Results 1 to 5 of 5

Thread: 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)

    $\displaystyle
    \quad \sin \left( {2\theta } \right) = 2\cos \left( \theta \right)\sin \theta $
    $\displaystyle

    \begin{gathered}
    \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 \\
    = 1.2\cos \left( {\sin ^{ - 1} \left( {0.6} \right)} \right) \hfill \\
    \end{gathered} $


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

    $\displaystyle
    \Rightarrow 1.2\cos \left( {\sin ^{ - 1} \left( {0.6} \right)} \right) = 1.2\sqrt {1 - \sin ^2 \left( {\sin ^{ - 1} \left( {0.6} \right)} \right)} = 1.2\sqrt {1 - 0.6^2 } = 0.96$

    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 $\displaystyle \sin \left( {\sin ^{ - 1} \theta } \right) = \theta $ ?
    So,

    $\displaystyle \begin{array}{l}
    2\sin \left( {\sin ^{ - 1} 0.6} \right)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
    = 2(0.6)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
    = 1.2\cos \left( {\sin ^{ - 1} 0.6} \right) \\
    \end{array}$


    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.
    $\displaystyle \cos \theta = \sqrt {1 - \sin ^2 \theta } $ is derived from $\displaystyle \cos ^2 \theta + \sin ^2 \theta = 1$

    $\displaystyle \begin{array}{l}
    \cos ^2 \theta = 1 - \sin ^2 \theta \\
    \cos \theta = \sqrt {1 - \sin ^2 \theta} \\
    \end{array}$

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

    $\displaystyle \begin{array}{l}
    \cos 2\theta = 1 - 2\sin ^2 \theta \\
    \sin ^2 \theta = \frac{{1 - \cos 2\theta}}{2} \\
    \sin \theta = \pm \sqrt {\frac{{1 - \cos 2\theta}}{2}} \\
    \sin \frac{\theta}{2} = \pm \sqrt {\frac{{1 - \cos \theta}}{2}} \\
    \end{array}$
    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 $\displaystyle \sin \left( {\sin ^{ - 1} \theta } \right) = \theta $ ?
    So,

    $\displaystyle \begin{array}{l}
    2\sin \left( {\sin ^{ - 1} 0.6} \right)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
    = 2(0.6)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
    = 1.2\cos \left( {\sin ^{ - 1} 0.6} \right) \\
    \end{array}$


    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.
    $\displaystyle \cos \theta = \sqrt {1 - \sin ^2 \theta } $ is derived from $\displaystyle \cos ^2 \theta + \sin ^2 \theta = 1$

    $\displaystyle \begin{array}{l}
    \cos ^2 \theta = 1 - \sin ^2 \theta \\
    \cos \theta = \sqrt {1 - \sin ^2 \theta} \\
    \end{array}$

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

    $\displaystyle \begin{array}{l}
    \cos 2\theta = 1 - 2\sin ^2 \theta \\
    \sin ^2 \theta = \frac{{1 - \cos 2\theta}}{2} \\
    \sin \theta = \pm \sqrt {\frac{{1 - \cos 2\theta}}{2}} \\
    \sin \frac{\theta}{2} = \pm \sqrt {\frac{{1 - \cos \theta}}{2}} \\
    \end{array}$
    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: Mar 4th 2010, 04:00 PM
  2. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Jul 5th 2009, 10:00 PM
  3. Inverse trigonometric function Question 3
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 2nd 2009, 10:57 AM
  4. Inverse trigonometric Function Question 2
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 2nd 2009, 05:43 AM
  5. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Apr 28th 2009, 10:31 AM

Search Tags


/mathhelpforum @mathhelpforum