Results 1 to 2 of 2

Math Help - Trignometric identities - Prove that the following is an identity.

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    83

    Trignometric identities - Prove that the following is an identity.

    Hello, I am having troubles with a question involving trigonometric identities. I have always had trouble with these types of problems, so I need as much help as possible.
    To the problem:

    "Prove that the following is an identity:  \frac{cos2\theta}{1 + sin2\theta} = \frac{cot \theta - 1}{cot \theta +1} "

    Here is my attempt:

    R.H.S
    \frac{\frac{1}{tan\theta} - 1}{\frac{1}{tan\theta} + 1}
    \frac{\frac{cos\theta}{sin\theta} -1}{\frac{cos\theta}{sin\theta} +1}

    I don't know what to do from here. On the left hand start I can't see where to even start. Any help at all is very much appreciated. I really do not understand these problems.

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    Quote Originally Posted by Kakariki View Post
    Hello, I am having troubles with a question involving trigonometric identities. I have always had trouble with these types of problems, so I need as much help as possible.
    To the problem:

    "Prove that the following is an identity:  \frac{cos2\theta}{1 + sin2\theta} = \frac{cot \theta - 1}{cot \theta +1} "

    Here is my attempt:

    R.H.S
    \frac{\frac{1}{tan\theta} - 1}{\frac{1}{tan\theta} + 1}
    \frac{\frac{cos\theta}{sin\theta} -1}{\frac{cos\theta}{sin\theta} +1}

    I don't know what to do from here. On the left hand start I can't see where to even start. Any help at all is very much appreciated. I really do not understand these problems.

    Thank you.
    \frac{\cot{\theta} - 1}{\cot{\theta} + 1} = \frac{\frac{\cos{\theta}}{\sin{\theta}} - 1}{\frac{\cos{\theta}}{\sin{\theta}} + 1}

    = \frac{\frac{\cos{\theta} - \sin{\theta}}{\sin{\theta}}}{\frac{\cos{\theta} + \sin{\theta}}{\sin{\theta}}}

     = \frac{\cos{\theta} - \sin{\theta}}{\cos{\theta} + \sin{\theta}}

     = \frac{(\cos{\theta} - \sin{\theta})^2}{(\cos{\theta} + \sin{\theta})(\cos{\theta} - \sin{\theta})}

     = \frac{\cos^2{\theta} - 2\cos{\theta}\sin{\theta} + \sin^2{\theta}}{\cos^2{\theta} - \sin^2{\theta}}

     = \frac{1 - \sin{(2\theta)}}{\cos{(2\theta)}}

     = \frac{[1 - \sin{(2\theta)}][1 + \sin{(2\theta)}]}{\cos{(2\theta)}[1 + \sin{(2\theta)}]}

     = \frac{1 - \sin^2{(2\theta)}}{\cos{(2\theta)}[1 + \sin{(2\theta)}]}

     = \frac{\cos^2{(2\theta)}}{\cos{(2\theta)}[1 + \sin{(2\theta)}]}

     = \frac{\cos{(2\theta)}}{1 + \sin{(2\theta)}}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. trignometric identities
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: November 24th 2011, 06:56 AM
  2. Trignometric Identities
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: September 22nd 2009, 03:44 AM
  3. lagrange's trignometric identity
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: July 7th 2009, 11:11 AM
  4. simplifying trignometric identity
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: June 4th 2009, 10:25 PM
  5. Trignometric Identity Questions
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: November 9th 2008, 04:42 PM

Search Tags


/mathhelpforum @mathhelpforum