Results 1 to 14 of 14
Like Tree9Thanks
  • 1 Post By Idea
  • 1 Post By Idea
  • 2 Post By MarkFL
  • 1 Post By MarkFL
  • 2 Post By Idea
  • 1 Post By SlipEternal
  • 1 Post By Idea

Thread: inverse trigonometric function

  1. #1
    Member
    Joined
    Aug 2011
    Posts
    102

    inverse trigonometric function

    The problem is

    Show that Sin-1(12/13) + Cos-1(4/5) + Tan-1(63/16) = Pi

    I converted Sin-1(12/13) as Tan-1(12/5)

    Cos-1(4/5) as Tan-1(3/4)

    Then I took the above problem as


    Tan-1(12/5) + Tan-1(3/4) = Tan-1{[12/5 + 3/4]/[1-(12/5)(3/4)]} = Tan-1(-63/16) = - Tan-1(63/16)

    The above problem becomes zero (0).

    How I could conclude that the answer is Pi?

    or What is wrong with my working.

    Kindly guide me.

    with warm regards,

    Aranga
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    843
    Thanks
    396

    Re: inverse trigonometric function

    $\displaystyle \tan ^{-1}a+\tan ^{-1}b\overset{\text{??}}{=}\tan ^{-1}\left(\frac{a+b}{1-a b}\right)$
    Thanks from arangu1508
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2011
    Posts
    102

    Re: inverse trigonometric function

    I could infer from your reply they are not equal.

    Thanks. Now I revisited the formula given in the book.

    They have mentioned that

    tan-1x + tan-1y = tan-1{[x + y]/[1-xy]} only when xy<1.

    Why such condition they have given I did not understand.

    in this case, xy is greater than 1. So it fails. That much I understand.

    Now I do not know how to solve this problem or how to proceed further.

    Kindly guide me.

    with warm regards,

    Aranga
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    843
    Thanks
    396

    Re: inverse trigonometric function

    if $\displaystyle x y>1$

    $\displaystyle \tan ^{-1}x+\tan ^{-1}y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) + \pi $

    P.S. I am not sure if this is correct
    Last edited by Idea; Jul 9th 2018 at 04:07 AM.
    Thanks from arangu1508
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2011
    Posts
    102

    Re: inverse trigonometric function

    Thanks. As you said If I put I am getting the answer.

    Whether it is correct?

    If it is so, how it is correct?

    with warm regards,

    Aranga
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,069
    Thanks
    780

    Re: inverse trigonometric function

    I think I would arrange the equation as:

    $\displaystyle \arcsin\left(\frac{12}{13}\right)+ \arccos\left(\frac{4}{5}\right)= \pi-\arctan\left(\frac{63}{16}\right)$

    Take the sine of both sides (apply the angle sum identity for sine on the left and the identity $\displaystyle \sin(\pi-\theta)=\sin(\theta)$ on the right):

    $\displaystyle \sin\left(\arcsin\left(\frac{12}{13}\right)\right) \cos\left(\arccos\left(\frac{4}{5}\right)\right)+ \cos\left(\arcsin\left(\frac{12}{13}\right)\right) \sin\left(\arccos\left(\frac{4}{5}\right)\right)= \sin\left(\arctan\left(\frac{63}{16}\right)\right)$

    $\displaystyle \frac{12}{13}\cdot\frac{4}{5}+ \frac{5}{13}\cdot\frac{3}{5}=\frac{63}{65}$

    $\displaystyle \frac{48+15}{65}=\frac{63}{65}$

    $\displaystyle \frac{63}{65}=\frac{63}{65}\quad\checkmark$
    Thanks from topsquark and arangu1508
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2011
    Posts
    102

    Re: inverse trigonometric function

    please see the post of IDEA.

    Whether what was said is correct?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,069
    Thanks
    780

    Re: inverse trigonometric function

    Quote Originally Posted by arangu1508 View Post
    please see the post of IDEA.

    Whether what was said is correct?
    It looks correct to me, and I think we only require $\displaystyle xy\ne1$.
    Thanks from arangu1508
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2011
    Posts
    102

    Re: inverse trigonometric function

    thanks.

    It was very useful.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    843
    Thanks
    396

    Re: inverse trigonometric function

    Quote Originally Posted by MarkFL View Post
    I think I would arrange the equation as:

    $\displaystyle \arcsin\left(\frac{12}{13}\right)+ \arccos\left(\frac{4}{5}\right)= \pi-\arctan\left(\frac{63}{16}\right)$

    Take the sine of both sides (apply the angle sum identity for sine on the left and the identity $\displaystyle \sin(\pi-\theta)=\sin(\theta)$ on the right):

    $\displaystyle \sin\left(\arcsin\left(\frac{12}{13}\right)\right) \cos\left(\arccos\left(\frac{4}{5}\right)\right)+ \cos\left(\arcsin\left(\frac{12}{13}\right)\right) \sin\left(\arccos\left(\frac{4}{5}\right)\right)= \sin\left(\arctan\left(\frac{63}{16}\right)\right)$

    $\displaystyle \frac{12}{13}\cdot\frac{4}{5}+ \frac{5}{13}\cdot\frac{3}{5}=\frac{63}{65}$

    $\displaystyle \frac{48+15}{65}=\frac{63}{65}$

    $\displaystyle \frac{63}{65}=\frac{63}{65}\quad\checkmark$
    This argument is not valid.

    To see this, start with the following equation which is obviously not correct

    $\displaystyle \sin ^{-1}\frac{12}{13}+\cos ^{-1}\frac{4}{5}=\tan ^{-1}\frac{63}{16}$

    and apply $sin$ to both sides to get

    $\displaystyle \frac{63}{65}=\frac{63}{65}$

    so taking sin of both sides and getting equal answers does not prove the original assertion
    Thanks from SlipEternal and MarkFL
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,523
    Thanks
    1401

    Re: inverse trigonometric function

    Three angles in $(0,\pi)$ add up to $\pi$ implies there exists a triangle with those angles as measures. Can that fact be used to prove the assertion?
    Last edited by SlipEternal; Jul 10th 2018 at 04:50 AM.
    Thanks from arangu1508
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,069
    Thanks
    780

    Re: inverse trigonometric function

    Yes...I bungled all over the place in this thread.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    843
    Thanks
    396

    Re: inverse trigonometric function

    if x > 0 , y > 0 , and xy > 1 then

    $\displaystyle \tan ^1x+\tan ^{-1}y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)+\pi $

    To prove this let $\displaystyle a=\tan ^{-1}x$, $\displaystyle b=\tan ^{-1}y$ where $\displaystyle 0<a,b<\pi /2$

    $\displaystyle \tan (a+b)=\frac{x+y}{1-x y}<0$

    therefore $\displaystyle \pi /2 < a+b < \pi $

    this gives

    $\displaystyle \tan ^{-1 }(\tan (a+b))=a+b-\pi $

    done.

    Another proof, using calculus of two variables, let

    $\displaystyle g(x,y)=\tan ^1x+\tan ^{-1}y-\tan ^{-1}\left(\frac{x+y}{1-x y}\right) $ , $\displaystyle x y\neq 1 $

    verify that both partial derivatives are zero in each of the three regions where $\displaystyle g$ is defined.

    therefore $\displaystyle g$ is constant. Check one point in each region to find out that the constants are $\displaystyle 0,\pi ,- \pi $
    Thanks from arangu1508
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Aug 2011
    Posts
    102

    Re: inverse trigonometric function

    Thank you very much. It is very useful. I am able to follow the first proof.

    So, In my case, x > 0, y > 0 and xy > 1,

    I can straight away conclude that

    tan-1x + tan-1y = tan-1{[x + y]/[1-xy]} + pi

    with warm regards,

    Aranga
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse Trigonometric Function
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Dec 26th 2016, 01:55 AM
  2. Replies: 2
    Last Post: Jan 16th 2014, 01:57 AM
  3. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Jul 5th 2009, 10:00 PM
  4. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Apr 28th 2009, 10:31 AM
  5. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Oct 15th 2008, 05:08 AM

/mathhelpforum @mathhelpforum