# Exam question

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• Jun 17th 2008, 05:47 AM
MathMatters
Exam question
Hi everyone,

This was one of the questions on my exam. Nobody was able to solve it, so I'm starting to wonder if there is a solution to this problem. Please have a look at it:

Quote:

sinx .(1/sinx . 1/sin2x + 1/sin2x . 1/sin3x) = cot (ax) + cot (bx)

What's the value of a and b?

I hope for you people who put effort in this that there is a solution to this question. Would be kind of a stupid exam question if there isn't, wouldn't it?

Thanks
• Jun 18th 2008, 07:59 AM
TKHunny
Try a = -3 and b = 1. I really don't know how humans are supposed to find that.
• Jun 18th 2008, 09:45 AM
flyingsquirrel
Hello

$\sin x\left(\frac{1}{\sin x}\cdot\frac{1}{\sin (2x)}+\frac{1}{\sin (2x)}\cdot\frac{1}{\sin (3x)}\right)=\frac{\cos(ax)}{\sin(ax)}+\frac{\cos( bx)}{\sin(bx)}$

Let's factor $\frac{1}{\sin(2x)}$ :

$\frac{\sin x}{\sin (2x)}
\left( \frac{1}{\sin x}+\frac{1}{\sin (3x)}\right)=\frac{\cos(ax) \sin(bx)+\sin(ax)\cos(bx)}{\sin(ax)\sin(bx)}$

Using $\sin[(a+b)x]=\cos(ax) \sin(bx)+\sin(ax)\cos(bx)$ :

$\frac{\sin x}{\sin (2x)}\cdot\frac{\sin (3x)+\sin x}{\sin(3x)\sin x}=\frac{\sin[(a+b)x]}{\sin(ax)\sin(bx)}$

Notice that
\begin{aligned}\sin (3x)+\sin x
&=\sin (2x+x)+\sin(2x-x)\\
&=\sin(2x)\cos x+\sin x \cos(2x)+\sin(2x)\cos x-\sin x\cos(2x)\\
&=2\sin(2x)\cos x
\end{aligned}

thus the equality becomes

$\frac{2\sin (2x)\cos x}{\sin(3x)\sin(2x)}=\frac{\sin[(a+b)x]}{\sin(ax)\sin(bx)}$

which can be written
$\sin(ax)\times 2\sin(bx)\cos x=\sin(3x)\times\sin[(a+b)x]$

From this last equality one can see more easily that the solution given by TKHunny is correct... but finding the values of $a$ and $b$ without knowing them seems to be a bit more difficult.
• Jun 18th 2008, 10:23 AM
MathMatters
Thank you very much for the help. It's much appreciated.

This forum rules (Rock)

By the way, there are two possible solutions to this problem:

a=1 and b=-3
or
a=-3 and b=1