Results 1 to 8 of 8

Math Help - Solving trig equation with identies

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    131

    Solving trig equation with identies

    If sin(2x) =0 tanx - cos 2x, then sinx - cosx =?

    I know of all the double angle identies for sin and cosine, and I know tan=sin/cos.
    I still can't manage to find sinx-cosx
    Perhaps you can get me in the right direction.. I tried subbing a few identites, but apparently not the right ones. And note that this problem isn't for class.... This is from Princeton Review math lvl 2 review.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,694
    Thanks
    450

    Re: Solving trig equation with identies

    Quote Originally Posted by benny92000 View Post
    If sin(2x) =0 tanx - cos 2x, then sinx - cosx =?

    I know of all the double angle identies for sin and cosine, and I know tan=sin/cos.
    I still can't manage to find sinx-cosx
    Perhaps you can get me in the right direction.. I tried subbing a few identites, but apparently not the right ones. And note that this problem isn't for class.... This is from Princeton Review math lvl 2 review.
    makes no sense ... is that 0 times tanx ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2008
    Posts
    131

    Re: Solving trig equation with identies

    I apologize. There is no 0.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,694
    Thanks
    450

    Re: Solving trig equation with identies

    If sin(2x) = tanx - cos 2x, then sinx - cosx =?
    \sin(2x) = \tan{x}-\cos(2x)

    \cos(2x) = \tan{x} - \sin(2x)

    \cos^2{x} - \sin^2{x} = \tan{x} - \sin(2x)

    (\cos{x}-\sin{x})(\cos{x}+\sin{x}) = \tan{x} - \sin(2x)

    \cos{x} - \sin{x} = \frac{\tan{x} - \sin(2x)}{\cos{x}+\sin{x}}

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

  5. #5
    Member
    Joined
    Oct 2008
    Posts
    131

    Re: Solving trig equation with identies

    Makes sense. They are looking for a number value, however. Choices are negative square root of 2, 0, 1, 2 times the square root of 2, or not enough information.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644

    Re: Solving trig equation with identies

    Hello, benny92000!

    \text{If }\sin 2x \:=\: \tan x - \cos 2x,\,\text{ then }\,\sin x - \cos x \:=

    . . (a)\;\text{-}\sqrt{2} \qquad (b)\;0 \qquad (c)\;1 \qquad (d)\;2\sqrt{2} \qquad (e)\text{ not enough information}
    Note that: . x \,\ne\,\tfrac{\pi}{2},\:\tfrac{3\pi}{2}


    We have: . \sin2x + \cos2x \:=\:\tan x

    . 2\sin x\cos x + 1 - 2\sin^2x \:=\:\tan x

    . . . . 2\sin^2\!x - 2\sin x\cos x \:=\:1 - \tan x

    . . . . 2\sin x(\sin x - \cos x) \:=\: 1 - \frac{\sin x}{\cos x}

    . . . . 2\sin x(\sin x - \cos x) \;=\;\frac{\cos x - \sin x}{\cos x}


    Multiply by \cos x\!:
    n . . . . . . . . . . . . 2\sin x\cos x(\sin x - \cos x) \;=\;\cos x - \sin x

    . . 2\sin x\cos x(\sin x - \cos x) - (\sin x - \cos x) \;=\;0

    n . . . . . . . . (\sin x - \cos x)(2\sin x\cos x - 1) \;=\;0


    And we have two equations: . \begin{Bmatrix}\sin x - \cos x &=& 0 \\ 2\sin x\cos x -1 &=& 0 \end{Bmatrix}

    Both equations have the solution: . x \:=\:\tfrac{\pi}{4} + \pi n


    Therefore: . \sin x - \cos x \;=\;0\;\;\;\text{ answer (b)}

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2008
    Posts
    131

    Re: Solving trig equation with identies

    That's pretty detailed. Thank you. How did you immediately eliminate pi/2 and 3pi/2 for x?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,549
    Thanks
    1418

    Re: Solving trig equation with identies

    Quote Originally Posted by benny92000 View Post
    That's pretty detailed. Thank you. How did you immediately eliminate pi/2 and 3pi/2 for x?
    Because tan(x) = sin(x)/cos(x) and we can't have cos(x) = 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 07:07 AM
  2. Solving a trig equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 8th 2009, 12:09 PM
  3. Replies: 3
    Last Post: April 30th 2009, 07:41 AM
  4. Trig identies and integrals
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 7th 2008, 10:38 AM
  5. Trig identies
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: April 13th 2008, 03:24 AM

Search Tags


/mathhelpforum @mathhelpforum