Results 1 to 10 of 10

Math Help - trig help....

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    6

    trig help....

    i have two problems that i can't figure out how to do...


    1. Give the exact real number value for Tan inverse [tan 7pi/12]





    and 2. Evaluate: (show your work. do not give a decimal approximation)

    Sin[cos inverse(1/3) + sin inverse(-2/5)]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by ilaura View Post
    i have two problems that i can't figure out how to do...


    1. Give the exact real number value for Tan inverse [tan 7pi/12]





    and 2. Evaluate: (show your work. do not give a decimal approximation)

    Sin[cos inverse(1/3) + sin inverse(-2/5)]
    On a restriced interval \arctan(x) is an inverse function of \tan(x) so \tan(\arctan(x))=x and for the second one use the fact that \sin(\arccos(x))=\sqrt{1-x^2} and \sin(\arcsin(x))=x after using the identity \sin\left(A+B\right)=\sin\left(A\right)\cos\left(B  \right)+\sin\left(B\right)\cos\left(A\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2008
    Posts
    6
    i'm still really confused.....


    i tried doing something with the second one,


    i don't think its right but i got:


    6(square root of 2) + 3(square root of 21) -1
    ____________________
    15
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by ilaura View Post
    i'm still really confused.....


    i tried doing something with the second one,


    i don't think its right but i got:


    6(square root of 2) + 3(square root of 21) -1
    ____________________
    15
    \begin{aligned}\sin\left(\arccos\left(\frac{1}{3}\  right)\right)+\arcsin\left(\frac{-2}{5}\right)&=\sin\left(\arccos\left(\frac{1}{3}\r  ight)\right)\cos\left(\arcsin\left(\frac{-2}{5}\right)\right)\\<br />
&+\sin\left(\arcsin\left(\frac{-2}{5}\right)\right)\cos\left(\arccos\left(\frac{1}  {3}\right)\right)\\<br />
&=\sqrt{1-\left(\frac{1}{3}\right)^2}\sqrt{1-\left(\frac{-2}{5}\right)^2}+\frac{1}{2}\cdot\frac{-2}{5}\\<br />
&=\frac{2\sqrt{42}-2}{15}<br />
\end{aligned}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2008
    Posts
    6
    Quote Originally Posted by Mathstud28 View Post
    ......and for the second one use the fact that \sin(\arccos(x))=\sqrt{1-x^2} and \sin(\arcsin(x))=x



    i don't understand where these come from....
    are they some kind of identities?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by ilaura View Post
    i don't understand where these come from....
    are they some kind of identities?
    Yes List of trigonometric identities - Wikipedia, the free encyclopedia
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by ilaura View Post
    [snip]
    and 2. Evaluate: (show your work. do not give a decimal approximation)

    Sin[cos inverse(1/3) + sin inverse(-2/5)]
    Let \cos^{-1} \left( \frac{1}{3} \right)= \alpha \Rightarrow \cos \alpha = \frac{1}{3}.

    Let \sin^{-1} \left(-\frac{2}{5}\right) = \beta \Rightarrow \sin \beta = - \frac{2}{5}.

    Then \sin (\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha = \, ....
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Nov 2008
    Posts
    6
    Quote Originally Posted by Mathstud28 View Post
    \begin{aligned}\sin\left(\arccos\left(\frac{1}{3}\  right)\right)+\arcsin\left(\frac{-2}{5}\right)&=\sin\left(\arccos\left(\frac{1}{3}\r  ight)\right)\cos\left(\arcsin\left(\frac{-2}{5}\right)\right)\\<br />
&+\sin\left(\arcsin\left(\frac{-2}{5}\right)\right)\cos\left(\arccos\left(\frac{1}  {3}\right)\right)\\<br />
&=\sqrt{1-\left(\frac{1}{3}\right)^2}\sqrt{1-\left(\frac{-2}{5}\right)^2}+\frac{1}{2}\cdot\frac{-2}{5}\\<br />
&=\frac{2\sqrt{42}-2}{15}<br />
\end{aligned}
    where did you get the 1/2? is it part of the identity?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by ilaura View Post
    where did you get the 1/2? is it part of the identity?
    Clearly it's a typo and should be 1/3.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by ilaura View Post
    My homework says:


    1. Give the exact real number value for arctan [tan (7pi/12)]

    steps would be appreciated. thanks
    Note that arctangent and tangent are inverses of each other.

    Thus, if we have \tan^{-1}\left(\tan(\varphi)\right), it just reduces to \varphi. The same thing applies to \tan\left(\tan^{-1}(\varphi)\right). It also equals \varphi.

    Knowing this, what do you think the answer is?
    Last edited by Chris L T521; November 30th 2008 at 12:22 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 8th 2011, 08:00 PM
  2. Replies: 7
    Last Post: April 15th 2010, 09:12 PM
  3. Replies: 6
    Last Post: November 20th 2009, 05:27 PM
  4. Replies: 1
    Last Post: July 24th 2009, 03:29 AM
  5. Trig Equations with Multiple Trig Functions cont.
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 7th 2008, 06:50 PM

Search Tags


/mathhelpforum @mathhelpforum