How do you find cot(sin^-1(p/q))cot(sin1(pq))?

4 Answers
Sep 6, 2016

+-sqrt(q^2-p^2)/p±q2p2p

Explanation:

Here, abs(p/q) <=1pq1, and so, abs(p)<= abs(q)|p||q|.

Let a=sin^(-1)(p/q) in Q1a=sin1(pq)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 acota

= cos a/sin a=cosasina

=sqrt(1-sin^2a)/sin a=1sin2asina

= sqrt(1 -p^2/q^2)/(p/q)=1p2q2pq

=+-sqrt(q^2-p^2)/p=±q2p2p

< 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

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

Sep 6, 2016

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...

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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

Oct 13, 2016

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