Results 1 to 5 of 5

Math Help - Trigonometric Function

  1. #1
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1

    Trigonometric Function

    Hello, I need help on this question:

    Find the value, or values, of x for which 4sinhx - 3coshx = 5 giving your answer, or answers, to 3 significant figures.

    , Thank you in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,888
    Thanks
    326
    Awards
    1
    Quote Originally Posted by Air View Post
    Hello, I need help on this question:

    Find the value, or values, of x for which 4sinhx - 3coshx = 5 giving your answer, or answers, to 3 significant figures.

    , Thank you in advance.
    There are a couple of ways. Here's one that uses the basic definitions:
    4sinh(x) - 3cosh(x) = 5

    4 \cdot \frac{e^x - e^{-x}}{2} - 3 \cdot \frac{e^x + e^{-x}}{2} = 5

    Now for a little grouping work:
    2(e^x - e^{-x}) - \frac{3}{2}(e^x + e^{-x}) = 5

    \left ( 2 - \frac{3}{2} \right ) e^x + \left (-2 - \frac{3}{2} \right ) = 5

    \frac{1}{2}e^x - \frac{7}{2}e^{-x} = 5

    Multiply both sides by 2e^x:
    e^{2x} - 7 = 10e^x

    e^{2x} - 10e^x - 7 = 0

    This is a quadratic in e^x so you can finish this. If that is hard for you to see then:

    Let y = e^x. The equation becomes:
    y^2 - 10y - 7 = 0

    Solve this with the quadratic formula (or completing the square) then take the natural log of it to get x.

    For reference I got x \approx 2.3662.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, Air!

    Solve: . 4\sinh x - 3\cosh x \:= \:5 . to 3 significant figures.
    I would use the defintions of \sinh x and \cosh x

    . . 4\overbrace{\left(\frac{e^x - e^{-x}}{2}\right)}^{\sinh x} \:-\; 3\overbrace{\left(\frac{e^x + e^{-x}}{2}\right)}^{\cosh x} \;=\;5


    Multiply by 2: . 4(e^x-e^{-x}) - 3(e^x+e^{-x}) \:=\:10

    This simplifies to: . e^x - 10 - 7e^{-x} \:=\:0

    Multiply by e^x\!:\;\;e^{2x} - 10e^x - 7 \:=\:0 \quad\Rightarrow\quad(e^x)^2 - 10(e^x) - 7 \:=\:0 . . . a quadratic

    Quadratic Formula: . e^x \;=\;\frac{10 \pm\sqrt{10^2 - 4(1)(-7)}}{2(1)} \;=\;\frac{10 \pm\sqrt{128}}{2}\;=\;\frac{10 \pm8\sqrt{2}}{2}

    Therefore: . e^x \:=\:5 + 4\sqrt{2}\quad\Rightarrow\quad x \;=\;\ln(5 + 4\sqrt{2}) \;\approx\;2.37

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Quote Originally Posted by Soroban View Post
    Hello, Air!

    I would use the defintions of \sinh x and \cosh x

    . . 4\overbrace{\left(\frac{e^x - e^{-x}}{2}\right)}^{\sinh x} \:-\; 3\overbrace{\left(\frac{e^x + e^{-x}}{2}\right)}^{\cosh x} \;=\;5


    Multiply by 2: . 4(e^x-e^{-x}) - 3(e^x+e^{-x}) \:=\:10

    This simplifies to: . e^x - 10 - 7e^{-x} \:=\:0

    Multiply by e^x\!:\;\;e^{2x} - 10e^x - 7 \:=\:0 \quad\Rightarrow\quad(e^x)^2 - 10(e^x) - 7 \:=\:0 . . . a quadratic

    Quadratic Formula: . e^x \;=\;\frac{10 \pm\sqrt{10^2 - 4(1)(-7)}}{2(1)} \;=\;\frac{10 \pm\sqrt{128}}{2}\;=\;\frac{10 \pm8\sqrt{2}}{2}

    Therefore: . e^x \:=\:5 + 4\sqrt{2}\quad\Rightarrow\quad x \;=\;\ln(5 + 4\sqrt{2}) \;\approx\;2.37


    Edit: Too fast for me, Dan!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,888
    Thanks
    326
    Awards
    1
    Quote Originally Posted by Soroban View Post
    Edit: Too fast for me, Dan!
    Yeah, but you make fewer mistakes than I do. I'm always happy when I see that I got the same answer you did.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Max value of trigonometric function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 7th 2011, 12:33 PM
  2. Trigonometric Function
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: August 2nd 2010, 04:42 PM
  3. Trigonometric function
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: November 18th 2008, 09:32 PM
  4. Trigonometric Function
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: November 7th 2007, 05:18 AM
  5. Trigonometric Function
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 4th 2007, 01:37 PM

Search Tags


/mathhelpforum @mathhelpforum