Results 1 to 7 of 7

Math Help - proving identities

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    4

    proving identities

    I need help wit a problem....

    1- cos = Sin
    1+sin


    Having trouble after i multipying for common denominator
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    May 2010
    Posts
    4
    Quote Originally Posted by dmidget View Post
    I need help wit a problem....

    1- cos(squared) = Sin
    ' 1+sin


    Having trouble after i multipying for common denominator
    revised
    Follow Math Help Forum on Facebook and Google+

  3. #3
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by dmidget View Post
    revised
    Very confusing...

    Do you mean \frac{1-\cos^2 \theta}{1+\sin \theta} = \sin \theta
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2010
    Posts
    4
    Quote Originally Posted by e^(i*pi) View Post
    Very confusing...

    Do you mean \frac{1-\cos^2 \theta}{1+\sin \theta} = \sin \theta
    yeah how did you get it too show like that??
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by dmidget View Post
    yeah how did you get it too show like that??
    I used LaTeX, you can click the image to see how I done it - you can find out more here: http://www.mathhelpforum.com/math-help/latex-help/

    I'm not sure how to try it, I've done the difference of two squares but that only gives \frac{(1-\cos \theta)(1+\cos \theta)}{1+\sin \theta}

    Using the Pythagorean identity gives only \frac{\sin^2 \theta}{1+\sin \theta}


    Edit: this doesn't appear to be an identity - counterexample \theta = \frac{\pi}{4}


    \frac{1-\cos^2 \left(\frac{\pi}{4}\right)}{1+\sin \left(\frac{\pi}{4}\right)} = \frac{1-\frac{1}{2}}{1+\frac{\sqrt2}{2}} = \frac{\frac{1}{2}}{\frac{2+\sqrt2}{2}} = \frac{1}{2+\sqrt2} = \frac{2-\sqrt2}{2}

    \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2} \neq \frac{2-\sqrt2}{2}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2010
    Posts
    4
    Quote Originally Posted by e^(i*pi) View Post
    I used LaTeX, you can click the image to see how I done it - you can find out more here: http://www.mathhelpforum.com/math-help/latex-help/

    I'm not sure how to try it, I've done the difference of two squares but that only gives \frac{(1-\cos \theta)(1+\cos \theta)}{1+\sin \theta}

    Using the Pythagorean identity gives only \frac{\sin^2 \theta}{1+\sin \theta}


    Edit: this doesn't appear to be an identity - counterexample \theta = \frac{\pi}{4}


    \frac{1-cos^2 \left(\frac{\pi}{4}\right)}{1+\sin \left(\frac{\pi}{4}\right)} = \frac{1-\frac{1}{2}}{1+\frac{\sqrt2}{2}} = \frac{\frac{1}{2}}{\frac{2+\sqrt2}{2}} = \frac{1}{2+\sqrt2} = \frac{2-\sqrt2}{2}

    \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2} \neq \frac{2-\sqrt2}{2}
    i got the problem wrong., its one minus cos(squared) over 1 plus sin= sin
    Follow Math Help Forum on Facebook and Google+

  7. #7
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by dmidget View Post
    i got the problem wrong., its one minus cos(squared) over 1 plus sin= sin
    1-\frac{\cos^2 \theta}{1+\sin \theta}

    Use the Pythagorean identity and then the difference of two squares
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: June 23rd 2010, 12:59 AM
  2. Proving Identities
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 24th 2010, 06:00 AM
  3. Proving identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 4th 2010, 09:19 PM
  4. Help with proving identities
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 11th 2010, 02:05 AM
  5. proving identities
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 6th 2008, 05:55 AM

Search Tags


/mathhelpforum @mathhelpforum