Results 1 to 3 of 3

Math Help - solving trig equations

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    1

    Red face solving trig equations

    If anyone could help me out here it would be greatl appreciated.(:

    find sin, cos, tan of:
    255 degrees
    17pi/12

    find value of:
    cos(x+ pi/6) - cos(x - pi/6)=1
    sin(x+ pi/6) - sin(x - pi/6)=1/2


    thank you sooo much
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by paramorechick99 View Post

    find sin, cos, tan of:
    255 degrees
    17pi/12
    Are you allowed to use a calculator?

    if not consider,

    \sin\left(\frac{17\pi}{12}\right) = \sin\left(\frac{12\pi}{12}+\frac{5\pi}{12}\right)

    Now apply the rules

    \sin(a+b) = \sin(a)\cos(b)+\sin(b)\cos(a)

    and

    \sin(a) = \sqrt{\frac{1-\cos(a)}{2}} and \cos(a) = \sqrt{\frac{1+\cos(a)}{2}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by paramorechick99 View Post
    If anyone could help me out here it would be greatl appreciated.(:

    find sin, cos, tan of:
    255 degrees
    17pi/12

    find value of:
    cos(x+ pi/6) - cos(x - pi/6)=1
    sin(x+ pi/6) - sin(x - pi/6)=1/2


    thank you sooo much
    the three sum difference formulas ...

    \cos(a \pm b) = \cos{a}\cos{b} \mp \sin{a}\sin{b}

    \sin(a \pm b) = \sin{a}\cos{b} \pm \cos{a}\sin{b}

    \tan(a \pm b) = \frac{\tan{a} \pm \tan{b}}{1 \mp \tan{a} \tan{b}}

    note that 255 = (225+30) or (300-45) ... unit circle values

    \frac{17\pi}{12} = \frac{5\pi}{3} - \frac{\pi}{4} or \frac{3\pi}{4} + \frac{2\pi}{3} ... unit circle values

    plug and chug.




    \cos\left(x + \frac{\pi}{6} \right) - \cos\left(x - \frac{\pi}{6} \right) = 1

    \cos{x} \cdot \frac{\sqrt{3}}{2} - \sin{x} \cdot \frac{1}{2} - \left[\cos{x} \cdot \frac{\sqrt{3}}{2} + \sin{x} \cdot \frac{1}{2}\right] = 1

    -\sin{x} = 1<br />

    \sin{x} = -1

    x = \frac{3\pi}{2}


    you do the second one.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving Trig. Equations
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 3rd 2010, 10:14 PM
  2. Solving Trig Equations Help
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: March 14th 2010, 08:47 PM
  3. Solving trig equations
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: March 4th 2010, 02:23 PM
  4. solving trig equations
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 14th 2008, 05:03 PM
  5. Solving Trig Equations
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 21st 2007, 07:18 PM

Search Tags


/mathhelpforum @mathhelpforum