Results 1 to 6 of 6

Math Help - sinx+cosx=0

  1. #1
    Newbie
    Joined
    Jul 2011
    Posts
    2

    sinx+cosx=0

    Solve for 0 <x< 360 degrees

    Okay this is probably a stupid question but I'm really not doing so good with this unit so far.

    Apparently to get the answer you have to take a squaring approach

    sinx=-cosx
    Square both sides
    sin^2x=cox^2x
    sin^2x=1-sin^2x
    2sin^2x=1
    sin^2x=1/2
    sinx=+ or - (1/sqrt(2))
    45, 135, 225, 315 degrees

    BUT what I would have done is

    sinx=-cosx
    -sinx/cosx=cosx/cosx
    -tanx=1
    tanx=-1
    135, 315 degrees

    I need someone to tell me why my thinking process would or would not work. Thanks!

    EDIT: Similar question I need help with too

    sinx-cosx=1

    Here's my approach
    (sinx)^2=(1+cosx)^2
    sin^2x=1+2cosx+cos^2x
    1-cos^2x=1+2cosx+cos^2x
    0=2cosx+2cos^2x
    0=2cosx(1+cosx)
    cosx=0,-1
    90, 180, 270 degrees

    However 270 is not an answer

    I feel like I understand the concept just not getting the same answers. Any help would be GREATLY appreciated
    Last edited by OhCurtis; July 23rd 2011 at 06:30 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2009
    Posts
    63

    Re: sinx+cosx=0

    Quote Originally Posted by OhCurtis View Post
    Solve for 0 <x< 360 degrees

    Okay this is probably a stupid question but I'm really not doing so good with this unit so far.

    Apparently to get the answer you have to take a squaring approach

    sinx=-cosx
    Square both sides
    sin^2x=cox^2x
    sin^2x=1-sin^2x
    2sin^2x=1
    sin^2x=1/2
    sinx=+ or - (1/sqrt(2))
    45, 135, 225, 315 degrees

    BUT what I would have done is

    sinx=-cosx
    -sinx/cosx=cosx/cosx
    -tanx=1
    tanx=-1
    135, 315 degrees

    I need someone to tell me why my thinking process would or would not work. Thanks!

    EDIT: Similar question I need help with too

    sinx-cosx=1

    Here's my approach
    (sinx)^2=(1+cosx)^2
    sin^2x=1+2cosx+cos^2x
    1-cos^2x=1+2cosx+cos^2x
    0=2cosx+2cos^2x
    0=2cosx(1+cosx)
    cosx=0,-1
    90, 180, 270 degrees

    However 270 is not an answer

    I feel like I understand the concept just not getting the same answers. Any help would be GREATLY appreciated
    hmm... remember the concept

    Quadrant I ----> sin, cos, and tan (+)
    Q II -----> sin (+)
    Q III -----> tan (+)
    Q IV -----> cos (+)

    and try again... good luck...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,115
    Thanks
    992

    Re: sinx+cosx=0

    Quote Originally Posted by OhCurtis View Post
    Solve for 0 <x< 360 degrees

    Okay this is probably a stupid question but I'm really not doing so good with this unit so far.

    Apparently to get the answer you have to take a squaring approach

    sinx=-cosx
    Square both sides
    sin^2x=cox^2x
    sin^2x=1-sin^2x
    2sin^2x=1
    sin^2x=1/2
    sinx=+ or - (1/sqrt(2))
    45, 135, 225, 315 degrees

    BUT what I would have done is

    sinx=-cosx
    -sinx/cosx=cosx/cosx
    -tanx=1
    tanx=-1
    135, 315 degrees

    I need someone to tell me why my thinking process would or would not work. Thanks!

    using tangent works fine if you remember that cosx cannot = 0

    EDIT: Similar question I need help with too

    sinx-cosx=1

    Here's my approach
    (sinx)^2=(1+cosx)^2
    sin^2x=1+2cosx+cos^2x
    1-cos^2x=1+2cosx+cos^2x
    0=2cosx+2cos^2x
    0=2cosx(1+cosx)
    cosx=0,-1
    90, 180, 270 degrees

    However 270 is not an answer

    solving an equation by squaring both sides sometimes introduces extraneous (invalid) solutions ... that's why you need to check all solutions when you use this technique. the steps you took were fine ... just remember to check the solutions.

    I feel like I understand the concept just not getting the same answers. Any help would be GREATLY appreciated
    ...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: sinx+cosx=0

    In general you can say for trigoniometric equations:
    \sin(x)=\sin(A)
    Two solutions:
    - x=A + 2k\pi
    - x=\pi - A + 2k\pi

    \cos(x)=\cos(A)
    Two solutions:
    - x=A + 2k\pi
    - x=-A+2k\pi

    \tan(x)=\tan(A)
    one solution
    - x=A+k\pi

    Where k \in \mathbb{Z}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1

    Re: sinx+cosx=0

    As skeeter stated, after squaring both sides you must check the solutions.
    For example, in the first quadrant, both Sinx and Cosx are non-negative
    and hence they cannot sum to zero.
    Cosx is zero at multiples of 90 degrees and Sinx is zero at 0 degrees and multiples of 180 degrees,
    so they are never zero for the same x.

    Solutions to Cosx+Sinx=0 require they be opposite signs,
    which occurs in quadrants 2 and 4.
    Also, their numerical magnitudes (moduli) must be equal in order to cancel and get zero,
    and hence they lie at the point of intersection of the line of slope -1 going through the origin
    and the circumference of the unit circle.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2011
    Posts
    2

    Re: sinx+cosx=0

    All of that explained it so much better than my book. Thanks for all the help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sin a sinx+ b cosx
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 23rd 2009, 08:48 AM
  2. sin a sinx+ b cosx
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 19th 2009, 08:35 AM
  3. Verify that √((1-cosx)/(1+cosx)) = (1-cosx)/|sinx|
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 10th 2008, 09:39 PM
  4. cosx and sinx when x= 0
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 1st 2008, 06:26 PM
  5. cosx-sinx
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: July 29th 2008, 02:26 PM

Search Tags


/mathhelpforum @mathhelpforum