Results 1 to 4 of 4

Math Help - Trigonometric Equations

  1. #1
    Newbie
    Joined
    May 2005
    Posts
    7

    Trigonometric Equations

    Thru what method would you solve this problem?

    cos^2x = sin^2x +1

    Linear Method
    Factoring
    Quadratic Formula
    Squaring
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Are we limited to those four methods?

    If not, then I will use substitution as the first step, and then I will see the resulting equation. It could then be by factoring or by using the quadratic formula, or by whatever.

    cos^2x = sin^2x +1
    Rewriting that,
    cos^2(x) = sin^2(x) +1 ----(i)

    First, I will use the trig identity
    sin^2(x) +cos^2(x) = 1
    to transform Eq.(i) into sines only.

    From that trig identity,
    cos^2(x) = 1 -sin^2(x)

    Substitute that into (i),

    1 -sin^2(x) = sin^2(x) +1
    Collect like terms,
    -sin^2(x) -sin^2(x) = 1 -1
    -2sin^2(x) = 0
    Divide both sides by -2,
    sin^2(x) = 0
    Take the square roots of both sides,
    sin(x) = 0
    Take the arcsine of both sides,
    x = arcsin(0) ----***

    If we limit the domain of x from 0 to 360 degrees only,
    x = 0, 180, or 360 degrees. ----answer.

    ---------------
    Another way.

    We transform Eq.(i) into cosines only.

    From the same trig identity we used above,
    sin^2(x) = 1 -cos^2(x)

    Substitute that into (i),

    cos^2(x) = [1 -cos^2(x)] +1
    cos^2(x) = 1 -cos^2(x) +1
    cos^2(x) +cos^2(x) = 1 +1
    2cos^2(x) = 2
    Umm, can be by factoring.

    2cos^2(x) = 2
    2cos^2(x) -2 = 0
    2(cos^2(x) -1) = 0
    Divide both sides by 2,
    cos^2(x) -1 = 0
    That quadratic equation is a difference of two squares,
    factoring that,
    (cos(x) +1)*(cos(x) -1) = 0

    cos(x) +1 = 0
    cos(x) = -1
    x = arccos(-1)
    x = 180 degrees

    cos(x) -1 = 0
    cos(x) = 1
    x = arccos(1)
    x = 0, or 360 degrees

    Therefore, if x is from 0 to 360 degrees only, then,
    x = 180, 0, or 360 degrees. ---same as in the first way above.
    Last edited by ticbol; May 14th 2005 at 12:21 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    hpe
    hpe is offline
    Member hpe's Avatar
    Joined
    Apr 2005
    Posts
    158

    Talking

    Quote Originally Posted by jp25111
    Thru what method would you solve this problem?

    cos^2x = sin^2x +1

    Linear Method
    Factoring
    Quadratic Formula
    Squaring
    You can also use a double angle formula :

    cos^2(x) - sin^2(x) = 1
    cos(2x) = 1
    so 2x = 0 or 2pi or 4pi ...
    so x must be a multiple of pi - as in ticbol's solution.

    Check whether this always gives a solution (always a good idea for trig. equations):

    cos (k*pi) = +1 or -1, sin(*pi) = 0,
    so these are indeed solutions of the problem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2005
    Posts
    7
    You guys are awesome. Thanks soooooooo much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometric equations?
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 11th 2011, 04:19 AM
  2. Six trigonometric equations
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: July 21st 2010, 09:17 PM
  3. 3 Trigonometric Equations
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 28th 2009, 12:14 PM
  4. Trigonometric Equations Help Please!
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: March 25th 2009, 06:10 AM
  5. trigonometric equations
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: July 22nd 2007, 01:28 PM

Search Tags


/mathhelpforum @mathhelpforum