4
$\begingroup$

To prove: $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$

My attempt at the solution: \begin{gather}\frac{\cos(x)\cos(2x)}{\sin(x)\sin(2x)}-\frac{\cos(2x)\cos(3x)}{\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\ \frac{\cos(x)\cos(2x)\sin(3x)-\cos(2x)\cos(3x)\sin(x)}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\ \frac{ \cos(2x)[ \cos(x)\sin(3x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\ \frac{\cos(2x)[\sin(4x)\sin(2x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\ \frac{\cos(2x)[2\sin(4x)\sin(2x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\ \frac{2\cos(2x)\sin(4x)}{\sin(x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\ \frac{2\cos(2x)\sin(4x)}{\sin(x)\sin(3x)}-\frac{\cos(4x)\cos(2x)}{2\sin(3x)\sin(x)}\end{gather}

The problem is, I don't know where to go from here (and due to so many calculations involved, I'm also not sure of the above result).

Also, if you see a more elegant way to solve this, please provide a hint (not the complete solution).

$\endgroup$
1
0
$\begingroup$

Here are your work upto the line I considered a problem occured: $$\frac{\cos(x)\cos(2x)}{\sin(x)\sin(2x)}-\frac{\cos(2x)\cos(3x)}{\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$

$$\frac{\cos(x)\cos(2x)\sin(3x)-\cos(2x)\cos(3x)\sin(x)}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$

$$\frac{ \cos(2x)[ \cos(x)\sin(3x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$

This line is incorrect to me. $$\frac{\cos(2x)[\sin(4x)\sin(2x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$

Using $\sin(2x) = \sin(3x -x) = \sin(3x)\cos(x) - \sin(x)\cos(3x)$, It should be $$\frac{\cos(2x)\sin(2x)}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$

Then cancle $sin2x$ $$\frac{\cos(2x)}{\sin(x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$

One more merge: $$\frac{\cos(2x) - \cos(3x)\cos(x)}{\sin(x)\sin(3x)}$$

Using $\cos(2x) = \cos(3x -x) = \cos(3x)\cos(x) + \sin(3x)\sin(x)$, the last one become $$\frac{\sin(3x)\sin(x)}{\sin(3x)\sin(x)} = 1$$

$\endgroup$
2
  • $\begingroup$ sorrry but you have literally copied this answer $\endgroup$ – Bhaskara-III Feb 10 '16 at 11:47
  • $\begingroup$ Copied from where? Both mine and Harish's answers are significantly different. Mine has an error in it, ans Harish uses a different equation and hence diverges from the solution that i was trying to continue. $\endgroup$ – Aayush Agrawal Feb 10 '16 at 13:55
3
$\begingroup$

Do it this way- Expand $\cot(3x-2x-x)$ in the

$$ \cot(A+B+C) = \dfrac{\cot(A)+\cot(B)+\cot(C)-3\cot(A)\cot(B)\cot(C)}{ 1-\cot(A)\cot(B)-\cot(B)\cot(C)-\cot(C)\cot(A)}$$

We know $\cot(0) = \infty$.
The denominator is zero.
So, ...you got it already.

$\endgroup$
4
1
$\begingroup$

Hints:

(1) Factor out $$\frac{\cos(2x)}{\sin(x)\sin(3x)}$$ from your last expression.

(2) Simplify $$2\sin(4x) - \frac{1}{2}\cos(4x).$$

Can you take it from here?

$\endgroup$
1
  • 2
    $\begingroup$ I'm sorry but i don't know what to do after multiplying $2\sin(4x)$ by 2 and taking out the $\frac{1}{2}$. $\endgroup$ – Aayush Agrawal Feb 10 '16 at 10:39
1
$\begingroup$

$$\cot(A+B)=\dfrac{1-\tan A\tan B}{\tan A+\tan B}=\dfrac{\cot A\cot B-1}{\cot B+\cot A}$$

$$\iff\cot A\cot B=1+\cot(A+B)[\cot B+\cot A]$$

Set $A=x, B=2x$

$$\cot(A-B)=\dfrac{1+\tan A\tan B}{\tan A-\tan B}=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$$

$$\iff\cot A\cot B=\cot(A-B)[\cot B-\cot A]-1$$

Set $A=3x,B=2x$

and $A=3x,B=x$

$\endgroup$
1
$\begingroup$

Notice, here is right approach followed by OP., $$LHS=\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x)$$ $$=\frac{\cos(x)\cos(2x)}{\sin(x)\sin(2x)}-\frac{\cos(2x)\cos(3x)}{\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$ $$=\frac{\cos(2x)[\sin(3x)\cos(x)-\sin(x)\cos(3x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$ using $\color{blue}{\sin A\cos B-\sin B\cos A=\sin(A-B)}$, $$=\frac{\cos(2x)[\sin(3x-x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$ $$=\frac{\cos(2x)}{\sin(x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$ $$=\frac{\cos(2x)-\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$ $$=\frac{\cos(3x-x)-\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$ using $\color{blue}{\cos(A-B)=\cos A \cos B+\sin A\sin B}$, $$=\frac{\cos(3x)\cos(x)+\sin(3x)\sin(x)-\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$ $$=\frac{\sin(3x)\sin(x)}{\sin(3x)\sin(x)}$$ $$=1=RHS$$

$\endgroup$
1
$\begingroup$

Its surprising that everyone else missed this out, but there's actually a very simple and elegant solution to this proof. So although this question is more than a year old, I still decided to post my proof.

I begin with a simple equation:

$$\cot(2x+x)=\cot{3x}$$

Now applying the formula: $\cot(A+B)=\frac{\cot{A}\cot{B}-1}{\cot{A}+\cot{B}}$, we get:

$$\frac{\cot{2x}\cot{x}-1}{\cot{2x}+\cot{x}}=\cot{3x}$$ $$\cot{2x}\cot{x}-1=\cot{2x}\cot{3x}+\cot{3x}\cot{x}$$ $$\cot{x}\cot{2x}-\cot{2x}\cot{3x}-\cot{3x}\cot{x}=1$$

and you're done.

$\endgroup$
0
$\begingroup$

Go slowly: \begin{align} \cot x\cot2x-\cot2x\cot3x &= \cot2x\left(\frac{\cos x}{\sin x}-\frac{\cos3x}{\sin3x}\right)\\[6px] &=\frac{\cos2x}{\sin2x}\frac{\sin3x\cos x-\cos3x\sin x}{\sin x\sin 3x} \\[6px] &=\frac{\cos2x}{\sin2x}\frac{\sin2x}{\sin x\sin 3x}\\[6px] &=\frac{\cos2x}{\sin x\sin 3x} \end{align} So you want to compute $$ \frac{\cos2x}{\sin x\sin 3x}-\frac{\cos3x}{\sin3x}\frac{\cos x}{\sin x} =\frac{\cos2x-\cos3x\cos x}{\sin x\sin3x} $$ Now $$ \cos2x=\cos(3x-x)=\cos3x\cos x+\sin3x\sin x $$ and you are done.

$\endgroup$
2
  • $\begingroup$ oopss... this is already answered/known, any other tip to prove? $\endgroup$ – Bhaskara-III Feb 10 '16 at 11:51
  • 1
    $\begingroup$ @Bhaskara-III The tip is to do the computation by examining a piece at a time, which avoids errors and better shows what's being used. $\endgroup$ – egreg Feb 10 '16 at 12:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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