Results 1 to 4 of 4

Math Help - Values of Trig Functions

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    2

    Values of Trig Functions

    Hello, 1st post.
    A little help needed.
    We're not allowed to use calculators, so if you could provide answers with working, great!

    Sinθ = - 2/5 (negative two fifths) Range: π ≤ θ ≤ 3π/2 (π=pi)

    Find Values for:
    Cosθ
    Tanθ

    Im told i am to use the triangles, but i dont get how they work.
    Help is appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Privatealpha View Post
    Hello, 1st post.
    A little help needed.
    We're not allowed to use calculators, so if you could provide answers with working, great!

    Sinθ = - 2/5 (negative two fifths) Range: π ≤ θ ≤ 3π/2 (π=pi)

    Find Values for:
    Cosθ
    Tanθ

    Im told i am to use the triangles, but i dont get how they work.
    Help is appreciated.
    The sine function is negative in Quadrants III and IV

    but we only want the Values in Quadrant III becuase of the range given.

    Values of Trig Functions-capture.jpg

    Using the pythagorean theorem we can solve for the other leg of the triangle

    (-2)^2+x^2=5^2 \iff 4+x^2=25 \iff x=\sqrt{21}

    Note that since we are in quadrant III the x coordinate is negative.

    so we get -\sqrt{21}

    \tan(\theta)=\frac{opposite}{adjacent}=\frac{-2}{-\sqrt{21}}=\frac{2\sqrt{21}}{21}

    \cos(\theta)=\frac{adjacent}{hypotenuse}=\frac{-\sqrt{21}}{5}<br /> <br />
    I hope this helps.

    Good luck.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    2
    Thanks for that.
    Working through my assignment, it goes on to ask:
    Evaluate Sin2 and Cos2 with the values above.

    Ive come up with:
    Sin2 = 2sin cos
    => 4√21 / 25

    Cos2 = Cos▓ - Sin▓
    => 1

    Could someone confirm or correct my answers?
    Last edited by Privatealpha; May 25th 2008 at 06:19 AM. Reason: numeric error
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Quote Originally Posted by Privatealpha View Post
    Thanks for that.
    Working through my assignment, it goes on to ask:
    Evaluate Sin2 and Cos2 with the values above.

    Ive come up with:
    Sin2 = 2sin cos
    => 4√21 / 25

    Cos2 = Cos▓ - Sin▓
    => 1

    Could someone confirm or correct my answers?
    Ok for \sin 2x


    \cos 2x=\cos^2 x-\sin^2 x=\left(\frac{-\sqrt{21}}{5}\right)^2-\left(\frac{-2}{5}\right)^2=\frac{21}{25}-\frac{4}{25} \neq 1


    Also, you can remember the two other formulas for \cos 2x : 2 \cos^2 x-1=1-2 \sin^2 x

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help finding values of trig functions.
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: January 13th 2010, 01:59 PM
  2. exact values of trig functions
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: September 30th 2009, 12:10 PM
  3. trig functions and exact values
    Posted in the Trigonometry Forum
    Replies: 11
    Last Post: July 2nd 2008, 04:58 AM
  4. help with hw-trig functions and values
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: September 24th 2007, 06:28 PM
  5. Values of Trig Functions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 17th 2006, 07:36 AM

Search Tags


/mathhelpforum @mathhelpforum