Results 1 to 2 of 2

Thread: could you please solve these problem

  1. #1
    Newbie
    Joined
    Apr 2018
    From
    saudi arabia
    Posts
    1

    could you please solve these problem

    Write cos(for the power of 4 ) (x) as a function in terms of cos 2x and cos 4x
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,358
    Thanks
    2732

    Re: could you please solve these problem

    let's see...

    $\cos(2x)=\cos^2(x) - \sin^2(x) =\cos^2(x) - (1-\cos^2(x))= 2\cos^2(x)-1$

    so

    $\cos^2(x) = \dfrac{1+\cos(2x)}{2}$

    Spoiler:
    $\begin{align*}
    &\cos^4(x) = \\ \\

    &(\cos^2(x))^2 = \\ \\

    &\dfrac{(1+\cos(2x))^2}{4} = \\ \\

    &\dfrac 1 4\left(\cos^2(2x)+2\cos(2x)+1\right)

    \end{align*}$

    From above letting $x \to 2x$

    $\cos^2(2x) = \dfrac{1+\cos(4x)}{2}$

    So combining these we get

    $\begin{align*}
    &\cos^4(x) = \\ \\

    &\dfrac 1 4\left(\cos^2(2x)+2\cos(2x)+1\right) = \\ \\

    &\dfrac 1 4\left(\dfrac{1+\cos(4x)}{2}+2\cos(2x) + 1\right)

    \end{align*}$

    and you can complete the remaining algebra
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How do I solve this problem?
    Posted in the Statistics Forum
    Replies: 3
    Last Post: Apr 21st 2015, 07:21 PM
  2. Solve Problem
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Sep 7th 2010, 09:22 AM
  3. how to solve for x in this problem
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Aug 2nd 2010, 05:40 AM
  4. Solve Problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jun 26th 2008, 10:12 AM
  5. How do I solve this ln problem?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 10th 2008, 12:08 PM

/mathhelpforum @mathhelpforum