Results 1 to 2 of 2

Math Help - Proof there is no real roots....

  1. #1
    Senior Member
    Joined
    Jul 2008
    Posts
    347

    Question Proof there is no real roots....

    Hi

    Given that a^2 + b^2 < c^2, show that the equation acosx + bsin x = c has no real roots.

    SOLUTION:

    acosx + bsiny = c
    as cos(x-y) = c/sqrt(a^2 + b^2)
    where cos y = a/sqrt(a^2+b^2) and siny = b/sqrt(a^2 + b^2)
    Since a^2 + b^2 < c^2
    cos(x-y)>1
    Hence, no real solution

    Could someone please explain to me how the solution works? I'm confused.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    a \cos x + b \sin x = c

    Divide by \sqrt{a^2+b^2}

    \frac{a}{\sqrt{a^2+b^2}}\: \cos x + \frac{b}{\sqrt{a^2+b^2}}\:  \sin x = \frac{c}{\sqrt{a^2+b^2}}

    \frac{a}{\sqrt{a^2+b^2}} and \frac{b}{\sqrt{a^2+b^2}} are two reals such that the sum of their square is equal to 1

    \left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = \frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2} = 1

    Therefore there exists y such that

    \cos y = \frac{a}{\sqrt{a^2+b^2}} and \sin y = \frac{b}{\sqrt{a^2+b^2}}

    The equation becomes

    \cos x \:\cos y +  \sin x \:\sin y = \frac{c}{\sqrt{a^2+b^2}}

    \cos (x-y) = \frac{c}{\sqrt{a^2+b^2}}

    And since \cos (x-y) \leq 1 and \frac{c}{\sqrt{a^2+b^2}} > 1 there are no solution
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Real roots
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 23rd 2011, 09:04 AM
  2. Real and non-real roots to a quintic equation
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: February 14th 2011, 08:42 PM
  3. Replies: 7
    Last Post: February 10th 2011, 08:34 PM
  4. real roots
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 17th 2010, 05:19 AM
  5. Replies: 8
    Last Post: April 7th 2009, 12:15 PM

Search Tags


/mathhelpforum @mathhelpforum