Results 1 to 7 of 7

Thread: Trig

  1. #1
    Member srirahulan's Avatar
    Joined
    Apr 2012
    From
    Srilanka
    Posts
    180

    Arrow Trig

    $\displaystyle 5cos^2{\theta}+18cos{\theta}sin{\theta}+29sin^2{\t heta}$ Simplify this into $\displaystyle a+bcos(2{\theta}+{\alpha})$ a,b are constant and $\displaystyle \alpha$ is completely different from $\displaystyle \theta$
    Last edited by srirahulan; Oct 22nd 2013 at 07:46 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,387
    Thanks
    1345

    Re: Trig

    Show what work you have so far.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member srirahulan's Avatar
    Joined
    Apr 2012
    From
    Srilanka
    Posts
    180

    Re: Trig

    I write that like this $\displaystyle 2cos^2{\theta}-1+1+3cos^2{\theta}+9sin2{\theta}+29sin^2{\theta}$ But then i could not find out the $\displaystyle \alpha$ It is very difficult to arrange them

    A correction--- that$\displaystyle 29sin^2{\theta}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member srirahulan's Avatar
    Joined
    Apr 2012
    From
    Srilanka
    Posts
    180

    Re: Trig

    How can i find out the alpha and simplify like this...... please help??
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,387
    Thanks
    1345

    Re: Trig

    Try using the sum of angles formula for $\displaystyle \cos$ and see what you get: $\displaystyle a+b\cos(2\theta+\alpha) = a+b[\cos(2\theta)\cos(\alpha)-\sin(2\theta)\sin(\alpha)]$... Keep trying manipulations until you get something close to what you started with.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jul 2012
    From
    INDIA
    Posts
    863
    Thanks
    220

    Re: Trig

    Trig-25-oct-13-3.png
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848

    Re: Trig

    Hello, srirahulan!

    $\displaystyle 5\cos^2\theta+18\cos\theta\sin\theta+29\sin^2\thet a$

    $\displaystyle \text{Simplify to }\,a+b\cos(2\theta+\alpha),\:\text{ where }a,b\text{ are constants and }\alpha \ne \theta$

    $\displaystyle 5\cos^2\theta + 18\cos\theta\sin\theta + 29\sin^2\theta$

    . . $\displaystyle =\;5\left(\frac{1+\cos2\theta}{2}\right) + 9\sin2\theta + 29\left(\frac{1-\cos2\theta}{2}\right)$

    . . $\displaystyle =\;\tfrac{1}{2}\big(5 + 5\cos2\theta + 18\sin2\theta + 29 - 29\cos2\theta\big)$

    . . $\displaystyle =\;\tfrac{1}{2}\big[34 - 24\cos2\theta + 18\sin2\theta\big]$

    . . $\displaystyle =\;\tfrac{1}{2}\bigg[34 - 6\big(4\cos2\theta - 3\sin2\theta\big)\bigg]$

    . . $\displaystyle =\;\tfrac{1}{2}\bigg[34 - 30\left(\tfrac{4}{5}\cos2\theta - \tfrac{3}{5}\sin2\theta\right)\bigg]$


    $\displaystyle \text{Let }\cos\alpha = \tfrac{4}{5},\;\sin\alpja = \tfrac{3}{5}$

    $\displaystyle \text{We have: }\:\tfrac{1}{2}\bigg[34 - 30\left(\cos\alpha\cos2\theta - \sin\alpha\sin2\theta\right)\bigg]$

    . . . . .$\displaystyle =\;\tfrac{1}{2}\bigg[34 - 30\cps(2\theta + \alpha)\bigg] $

    . . . . .$\displaystyle =\;17 - 15\cos(2\theta + \alpha)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Sep 8th 2011, 07:00 PM
  2. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: Mar 14th 2011, 07:07 AM
  3. Replies: 7
    Last Post: Apr 15th 2010, 08:12 PM
  4. Replies: 6
    Last Post: Nov 20th 2009, 04:27 PM
  5. Replies: 2
    Last Post: Apr 21st 2006, 03:04 PM

Search Tags


/mathhelpforum @mathhelpforum