Question

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

Answer 1

`+-sqrt(q^2-p^2)/p`

Explanation:

Here, `abs(p/q) <=1`, and so, `abs(p)<= abs(q)`.

Let `a=sin^(-1)(p/q) in Q1`, 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 a/sin a`

`=sqrt(1-sin^2a)/sin a`

`= sqrt(1 -p^2/q^2)/(p/q)`

`=+-sqrt(q^2-p^2)/p`

`< 0, if p < 0 and q > 0 or p > 0 and q < 0`.

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

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

`sin a = = -1/2 and a = -pi/6`, and so, `cot a =cot(-pi/6) = - sqrt 3`

Answer 2

`sqrt(q^2-p^2)/p`

Explanation:

Let's start with the original:

`cot(sin^-1(p/q))`

First thing to do is to understand what `sin^-1(p/q)` is telling us. The sine function is a ratio with the Opposite side `-:` the Hypotenuse. So we know that Opp = p and Hyp = q.

The question is asking for the cotangent, which is Adjacent side `-:` 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="adj"/"opp"=sqrt(q^2-p^2)/p`

Answer 3

`cot(sin^(-1) (p/q)) = sqrt(1-p^2/q^2)/(p/q)`

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

`sqrt(q^2-p^2)/p`

Explanation:

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

where

`sin theta = "opposite"/"hypotenuse" = p/q`

`cot theta = "adjacent"/"opposite" = 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 [-1, 1]`, then `sin^(-1) s = theta` for some `theta in [-pi/2, pi/2]`

If `theta in [-pi/2, pi/2]` then `cos theta >= 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(sin^(-1) s) = cot theta = cos theta / sin theta = sqrt(1-s^2)/s`

Substituting `s = p/q`, we find:

`cot(sin^(-1) (p/q)) = sqrt(1-p^2/q^2)/(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(sin^(-1) (p/q)) = sqrt(1-p^2/q^2)/(p/q)`

`color(white)(cot(sin^(-1) (p/q))) = sqrt(1-p^2/q^2)/(p/q) * q/q`

`color(white)(cot(sin^(-1) (p/q))) = sqrt(q^2-p^2)/p`

Answer 4

`cot(sin^-1(p/q))->"possible for " q>p`

`=cot(csc^-1(q/p))`

`=cot(cot^-1sqrt((q^2/p^2-1)))`

`=sqrt(q^2-p^2)/p`