Results 1 to 9 of 9

Math Help - identity problem

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    62

    identity problem

    Show: \frac{\sin^2x}{1-\cos x} = \frac{1 + \sec}{\sec x}

    Thanks for your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    \frac{\sin^2x}{1-\cos x}=\frac{1-\cos^2x}{1-\cos x}=1+\cos x

    and \frac{1+\sec x}{\sec x}=\frac{1}{\sec x}+1=\cos x+1

    Follow Math Help Forum on Facebook and Google+

  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by Savior_Self View Post
    Show: \dfrac{\sin^2x}{1-\cos x} = \dfrac{1 + \sec x}{\sec x}

    Thanks for your help.
    Hi Savior_Self,

    I'm going to mess with both sides in order to prove this identity.

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

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

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

    \cos x+1=\cos x+1

    Moo hath beaten me to the punch!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2008
    Posts
    62
    ahhh, thank you. Your reply made it easier to understand Moo's, masters. So worry not. You both contributed much to my instruction. haha, thanks again.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2008
    Posts
    62
    Alright guys.. I asked for help with this problem a few weeks ago, and I just received the graded worksheet. I liked what you guys did with this, I thought it was clever. But, apparently, this wasn't the answer the lady was looking for because it's marked completely wrong. I'm about to start trying to figure it out myself, but I can always use some help from you guys. So my question is: Is there any way to work through this using only the left side? Starting with \frac{\sin^2 x}{1 - \cos x} and, changing only that, showing that it equals \frac{1 + \sec x}{\sec x}?

    Thanks! and sorry to ask again.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by Savior_Self View Post
    Alright guys.. I asked for help with this problem a few weeks ago, and I just received the graded worksheet. I liked what you guys did with this, I thought it was clever. But, apparently, this wasn't the answer the lady was looking for because it's marked completely wrong. I'm about to start trying to figure it out myself, but I can always use some help from you guys. So my question is: Is there any way to work through this using only the left side? Starting with \frac{\sin^2 x}{1 - \cos x} and, changing only that, showing that it equals \frac{1 + \sec x}{\sec x}?

    Thanks! and sorry to ask again.
    Hi Savior_Self,

    Yes, there is and your instructor should have given you better instructions about which side she wanted to work with. Here is just the left side manipulated:

    \frac{\sin^2 x}{1-\cos x}=

    \frac{1-\cos^2 x}{1-\cos x}=

    \frac{(1-\cos x)(1+\cos x)}{1-\cos x}=

    1+\cos x =

    1+\frac{1}{\sec x}=

    \frac{\sec x + 1}{\sec x}=\frac{1+\sec x}{\sec x}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2008
    Posts
    62
    Quote Originally Posted by masters View Post
    Hi Savior_Self,

    Yes, there is and your instructor should have given you better instructions about which side she wanted to work with. Here is just the left side manipulated:

    \frac{\sin^2 x}{1-\cos x}=

    \frac{1-\cos^2 x}{1-\cos x}=

    \frac{(1-\cos x)(1+\cos x)}{1-\cos x}=

    1+\cos x =

    1+\frac{1}{\sec x}=

    \frac{\sec x + 1}{\sec x}=\frac{1+\sec x}{\sec x}
    May I ask how you manipulated  1 + \frac{1}{\sec x} into \frac{\sec x + 1}{\sec x}?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by Savior_Self View Post
    May I ask how you manipulated  1 + \frac{1}{\sec x} into \frac{\sec x + 1}{\sec x}?
    1=\frac{\sec x}{\sec x}, so

    1+\frac{1}{\sec x}=\frac{\sec x}{\sec x}+\frac{1}{\sec x}=\frac{\sec x+1}{\sec x}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Nov 2008
    Posts
    62
    Quote Originally Posted by masters View Post
    1=\frac{\sec x}{\sec x}, so

    1+\frac{1}{\sec x}=\frac{\sec x}{\sec x}+\frac{1}{\sec x}=\frac{\sec x+1}{\sec x}
    Ahh, I'm an idiot. I'm so worried about trig identities and whatnot I forget basic rational addition.

    But thanks, masters. I really appreciate it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Identity Problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 2nd 2012, 04:51 PM
  2. Identity problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 23rd 2010, 05:33 AM
  3. identity problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 1st 2009, 04:00 PM
  4. Another Identity problem
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: October 27th 2009, 07:01 AM
  5. Trigonometric Identity Problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: September 2nd 2008, 05:00 PM

Search Tags


/mathhelpforum @mathhelpforum