Results 1 to 7 of 7

Math Help - verifying,simplifying, factoring trigs

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    16

    verifying,simplifying, factoring trigs

    I think I've tried every way possible, except the correct way.
    Write the identity algebraically
    1) (1+csc x)/(cot x + cos x)=sec x

    factor the expression and use the fundamental identities to simplify
    2)sec^4 x - tan^4 x

    rewrite the expression so that it is not in fractional form.
    3)tan^2 x/csc x + 1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,547
    Thanks
    1418
    Quote Originally Posted by dmak263 View Post
    I think I've tried every way possible, except the correct way.
    Write the identity algebraically
    1) (1+csc x)/(cot x + cos x)=sec x

    factor the expression and use the fundamental identities to simplify
    2)sec^4 x - tan^4 x

    rewrite the expression so that it is not in fractional form.
    3)tan^2 x/csc x + 1
    1. \frac{1 + \csc{x}}{\cot{x} + \cos{x}} = \frac{1 + \frac{1}{\sin{x}}}{\frac{\cos{x}}{\sin{x}} + \cos{x}}

     = \frac{\frac{\sin{x} + 1}{\sin{x}}}{\frac{\cos{x} + \cos{x}\sin{x}}{\sin{x}}}

     = \frac{\sin{x} + 1}{\sin{x}} \times \frac{\sin{x}}{\cos{x} + \cos{x}\sin{x}}

     = \frac{\sin{x} + 1}{\cos{x} + \cos{x}\sin{x}}

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

     = \frac{1}{\cos{x}}

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

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,547
    Thanks
    1418
    Quote Originally Posted by dmak263 View Post
    I think I've tried every way possible, except the correct way.
    Write the identity algebraically
    1) (1+csc x)/(cot x + cos x)=sec x

    factor the expression and use the fundamental identities to simplify
    2)sec^4 x - tan^4 x

    rewrite the expression so that it is not in fractional form.
    3)tan^2 x/csc x + 1
    2. \sec^4{x} - \tan^4{x} = (\sec^2{x})^2 - (\tan^2{x})^2

     = (\sec^2{x} + \tan^2{x})(\sec^2{x} - \tan^2{x})

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

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

     = \left(\frac{1 + \sin^2{x}}{\cos^2{x}}\right)\left(\frac{\cos^2{x}}  {\cos^2{x}}\right)

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

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

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

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

     = 2\sec^2{x} - 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,547
    Thanks
    1418
    Quote Originally Posted by dmak263 View Post
    I think I've tried every way possible, except the correct way.
    Write the identity algebraically
    1) (1+csc x)/(cot x + cos x)=sec x

    factor the expression and use the fundamental identities to simplify
    2)sec^4 x - tan^4 x

    rewrite the expression so that it is not in fractional form.
    3)tan^2 x/csc x + 1
    3. Is this \frac{\tan^2{x}}{\csc{x} + 1} or \frac{\tan^2{x}}{\csc{x}} + 1?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2009
    Posts
    16
    for #3, it's the first equation.
    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 Prove It View Post
    3. Is this \frac{\tan^2{x}}{\csc{x} + 1} or \frac{\tan^2{x}}{\csc{x}} + 1?
    Quote Originally Posted by dmak263 View Post
    for #3, it's the first equation.

    Hi dmak263,

    \frac{\tan^2{x}}{\csc{x} + 1}

    Multiply top and bottom by \csc x - 1, and recall that \csc^2 x-1=\cot^2 x

    \frac{\tan^2 x(\csc x - 1)}{\csc^2 x - 1}=\frac{\tan^2 x(\csc x - 1)}{\cot^2 x}=\frac{\tan^2 x(\csc x - 1)}{\frac{1}{\tan^2 x}}=\tan^4 x(\csc x-1)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2009
    Posts
    16
    I'm lost for words realizing how easy that was. Didn't see it at all. Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help simplifying/factoring
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 6th 2011, 06:12 AM
  2. Simplifying by factoring
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 14th 2010, 01:02 AM
  3. Factoring and Simplifying Expressions.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 21st 2009, 03:58 AM
  4. HELP with factoring and simplifying
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 11th 2009, 12:26 AM
  5. Factoring a polynomial/simplifying
    Posted in the Algebra Forum
    Replies: 5
    Last Post: November 21st 2006, 04:47 AM

Search Tags


/mathhelpforum @mathhelpforum