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

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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.