# Exam question

• Jun 17th 2008, 04:47 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, 06:59 AM
TKHunny
Try a = -3 and b = 1. I really don't know how humans are supposed to find that.
• Jun 18th 2008, 08:45 AM
flyingsquirrel
Hello

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

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

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

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

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

Notice that
\displaystyle \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

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

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

From this last equality one can see more easily that the solution given by TKHunny is correct... but finding the values of $\displaystyle a$ and $\displaystyle b$ without knowing them seems to be a bit more difficult.
• Jun 18th 2008, 09:23 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