Results 1 to 6 of 6

Math Help - help req/d proving an identity please?

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    17

    help req/d proving an identity please?

    hi there
    can anyone get me a little bit further with this question?
    prove the identity:

    cos 3x = 4cos^3x -3cosx cos 3x can also =cos(2x+x)

    so far i have used the formula

    cos(a+b) = cosa.cosb-sina.sinb

    so this i have equated to

    cos(2x+x)=cos2x.cosx-sin2x.sinx

    back to the orig equation cos2x.cosx-sin2x.sinx=4cos^3x-3cosx

    so thats where i end up and i am now confused as how to proceed next!

    any help appreciated.

    many thanks in advance sure its nearly there.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    You need to expand your cosine identities a bit. Use the one you stated to produce:

    \cos(2x) = \cos^{2}(x) - \sin^{2}(x) = 1 - 2\sin^{2}(x) = 2\cos^{2}(x) - 1

    This should get you where you are going.
    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
    You're on the right path, you need to use the double angle identities to expand further

    cos(3x) = cos(2x)cos(x)-sin(2x)sin(x). Useful identities are below

    cos(2x) = 2cos^2(x)-1

    sin(2x) = 2sin(x)cos(x)

    sin^2(x) = 1-cos^2(x)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2009
    Posts
    17

    many thanks

    hi again,firstly many thanks for the very prompt replies.

    i'm sorry to say i not quite getting there which is annoying me.

    i know the lhs needs to equal the rhs and that the identities can represent other identities and thats how it is all simplified but i cant seem to convert to get the appropriate identities and the theory is so simple!

    my ongoing attempt, is this getting there or not?

    cos2x.cosx-sin2x.sinx changes to the following?

    2cos^2-1.cosx-sin2x.sinx

    2cos^3x-1-sinx.sinx

    2cos^3-1-sin^2 2x?

    so its all going a bit pete tong as they say

    not seeing how this will equate to the rhs i'm afraid and i have honestly been trying for hours now.
    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 ads114 View Post
    hi again,firstly many thanks for the very prompt replies.

    i'm sorry to say i not quite getting there which is annoying me.

    i know the lhs needs to equal the rhs and that the identities can represent other identities and thats how it is all simplified but i cant seem to convert to get the appropriate identities and the theory is so simple!

    my ongoing attempt, is this getting there or not?

    cos2x.cosx-sin2x.sinx changes to the following?

    2cos^2-1.cosx-sin2x.sinx

    2cos^3x-1-sinx.sinx

    2cos^3-1-sin^2 2x?

    so its all going a bit pete tong as they say

    not seeing how this will equate to the rhs i'm afraid and i have honestly been trying for hours now.
    Take it one step at a time and see if you can follow my steps:

    cos(2x)cos(x)-sin(2x)sin(x)

    As we know cos(2x) = 2cos^2(x)-1 and sin(2x) = 2sin(x)cos(x)

    [2cos^2(x)-1]cos(x) - 2sin(x)cos(x)sin(x)

    Expand and simplify

    2cos^3(x)-cos(x) - 2cos(x)sin^2(x)

    As sin^2(x) = 1-cos^2(x)

    2cos^3(x)-cos(x)-2cos(x)[1-cos^2(x)]

    Expand again

    2cos^3(x)-cos(x)-2cos(x)+2cos^3(x)

    Collect like terms:  4cos^3(x)-3cos(x) = RHS
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2009
    Posts
    17

    many thanks

    it is actually incredibly straight forward isnt it?

    i think my problem was i was struggling with the multiplications believe it or not how stupid can you be.

    it is a great feeling when it all comes together and you can see where you went wrong ,thankyou again its very clear now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving an identity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: July 17th 2011, 01:23 PM
  2. Proving an identity.
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 12th 2011, 06:22 PM
  3. Proving an identity that's proving to be complex
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 21st 2009, 02:30 PM
  4. Proving identity #1
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 9th 2009, 03:33 PM
  5. Please help proving this identity
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 25th 2007, 12:36 PM

Search Tags


/mathhelpforum @mathhelpforum