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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).

gebruiker
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Aayush Agrawal
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  • Multiply out by $$\tan x\tan2x\tan3x$$ and use $$3x+(-2x)+(-x)=0\cdot\pi$$ and http://math.stackexchange.com/questions/477364/prove-that-tan-a-tan-b-tan-c-tan-a-tan-b-tan-c-abc-180-circ/477387 OR http://math.stackexchange.com/questions/477364/prove-that-tan-a-tan-b-tan-c-tan-a-tan-b-tan-c-abc-180-circ – lab bhattacharjee Feb 10 '16 at 10:35

7 Answers7

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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.

lab bhattacharjee
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Win Vineeth
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1

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?

Jose Arnaldo Bebita Dris
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$$\cot(A+B)=\dfrac{1-\tan A\tan B}{\tan A+\tan B}=\dfrac{\cot A\cot B-1}{\cot B+\cot A}$$

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

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 $A=3x,B=x$

lab bhattacharjee
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1

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 $\color{blue}{\cos(A-B)=\cos A \cos B+\sin A\sin B}$, $$=\frac{\cos(3x)\cos(x)+\sin(3x)\sin(x)-\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$ $$=\frac{\sin(3x)\sin(x)}{\sin(3x)\sin(x)}$$ $$=1=RHS$$

Harish Chandra Rajpoot
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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.

Community
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Raknos13
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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.

egreg
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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 $$\frac{\sin(3x)\sin(x)}{\sin(3x)\sin(x)} = 1$$

runaround
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  • sorrry but you have literally copied this answer – Bhaskara-III Feb 10 '16 at 11:47
  • 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. – Aayush Agrawal Feb 10 '16 at 13:55