# Thread: cot-1(x) in terms of sin-1

1. ## cot-1(x) in terms of sin-1

problem is cot-1(x) in terms of sin-1

My attempt at the solution.
drawing a right triangle where cosy = x, sin y = 1 and (x^2+1)^1/2= hypothenuse.
trig identity 1+ cot^2(y)=1/sin^2(y)
then just a cross multiply and change terms for coty which is x and get
siny= 1/(1+x^2)^1/2. inversing it into
sin-1(1/(1+x^2)^1/2))=y

Solution says cot−1 x = sin−1 (sgn x /(1+x^2)^1/2).
earlier they also state that

If y = cot−1 x, then x = cot y and 0 < y < pi/2.

What i dont get is the sgn x part. I know its |X|/x but i dont understand how i am supposed to get it in to my solution.

Thankful for any help.

2. Originally Posted by Zamzen
problem is cot-1(x) in terms of sin-1

My attempt at the solution.
drawing a right triangle where cosy = x, sin y = 1 and (x^2+1)^1/2= hypothenuse.
trig identity 1+ cot^2(y)=1/sin^2(y)
then just a cross multiply and change terms for coty which is x and get
siny= 1/(1+x^2)^1/2. inversing it into
sin-1(1/(1+x^2)^1/2))=y

Solution says cot−1 x = sin−1 (sgn x /(1+x^2)^1/2).
earlier they also state that

If y = cot−1 x, then x = cot y and 0 < y < pi/2.

What i dont get is the sgn x part. I know its |X|/x but i dont understand how i am supposed to get it in to my solution.

Thankful for any help.
Dear Zamzen,

Let, $\cot^{-1}x=\sin^{-1}y=z$

$x=\cot z=\dfrac{\cos z}{\sin z}=\dfrac{\sqrt{1-\sin^{2}z}}{\sin z}$----------(1)

$y=\sin z$------------(2)

From (1) and (2); $x=\dfrac{\sqrt{1-y^2}}{y}\Rightarrow{y=\pm\dfrac{1}{\sqrt{1+x^2}}}$

Therefore, $\cot^{-1}x=\sin^{-1}\left(\pm\dfrac{1}{\sqrt{1+x^2}}\right)$

Since $sgn~ x=\pm 1~ when~x\neq 0$

$\cot^{-1}x=\sin^{-1}\left(\dfrac{sgn~x}{\sqrt{1+x^2}}\right)~;~x\neq 0$

Hope you understood. When we are dealing with traingles we do not consider the negative sign for the length of the hypotenuse. That is why for you did not get the sign function (sgn x) for your answer.