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I'm trying to solve: $$\cot\frac{\pi}{12}$$

Given: $$\cot(\theta-\phi)=\frac{\cot\theta \cot\phi+1}{\cot\theta-\cot\phi}$$

And: $$\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}};\cot\frac{\pi}{4}=1$$

$$\frac{\frac{1}{\sqrt{3}}(1)+1}{\frac{1}{\sqrt{3}}-1}$$ $$\frac{1+\sqrt{3}}{1-\sqrt{3}}$$

The answer should be: $3.732(4sf)$ but I keep getting: $-3.732(4sf)$

Where am I going wrong?

Is it because: $\cot\theta = -\cot\theta$ ?

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4 Answers 4

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Your error is your formula. It is:

$$\cot(\theta-\phi)=\frac{\cot(\theta)\cot(\phi)+1}{\cot\phi-\cot\theta}$$

Notice the denominator order. You switched them, hence switching the sign.

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    $\begingroup$ Your formula is wrong! Try $\theta=\phi$ or $\theta=0$, but it's still $+1$. $\endgroup$
    – Michael Rozenberg
    Apr 12, 2017 at 19:22
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    $\begingroup$ Yeah, meant $\cot$ when I wrote $\cos$. @MichaelRozenberg $\endgroup$
    – Thomas Andrews
    Apr 12, 2017 at 19:29
  • $\begingroup$ @Thomas Andrews That checks out. $\endgroup$
    – Kantura
    Apr 12, 2017 at 19:34
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$\cot\frac{\pi}{12}=\frac{1+\cos\frac{\pi}{6}}{\sin\frac{\pi}{6}}=2+\sqrt3$

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  • $\begingroup$ Do you mind if I enlarge your fractions, Michael? By either double dollar-signs, or else $\dfrac$ for the teeny fractions (especially the $\frac{1+ \cos \frac \pi 6}{\sin \frac \pi 6}$ or $$\frac{1+ \cos \frac \pi 6}{\sin \frac \pi 6}$$ $\endgroup$
    – amWhy
    Apr 12, 2017 at 19:43
  • $\begingroup$ @amWhy Yes of course! $\endgroup$
    – Michael Rozenberg
    Apr 12, 2017 at 19:59
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Let's see where you might be going wrong...

$\cot(\theta-\phi)=\frac{\cos\theta \cos\phi+\sin\theta\sin\phi}{\sin\theta\sin\phi-\cos\theta\sin\phi}$ divide top and bottom by $\sin\theta\sin\phi$

$\cot(\theta-\phi)=\frac{\cot\theta \cot\phi+1}{\cot\phi-\cot\theta}$

Looks like you have the sign of the denominator flipped.

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Hint: use that $$\cot(2x)=\frac{1}{2}\left(\cot(x)-\tan(x)\right)$$ and now $$\cot\left(2\frac{\pi}{12}\right)=\frac{1}{2}\left(\left(\cot\frac{\pi}{12}\right)-\tan\left(\frac{\pi}{12}\right)\right)$$

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    $\begingroup$ I'm sincerely uncertain and curious here: how is this helpful, it will still require the computation of $\frac 12 \left(\cot \left(\frac{\pi}{24}\right) - \tan\left(\frac \pi{24}\right)\right)$. Perhaps I've worked too long today (really have), so I'm not seeing how this hint will work. But I'd be happy to get educated on this. $\endgroup$
    – amWhy
    Apr 12, 2017 at 19:36

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