Results 1 to 2 of 2

Math Help - cot-1(x) in terms of sin-1

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    32

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by Zamzen View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. in terms of X 10^-n
    Posted in the Algebra Forum
    Replies: 0
    Last Post: November 22nd 2009, 07:54 AM
  2. Replies: 2
    Last Post: November 15th 2009, 07:07 PM
  3. To sum n terms
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: December 19th 2008, 08:48 PM
  4. tan, cot in terms of p,q
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 6th 2008, 05:22 PM
  5. sum of 1st nth terms
    Posted in the Algebra Forum
    Replies: 9
    Last Post: January 25th 2008, 06:11 PM

Search Tags


/mathhelpforum @mathhelpforum