Results 1 to 5 of 5

Math Help - Trigonometry Problems [involves inverse functions]

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    15

    Trigonometry Problems [involves inverse functions]

    So I am having a bit of trouble studying for my trig test coming up....
    for the section on inverse trig functions I am a bit confused with some questions

    ex: sin[29arctanx)+arcos(2x)]

    ex: cos[arctan3 - arcsin(-0.5)]

    I understand the basics such as remembering the domains of the different functions and what not... but these questions seem to stump me.

    Can you maybe solve one but with a thorough description.... =)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,689
    Thanks
    448
    I assume the first expression has a typo, and should be

    \sin[2\arctan{x} + \arccos(2x)]<br />

    please confirm.


    for the second expression ...

    \cos[\arctan(3) - \arcsin(-0.5)]

    let \theta = \arctan(3) , and you should already know that \arcsin(-0.5) = -\dfrac{\pi}{6}

    using the sum identity for cosine ...

    \cos\left(\theta + \dfrac{\pi}{6}\right) =  \cos{\theta} \cdot \cos\left(\dfrac{\pi}{6}\right) - \sin{\theta} \cdot \sin\left(\dfrac{\pi}{6}\right)

    since \tan{\theta} = 3 , then \cos{\theta} = \dfrac{1}{\sqrt{10}} and \sin{\theta} = \dfrac{3}{\sqrt{10}}

    sub these values into the sum expression above ...

    \cos\left(\theta + \dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{10}} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{3}{\sqrt{10}} \cdot \dfrac{1}{2} = \dfrac{\sqrt{3} - 3}{2\sqrt{10}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    15
    So how is it you found the value of arctan(3)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,689
    Thanks
    448
    Quote Originally Posted by FallenStar117 View Post
    So how is it you found the value of arctan(3)?
    I didn't ... I said let \theta = \arctan(3) , which also says \tan{\theta} = 3 ... I used this fact to calculate \sin{\theta} and \cos{\theta} which was necessary evaluate the original expression.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2011
    Posts
    15
    So how would I start solving the first question?
    Is the idea the same as the second question...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse Trigonometry.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 15th 2009, 08:20 PM
  2. derivatives of inverse trigonometry functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 2nd 2009, 02:11 AM
  3. inverse trigonometry 2
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 31st 2009, 07:02 AM
  4. Inverse trigonometry?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 2nd 2009, 11:15 PM
  5. Inverse Problems Functions
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 15th 2008, 04:43 PM

Search Tags


/mathhelpforum @mathhelpforum