Results 1 to 6 of 6

Math Help - Simple Simplification

  1. #1
    Member
    Joined
    Oct 2008
    From
    Melbourne
    Posts
    166

    Cool Simple Simplification

    Hey,

    Anyone know if this can be simplified any further?

    [\sqrt{4(cos^2t) + 9(sin^2t)} ]^3

    Cheers
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,
    Quote Originally Posted by tsal15 View Post
    Anyone know if this can be simplified any further?

    [\sqrt{4(cos^2t) + 9(sin^2t)} ]^3
    Yes, it can be simplified : \left[\sqrt{4\cos^2t + 9\sin^2t} \right]^3=\left[\sqrt{4(\cos^2t +\sin^2t)+ 5\sin^2t} \right]^3 and remember that \cos^2t+\sin^2t=1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2008
    From
    Melbourne
    Posts
    166
    Quote Originally Posted by flyingsquirrel View Post
    Hi,

    Yes, it can be simplified : \left[\sqrt{4\cos^2t + 9\sin^2t} \right]^3=\left[\sqrt{4(\cos^2t +\sin^2t)+ 5\sin^2t} \right]^3 and remember that \cos^2t+\sin^2t=1.
    hey flyingsquirrel,

    Is this as far as I can go?

    mathematically speaking...

    \left[\sqrt{5 + 5sin^2t} \right]^3
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by tsal15 View Post
    Is this as far as I can go?

    mathematically speaking...

    \left[\sqrt{5 + 5sin^2t} \right]^3
    Mathematically speaking, you're going too far.

    \left[\sqrt{4\cos^2t + 9\sin^2t}\right]^3=\left[\sqrt{{\color{red}4} + 5\sin^2t}\right]^3

    I don't think this last expression can be further simplified.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2008
    From
    Singapore
    Posts
    160
    Sorry flyingsquirrel may i know why the equation cannot be simplified this way ? Learning along the way

    [\sqrt{4(cos^2t) + 9(sin^2t)} ]^3

    [2cost + 3sint]^3

    8cos^3t + 27sin^3t
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by tester85 View Post
    Sorry flyingsquirrel may i know why the equation cannot be simplified this way ? Learning along the way

    [\sqrt{4(cos^2t) + 9(sin^2t)} ]^3

    [2cost + 3sint]^3


    Wooo, according to you, it is correct to write something like 5=\sqrt{25}=\sqrt{9+16}= \sqrt{3^2+4^2}=3+4=7 ?!

    The problem is that, in general, \sqrt{a^2+b^2} does not equal a+b : you can't say that \sqrt{4 \cos^2t + 9\sin^2t}=2\cos t+3\sin t and you can't say that \sqrt{3^2+4^2}=3+4.

    However, it is correct to write  \left[\sqrt{x}\right]^2=x or \sqrt{x^2}=|x|... but these two identities can't be used here.

    [2cost + 3sint]^3

    8cos^3t + 27sin^3t


    Again, in general, (a+b)^3 does not equal a^3+b^3. Remember that (a+b)^3=(a+b)(a+b)(a+b). If you expand this, you'll see that (a+b)^3=a^3+3a^2b+3ab^2+b^3...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simple Exponent Transformation/simplification
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 29th 2011, 09:16 AM
  2. Stuck on a simple logic simplification
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: October 27th 2010, 04:20 PM
  3. simple simplification
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 5th 2009, 12:46 PM
  4. Simplification too simple?
    Posted in the Algebra Forum
    Replies: 8
    Last Post: September 25th 2007, 04:16 PM
  5. A simple simplification problem
    Posted in the Algebra Forum
    Replies: 5
    Last Post: December 7th 2006, 11:31 AM

Search Tags


/mathhelpforum @mathhelpforum