# Finding $\cot(\frac{\pi}{12})$

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: $\cot\theta = -\cot\theta$ ?

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

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

• Your formula is wrong! Try $\theta=\phi$ or $\theta=0$, but it's still $+1$. Apr 12, 2017 at 19:22
• Yeah, meant $\cot$ when I wrote $\cos$. @MichaelRozenberg Apr 12, 2017 at 19:29
• @Thomas Andrews That checks out. Apr 12, 2017 at 19:34

$\cot\frac{\pi}{12}=\frac{1+\cos\frac{\pi}{6}}{\sin\frac{\pi}{6}}=2+\sqrt3$

• 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 $$\frac{1+ \cos \frac \pi 6}{\sin \frac \pi 6}$$ Apr 12, 2017 at 19:43
• @amWhy Yes of course! Apr 12, 2017 at 19:59

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.

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

• 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. Apr 12, 2017 at 19:36