Results 1 to 8 of 8

Thread: Determining the exact value of each expression

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    4

    Determining the exact value of each expression

    Hi guys, I'm having trouble with obtaining the correct results for this expression.

    The question is: Determine the exact value of each expression.

    [cos(pi/4)tan(5pi/6)] / [sin(3pi/4)]

    Somehow, I obtained the answer square root of three... but the answer is supposed to be square root of 6 divided by 6.

    Thank you in advance for all of your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Henderson's Avatar
    Joined
    Dec 2007
    Posts
    127
    Thanks
    2
    Are you sure either answer is right, or that you copied the problem correctly?

    I get an answer of -\frac{\sqrt3}{3}.

    \frac{(\frac{\sqrt2}{2})\cdot(-\frac{\sqrt3}{3})}{(\frac{\sqrt2}{2})}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    Quote Originally Posted by Henderson View Post
    Are you sure either answer is right, or that you copied the problem correctly?

    I get an answer of -\frac{\sqrt3}{3}.

    \frac{(\frac{\sqrt2}{2})\cdot(-\frac{\sqrt3}{3})}{(\frac{\sqrt2}{2})}
    Ditto what Henderson said
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    4
    Oh no! I didn't copy it right! It's supposed to be cos(4pi/3) instead of cos(pi/4)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,155
    Thanks
    3672
    well, you should know that \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}

    sub it in for the \frac{\sqrt{2}}{2} in the numerator of the expressions given and re-evaluate.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2008
    Posts
    4
    how did you get square root of 2 over 2 as the denominator tho?

    -cos(pi/3) = -1/2
    -tan(pi/6) = - 1/ square root of 3
    sin(pi/4) = 1/ square root of 2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,155
    Thanks
    3672
    \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2008
    Posts
    4
    i mean.. how did you get your answer? the answer that i got wat 2/(2*square root of 3)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Aug 8th 2011, 02:50 PM
  2. Determining exact values for non-special angles
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Nov 2nd 2010, 08:38 PM
  3. Determining an exact expression for a length.
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: Dec 6th 2009, 03:14 PM
  4. Replies: 3
    Last Post: Nov 28th 2009, 07:06 PM
  5. Exact Value of Expression
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Mar 30th 2009, 05:48 AM

Search Tags


/mathhelpforum @mathhelpforum