Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By skeeter

Thread: Puzzling Inverse Trig Function question #2

  1. #1
    Newbie
    Joined
    Apr 2017
    From
    philadelphia
    Posts
    3

    Puzzling Inverse Trig Function question #2

    Puzzling Inverse Trig Function question #2-159.jpg

    My question here is how they turned the answer into a fraction.

    My process would go;

    sin ( 2arccsc (13/5)) = sin (2 arcsin (5/13)) = sin (2 arcsin (0.3846153)) = sin (2 (22.619)) = sin (45.2397) = 0.71005917

    How do you go about turning that into a fraction?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,191
    Thanks
    3686

    Re: Puzzling Inverse Trig Function question #2




    let $\theta = \text{ arccsc}\left(\dfrac{13}{5}\right)$, assume $0<\theta<\dfrac{\pi}{2}$

    $\sin(2\theta) = 2\sin{\theta} \cos{\theta}$

    sketch a reference triangle in quad I (see attach)

    $2 \cdot \dfrac{5}{13} \cdot \dfrac{12}{13} = \dfrac{120}{169}$

    ... have you studied identities yet?
    Attached Thumbnails Attached Thumbnails Puzzling Inverse Trig Function question #2-reftriangle.jpg  
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,485
    Thanks
    2730
    Awards
    1

    Re: Puzzling Inverse Trig Function question #2

    Quote Originally Posted by blakesgrave View Post
    Click image for larger version. 

Name:	159.jpg 
Views:	11 
Size:	30.6 KB 
ID:	37344
    My question here is how they turned the answer into a fraction.
    My process would go;
    sin ( 2arccsc (13/5)) = sin (2 arcsin (5/13)) = sin (2 arcsin (0.3846153)) = sin (2 (22.619)) = sin (45.2397) = 0.71005917
    How do you go about turning that into a fraction?
    As I said in another reply: you are not prepared to do these questions.
    You need to learn(memorize) these sum/difference and double angle formula.
    $\sin(2A)=2\sin(a)\cos(A)~\&~\cos(2A)=\left\{ \begin{array}{l}\cos^2(A)-\sin^2(A)\\1-2\sin^2(A)\\2\cos^2(A)-1\end{array} \right.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Puzzling Inverse Trig Function solution
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Apr 6th 2017, 05:32 PM
  2. Replies: 5
    Last Post: Oct 28th 2014, 05:07 AM
  3. Inverse Trig Function Question.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Aug 23rd 2011, 08:11 PM
  4. Replies: 5
    Last Post: Apr 26th 2011, 06:08 AM
  5. Inverse Trig Function Question
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Dec 7th 2007, 05:42 PM

/mathhelpforum @mathhelpforum