Results 1 to 4 of 4

Math Help - Double angle equations

  1. #1
    Newbie pilotguy's Avatar
    Joined
    Nov 2008
    From
    Oklahoma
    Posts
    3

    Double angle equations

    Hello all! I just registered and have a test over double angle and trigonometric addition formulas on Tuesday, but there are a few problems I can't figure out.

    1. sin4x=-2sin2x

    2. (sin2x+cos2x)^2=0

    3. 2sin2x=-tan2x

    Thanks in advance for any and all help!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,932
    Thanks
    763
    Quote Originally Posted by pilotguy View Post
    Hello all! I just registered and have a test over double angle and trigonometric addition formulas on Tuesday, but there are a few problems I can't figure out.
    1. \sin(4x)=-2\sin(2x)

    2\sin(2x)\cos(2x) + 2\sin(2x) = 0

    2\sin(2x)[\cos(2x) + 1] = 0

    set each factor equal to 0 and solve.


    2. [\sin(2x)+\cos(2x)]^2=0

    \sin(2x) + \cos(2x) = 0

    \sin(2x) = -\cos(2x)

    for what angles do cosine and sine have opposite values?


    3. 2\sin(2x)=-\tan(2x)

    2\sin(2x) + \tan(2x) = 0

    \sin(2x)[2 + \sec(2x)] = 0

    set each factor equal to 0 and solve.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie pilotguy's Avatar
    Joined
    Nov 2008
    From
    Oklahoma
    Posts
    3
    Quote Originally Posted by skeeter View Post
    1. \sin(4x)=-2\sin(2x)

    2\sin(2x)\cos(2x) + 2\sin(2x) = 0

    2\sin(2x)[\cos(2x) + 1] = 0

    set each factor equal to 0 and solve.


    2. [\sin(2x)+\cos(2x)]^2=0

    \sin(2x) + \cos(2x) = 0

    \sin(2x) = -\cos(2x)

    for what angles do cosine and sine have opposite values?


    3. 2\sin(2x)=-\tan(2x)

    2\sin(2x) + \tan(2x) = 0

    \sin(2x)[2 + \sec(2x)] = 0

    set each factor equal to 0 and solve.
    Thanks! But I'm still a little confused on 2. I don't really know off the top of my head that sine and cosine have opposite values at 3\pi/4 and 7\pi/4

    Is there another way to solve it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,932
    Thanks
    763
    Quote Originally Posted by pilotguy View Post
    Thanks! But I'm still a little confused on 2. I don't really know off the top of my head that sine and cosine have opposite values at 3\pi/4 and 7\pi/4

    Is there another way to solve it?
    another way to do it ...

    \sin(u) = -\cos(u)

    \frac{\sin(u)}{\cos(u)} = -1

    \tan(u) = -1

    or ...

    \sin(u) = -\cos(u)

    \sin^2(u) = \cos^2(u)

    \sin^2(u) = 1 - \sin^2(u)

    2\sin^2(u) - 1 = 0

    \sin^2(u) = \frac{1}{2}

    \sin(u) = \pm \frac{1}{\sqrt{2}}

    and you still need to know where cosine and sine have opposite values ... so, yes, you really need to know that info off the top of your head.
    Last edited by skeeter; November 24th 2008 at 07:42 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 1st 2010, 03:57 AM
  2. Simplify the Expression (Double-Angle or Half-Angle)
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 3rd 2009, 03:38 PM
  3. Help with double angle and half angle formulas
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 20th 2009, 05:30 PM
  4. Replies: 4
    Last Post: February 20th 2009, 08:24 PM
  5. Replies: 2
    Last Post: February 15th 2007, 10:16 PM

Search Tags


/mathhelpforum @mathhelpforum