Results 1 to 6 of 6

Math Help - Trig ratio help

  1. #1
    Member
    Joined
    Nov 2010
    Posts
    119

    Trig ratio help

    I feel terrible for making yet another thread, but I've been stuck on this for 30 mins:

    LS=cos(2x)/(1-sin2x)

    RS=(cosx+sinx)/(cosx-sinx)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Since you see double angles on the LS, convert them.

    \cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)

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

    It should be okay then
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2010
    Posts
    119
    Quote Originally Posted by Unknown008 View Post
    Since you see double angles on the LS, convert them.

    \cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)

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

    It should be okay then
    I don't suppose you could continue it. This is the only thing that's giving me trouble and I'm still kind of lost
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,905
    Thanks
    765
    Hello, youngb11!


    Another approach . . .


    \dfrac{\cos 2x}{1-\sin2x} \;=\;\dfrac{\cos x+\sin x}{\cos x-\sin x}

    Multiply the right side by \frac{\cos x - \sin x}{\cos x - \sin x}

    . . \displaystyle \frac{\cos x + \sin x}{\cos x - \sin x}\cdot\frac{\cos x - \sin x}{\cos x - \sin x} \;=\;\frac{\cos^2\!x - \sin^2\!x}{\cos^2\!x - 2\cos x\sin x + \sin^2\!x}

    . . . . \displaystyle =\;\frac{\overbrace{\cos^2\!x - \sin^2\!x}^{\text{This is }\cos2x}}{\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} - \underbrace{2\sin x\cos x}_{\text{This is }\sin2x}} \;=\;\frac{\cos2x}{1-\sin2x}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2010
    Posts
    119
    Thanks a lot! I never saw that approach and was intent on simplifying the LS to equal the RS. Thanks again!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Just use the identities I gave you otherwise.

    \dfrac{\cos 2x}{1-\sin2x} = \dfrac{\cos^2(x) - \sin^2(x)}{1 - 2\cos(x)\sin(x)}

    Here, the trick is to spot that the numerator is a difference of two squares and that the numerator can be a square by using the Pythagorean identity, \cos^2(x) + \sin^2(x) = 1

    This said, you get:

    \dfrac{\cos^2(x) - \sin^2(x)}{1 - 2\cos(x)\sin(x)} = \dfrac{(\cos(x) - \sin(x))(\cos(x) + sin(x))}{\cos^2(x) + \sin^2(x) - 2\cos(x)\sin(x)}

    You then factor the denominator;

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

    Some things cancel out, leaving the RS.

    You can also see that it's the reverse process that Soroban posted
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig ratio as fraction on calculator
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 15th 2011, 09:12 AM
  2. trig ratio
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 23rd 2009, 01:26 PM
  3. all possible angles of a trig ratio
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: July 1st 2009, 11:03 AM
  4. Trig. Ratio Word Problem
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: August 19th 2008, 04:29 AM
  5. Trig. with ratio question
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: March 14th 2008, 09:06 PM

Search Tags


/mathhelpforum @mathhelpforum