Find $\cot\left(\arccos\left(-\frac{1}{3}\right)-\pi\right)$

Question

Find $$\cot\left(\arccos\left(-\frac{1}{3}\right)-\pi\right)$$

This question is from my book. Is this question incorrect, because $\cot(\theta)$ is undefined at $\pi$?

Answer 1

$\arccos(-x)=\pi-\arccos{x}$.

Thus, $$\cot(\arccos(-\frac{1}{3})-\pi)=-\cot\arccos\frac{1}{3}=-\frac{\frac{1}{3}}{\sqrt{1-\frac{1}{9}}}=-\frac{1}{2\sqrt2}$$

Answer 2

As $\cot(n\pi+y)=\cot y$

$\cot\left(\arccos\left(-\dfrac13\right)-\pi\right)=\cot\left(\arccos\left(-\dfrac13\right)\right)$

Now if $\arccos\left(-\dfrac13\right)=y,\dfrac\pi2<y\le\pi$

$\implies\cos y=-\dfrac13,\sin y=+\sqrt{1-\cos^2y}=?$

Answer 3

Remember that $$ \cot(-\alpha)=-\cot\alpha \qquad \cot(\pi-\alpha)=-\cot\alpha $$ so, if $\alpha=\arccos(-1/3)$, you have $$ \cot(\alpha-\pi)=-\cot(\pi-\alpha)=\cot\alpha $$ Now express the cotangent in terms of the cosine, keeping in mind that $\pi/2<\alpha<\pi$.