{4\pi}3+A\right)\right)=?$$ – lab bhattacharjee Sep 18 '14 at 12:41
  • Thank you so much! Now I understood completely :) – Shubham Sep 18 '14 at 12:49