# How do you find cot(sin^-1(p/q))?

Sep 6, 2016

$\pm \frac{\sqrt{{q}^{2} - {p}^{2}}}{p}$

#### Explanation:

Here, $\left\mid \frac{p}{q} \right\mid \le 1$, and so, $\left\mid p \right\mid \le \left\mid q \right\mid$.

Let $a = {\sin}^{- 1} \left(\frac{p}{q}\right) \in Q 1$, when p and q have the same sign, or

Q4#, when p and q have opposite signs.

Here, in both quadrants, cosine is positive. So, cotangent and sine

have the same sign.

The given expression is

$\cot a$

$= \cos \frac{a}{\sin} a$

$= \frac{\sqrt{1 - {\sin}^{2} a}}{\sin} a$

$= \frac{\sqrt{1 - {p}^{2} / {q}^{2}}}{\frac{p}{q}}$

$= \pm \frac{\sqrt{{q}^{2} - {p}^{2}}}{p}$

$< 0 , \mathmr{if} p < 0 \mathmr{and} q > 0 \mathmr{and} p > 0 \mathmr{and} q < 0$.

In brief, the negative sign is necessary, despite p > 0.

For example, if p = 1 and $q = - 2$,

$\sin a = = - \frac{1}{2} \mathmr{and} a = - \frac{\pi}{6}$, and so, $\cot a = \cot \left(- \frac{\pi}{6}\right) = - \sqrt{3}$

$\frac{\sqrt{{q}^{2} - {p}^{2}}}{p}$

#### Explanation:

$\cot \left({\sin}^{-} 1 \left(\frac{p}{q}\right)\right)$

First thing to do is to understand what ${\sin}^{-} 1 \left(\frac{p}{q}\right)$ is telling us. The sine function is a ratio with the Opposite side $\div$ the Hypotenuse. So we know that Opp = p and Hyp = q.

The question is asking for the cotangent, which is Adjacent side $\div$ the Opposite side. So we need to find the Adjacent side.

We can find the Adjacent in terms of the Opposite and Hypotenuse using the Pythagorean Theorem:

${a}^{2} + {b}^{2} = {c}^{2}$

${p}^{2} + {b}^{2} = {q}^{2}$

$b = \sqrt{{q}^{2} - {p}^{2}}$

We now know all the sides so we can answer the cotangent part of the question:

$\cot = \text{adj"/"opp} = \frac{\sqrt{{q}^{2} - {p}^{2}}}{p}$

Sep 6, 2016

$\cot \left({\sin}^{- 1} \left(\frac{p}{q}\right)\right) = \frac{\sqrt{1 - {p}^{2} / {q}^{2}}}{\frac{p}{q}}$

If we are told that $q > 0$ then this simplifies to:

$\frac{\sqrt{{q}^{2} - {p}^{2}}}{p}$

#### Explanation:

If $p , q > 0$ then we can picture this as a triangle...

where

$\sin \theta = \text{opposite"/"hypotenuse} = \frac{p}{q}$

$\cot \theta = \text{adjacent"/"opposite} = \frac{\sqrt{{q}^{2} - {p}^{2}}}{p}$

If we cannot assume $p , q > 0$, then it may be better to deal with it algebraically...

Given Real number $s \in \left[- 1 , 1\right]$, then ${\sin}^{- 1} s = \theta$ for some $\theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

If $\theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ then $\cos \theta \ge 0$, so we can use the principal, non-negative square root:

$\cos \theta = \sqrt{1 - {\sin}^{2} \theta} = \sqrt{1 - {s}^{2}}$

So we find:

$\cot \left({\sin}^{- 1} s\right) = \cot \theta = \cos \frac{\theta}{\sin} \theta = \frac{\sqrt{1 - {s}^{2}}}{s}$

Substituting $s = \frac{p}{q}$, we find:

$\cot \left({\sin}^{- 1} \left(\frac{p}{q}\right)\right) = \frac{\sqrt{1 - {p}^{2} / {q}^{2}}}{\frac{p}{q}}$

This will have the correct sign, regardless of whether $p$ and or $q$ are positive or negative.

If we are additionally told that $q > 0$ then we find:

$\cot \left({\sin}^{- 1} \left(\frac{p}{q}\right)\right) = \frac{\sqrt{1 - {p}^{2} / {q}^{2}}}{\frac{p}{q}}$

$\textcolor{w h i t e}{\cot \left({\sin}^{- 1} \left(\frac{p}{q}\right)\right)} = \frac{\sqrt{1 - {p}^{2} / {q}^{2}}}{\frac{p}{q}} \cdot \frac{q}{q}$

$\textcolor{w h i t e}{\cot \left({\sin}^{- 1} \left(\frac{p}{q}\right)\right)} = \frac{\sqrt{{q}^{2} - {p}^{2}}}{p}$

Oct 13, 2016

$\cot \left({\sin}^{-} 1 \left(\frac{p}{q}\right)\right) \to \text{possible for } q > p$

$= \cot \left({\csc}^{-} 1 \left(\frac{q}{p}\right)\right)$

$= \cot \left({\cot}^{-} 1 \sqrt{\left({q}^{2} / {p}^{2} - 1\right)}\right)$

$= \frac{\sqrt{{q}^{2} - {p}^{2}}}{p}$