Results 1 to 8 of 8

Thread: prov trig indentity

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    631

    prov trig indentity

    prove that $\displaystyle cos4x = 8cos^{4}x - 8cos^{2}x - 1 $

    I have tried for quite a while, but am completely stuck now.

    Can anyone have ago?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Quote Originally Posted by Tweety View Post
    prove that $\displaystyle cos4x = 8cos^{4}x - 8cos^{2}x - 1 $

    I have tried for quite a while, but am completely stuck now.

    Can anyone have ago?

    Thanks.
    You should use the identity

    $\displaystyle \cos(2u)= 2\cos^2(u)-1$

    Now make $\displaystyle u = 2x$

    $\displaystyle \cos(2\times 2x)= 2\cos^2(2x)-1$

    $\displaystyle \cos(4x)= 2\cos^2(2x)-1$

    Now use it again

    $\displaystyle \cos(4x)= 2(2\cos^2(x)-1)^2-1$

    Now expand the RHS, you should have your answer.
    Last edited by pickslides; Jan 12th 2010 at 12:51 PM. Reason: bad latex
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Hi Tweety,

    first factorise..

    $\displaystyle 8Cos^{2}x(Cos^{2}x-1)-1=8Cos^{2}x(-Sin^{2}x)-1$

    $\displaystyle =-8Cos^{2}xSin^{2}x-1=-8\frac{1}{2}(1+cos2x)\frac{1}{2}(1-cos2x)-1$

    $\displaystyle =-2(1-Cos^{2}2x)-1=-2[1-\frac{1}{2}(1+Cos4x)]-1$

    $\displaystyle =-2+1+Cos4x-1$

    I guess you have a typo,
    i reckon your original expression should have +1 instead of -1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by pickslides View Post
    You should use the identity

    $\displaystyle \cos(2u)= 2\cos^2(u)-1$

    Now make $\displaystyle u = 2x$

    $\displaystyle \cos(2\times 2x)= 2\cos^2(2x)-1$

    $\displaystyle \cos(4x)= 2\cos^2(2x)-1$

    Now use it again

    $\displaystyle \cos(4x)= 2(2\cos^2(x)-1)^2-1$

    Now expand the RHS, you should have your answer.

    Thank you,

    Could you please explain the last step, what happened to 2x? How did it become just 'x'?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    $\displaystyle Cos2x=2Cos^{2}x-1$

    $\displaystyle Cos^{2}2x=(Cos2x)^2$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Quote Originally Posted by Tweety View Post
    Thank you,

    Could you please explain the last step, what happened to 2x? How did it become just 'x'?
    Archie Meade has explained the substitution I have used, is it clear?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by pickslides View Post
    Archie Meade has explained the substitution I have used, is it clear?
    Yes thank you!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    The method that pickslides used was very neat, Tweety.

    He used the same identity at each step.

    You may see the identity written as

    $\displaystyle cos^{2}A=\frac{1}{2}(1+cos2A).$

    Rearranging this, we get

    $\displaystyle 2cos^{2}A=1+cos2A$

    so

    $\displaystyle cos2A=2cos^{2}A-1$

    He simply used this, by letting A=2x to get cos4x
    and the powers of x follow.

    I started with the powers and worked back, eventually using that identity,
    but i also used

    $\displaystyle cos^{2}x+sin^{2}x=1$

    in writing

    $\displaystyle cos^{2}x-1=-sin^{2}x$

    The other identity i used is

    $\displaystyle sin^2x=\frac{1}{2}(1-cos2x)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Trig Indentity Limits
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Jun 2nd 2010, 10:20 PM
  2. trig indentity proof help,
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Dec 15th 2009, 12:58 PM
  3. please check my working, trig indentity proof
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Dec 14th 2009, 08:22 AM
  4. Trig Indentity Help (Proof)
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Mar 5th 2008, 03:21 PM
  5. Trig indentity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 2nd 2006, 01:30 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum