Results 1 to 4 of 4

Thread: Find all solutions for and exact value of...

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    11
    Awards
    1

    Find all solutions for and exact value of...

    1) Find all solutions θ
    0 ≤ θ < 360 to the equation cos(2θ) = cosθ


    2) Find the exact value of cot(795)




    I have no idea how to do these, any help or tips?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by alexcross View Post
    1) Find all solutions θ
    0 ≤ θ < 360 to the equation cos(2θ) = cosθ


    2) Find the exact value of cot(795)




    I have no idea how to do these, any help or tips?
    For 1. you have to use some Double Angle identities.

    $\displaystyle \cos{(2\theta)} = \cos^2{\theta} - \sin^2{\theta} = \cos^2{\theta} - (1 - \cos^2{\theta}) = 2\cos^2{\theta} - 1$.


    So if $\displaystyle \cos{(2\theta)} = \cos{\theta}$

    $\displaystyle 2\cos^2{\theta} - 1 = \cos{\theta}$

    $\displaystyle 2\cos^2{\theta} - \cos{\theta} - 1 = 0$

    This is a quadratic equation. Let $\displaystyle X = \cos{\theta}$

    $\displaystyle 2X^2 - X - 1 = 0$

    $\displaystyle 2X^2 - 2X + X - 1 = 0$

    $\displaystyle 2X(X - 1) + 1(X - 1) = 0$

    $\displaystyle (X - 1)(2X + 1) = 0$


    So Case 1:

    $\displaystyle X - 1 = 0 \implies \cos{\theta} - 1 = 0 \implies \cos{\theta} = 1$

    Case 2:

    $\displaystyle 2X + 1 = 0 \implies 2\cos{\theta} +1 = 0 \implies \cos{\theta} = -\frac{1}{2}$.


    Solve both cases for $\displaystyle \theta$ over the interval $\displaystyle 0 \leq \theta < 360^\circ$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by alexcross View Post
    1) Find all solutions θ
    0 ≤ θ < 360 to the equation cos(2θ) = cosθ


    2) Find the exact value of cot(795)




    I have no idea how to do these, any help or tips?
    2. $\displaystyle \cot{795^\circ} = \cot{(2\times 360^\circ + 75^\circ)} = \cot{75^\circ}$

    $\displaystyle \cot{75^\circ} = \frac{1}{\tan{75^\circ}}$

    $\displaystyle = \frac{1}{\tan{(45^\circ + 30^\circ)}}$


    To evaluate this use the sum formula for tangent.

    $\displaystyle \tan{(\alpha + \beta)} = \frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}}$.

    Can you go from here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    Quote Originally Posted by alexcross View Post
    1) Find all solutions θ
    0 ≤ θ < 360 to the equation cos(2θ) = cosθ


    2) Find the exact value of cot(795)




    I have no idea how to do these, any help or tips?
    1. change $\displaystyle \cos(2\theta)$ to $\displaystyle 2\cos^2{\theta} - 1$ ...

    $\displaystyle 2\cos^2{\theta} - 1 = \cos{\theta}$

    $\displaystyle 2\cos^2{\theta} - \cos{\theta} - 1 = 0$

    factor and solve for $\displaystyle \theta$


    2. 795 - 2(360) = 75 , a coterminal angle

    $\displaystyle \cot(75) = \frac{\cos(45+30)}{\sin(45+30)} $

    use your sum identities for cosine and sine, then evaluate.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Exact solutions with radians
    Posted in the Trigonometry Forum
    Replies: 11
    Last Post: Mar 15th 2012, 04:33 AM
  2. Finding Exact Solutions Trig Equations
    Posted in the Trigonometry Forum
    Replies: 11
    Last Post: Dec 11th 2011, 04:42 PM
  3. Finding exact solutions in the interval
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Apr 30th 2010, 12:05 AM
  4. Finding exact solutions
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Apr 12th 2010, 10:07 PM
  5. Find exact solutions algebraically?
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: Dec 1st 2009, 10:28 AM

Search Tags


/mathhelpforum @mathhelpforum