Results 1 to 5 of 5

Math Help - Inverse trigonometric function

  1. #1
    Member maybeline9216's Avatar
    Joined
    Sep 2008
    From
    Singapore~
    Posts
    108

    Inverse trigonometric function

    how to do these types of qns??any tips??
    Attached Thumbnails Attached Thumbnails Inverse trigonometric function-unresolvedqns.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    1. (i)

    <br />
\quad \sin \left( {2\theta } \right) = 2\cos \left( \theta  \right)\sin \theta
    <br /> <br />
\begin{gathered}<br />
  \sin \left( {2\sin ^{ - 1} (0.6)} \right) = 2\sin \left( {\sin ^{ - 1} \left( {0.6} \right)} \right)\cos \left( {\sin ^{ - 1} \left( {0.6} \right)} \right) =  \hfill \\<br />
   = 1.2\cos \left( {\sin ^{ - 1} \left( {0.6} \right)} \right) \hfill \\ <br />
\end{gathered}


    \cos \left( \theta  \right) = \sqrt {1 - \sin ^2 \left( \theta  \right)}

    <br />
 \Rightarrow 1.2\cos \left( {\sin ^{ - 1} \left( {0.6} \right)} \right) = 1.2\sqrt {1 - \sin ^2 \left( {\sin ^{ - 1} \left( {0.6} \right)} \right)}  = 1.2\sqrt {1 - 0.6^2 }  = 0.96

    all the other problems can be solved similarly using various trigonometric identities. So you better familiarize yourself with them. Here's a good place to start:

    List of trigonometric identities - Wikipedia, the free encyclopedia
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member maybeline9216's Avatar
    Joined
    Sep 2008
    From
    Singapore~
    Posts
    108


    How do u get this? the 1.2 how to get did u use any exact values of trigo functions? If so, wat did u use??


    Do u have any special way to rmb it?? cos it is unlike the half-angle formulae

    thks for the link but i am so "overwhelmed" by it...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    10
    Quote Originally Posted by maybeline9216 View Post


    How do u get this? the 1.2 how to get did u use any exact values of trigo functions? If so, wat did u use??


    Do u have any special way to rmb it?? cos it is unlike the half-angle formulae

    thks for the link but i am so "overwhelmed" by it...
    Do you know that \sin \left( {\sin ^{ - 1} \theta } \right) = \theta ?
    So,

    \begin{array}{l}<br />
 2\sin \left( {\sin ^{ - 1} 0.6} \right)\cos \left( {\sin ^{ - 1} 0.6} \right) \\ <br />
  = 2(0.6)\cos \left( {\sin ^{ - 1} 0.6} \right) \\ <br />
  = 1.2\cos \left( {\sin ^{ - 1} 0.6} \right) \\ <br />
 \end{array}


    There's no need to remember complicated trigo identities, just remember the original identities(which is given on your formula table during your O Level exam) and manipulate it.
    \cos \theta  = \sqrt {1 - \sin ^2 \theta } is derived from \cos ^2 \theta + \sin ^2 \theta = 1

    \begin{array}{l}<br />
 \cos ^2 \theta = 1 - \sin ^2 \theta \\ <br />
 \cos \theta = \sqrt {1 - \sin ^2 \theta}  \\ <br />
 \end{array}

    Similarly, for half-angle formula, consider \sin \frac{\theta }{2}

    \begin{array}{l}<br />
 \cos 2\theta = 1 - 2\sin ^2 \theta \\ <br />
 \sin ^2 \theta = \frac{{1 - \cos 2\theta}}{2} \\ <br />
 \sin \theta =  \pm \sqrt {\frac{{1 - \cos 2\theta}}{2}}  \\ <br />
 \sin \frac{\theta}{2} =  \pm \sqrt {\frac{{1 - \cos \theta}}{2}}  \\ <br />
 \end{array}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member maybeline9216's Avatar
    Joined
    Sep 2008
    From
    Singapore~
    Posts
    108
    Quote Originally Posted by NyRychVantel View Post
    Do you know that \sin \left( {\sin ^{ - 1} \theta } \right) = \theta ?
    So,

    \begin{array}{l}<br />
2\sin \left( {\sin ^{ - 1} 0.6} \right)\cos \left( {\sin ^{ - 1} 0.6} \right) \\ <br />
= 2(0.6)\cos \left( {\sin ^{ - 1} 0.6} \right) \\ <br />
= 1.2\cos \left( {\sin ^{ - 1} 0.6} \right) \\ <br />
\end{array}


    There's no need to remember complicated trigo identities, just remember the original identities(which is given on your formula table during your O Level exam) and manipulate it.
    \cos \theta = \sqrt {1 - \sin ^2 \theta } is derived from \cos ^2 \theta + \sin ^2 \theta = 1

    \begin{array}{l}<br />
\cos ^2 \theta = 1 - \sin ^2 \theta \\ <br />
\cos \theta = \sqrt {1 - \sin ^2 \theta} \\ <br />
\end{array}

    Similarly, for half-angle formula, consider \sin \frac{\theta }{2}

    \begin{array}{l}<br />
\cos 2\theta = 1 - 2\sin ^2 \theta \\ <br />
\sin ^2 \theta = \frac{{1 - \cos 2\theta}}{2} \\ <br />
\sin \theta = \pm \sqrt {\frac{{1 - \cos 2\theta}}{2}} \\ <br />
\sin \frac{\theta}{2} = \pm \sqrt {\frac{{1 - \cos \theta}}{2}} \\ <br />
\end{array}
    Thanks so much!!This is really useful!! =)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse Trigonometric Function Integral?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 4th 2010, 05:00 PM
  2. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 5th 2009, 11:00 PM
  3. Inverse trigonometric function Question 3
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 2nd 2009, 11:57 AM
  4. Inverse trigonometric Function Question 2
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 2nd 2009, 06:43 AM
  5. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 28th 2009, 11:31 AM

Search Tags


/mathhelpforum @mathhelpforum