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Thread: Proof there is no real roots....

  1. #1
    Senior Member
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    Jul 2008
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    Question Proof there is no real roots....

    Hi

    Given that $\displaystyle a^2 + b^2 < c^2$, 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.
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  2. #2
    MHF Contributor
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    Hi

    $\displaystyle a \cos x + b \sin x = c$

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

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

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

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

    Therefore there exists $\displaystyle y$ such that

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

    The equation becomes

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

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

    And since $\displaystyle \cos (x-y) \leq 1$ and $\displaystyle \frac{c}{\sqrt{a^2+b^2}} > 1$ there are no solution
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