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Thread: [SOLVED] Applying Identities

  1. #1
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    [SOLVED] Applying Identities

    Hello,

    Was hoping I could get some direction in this one, can't seem to get the desired results.

    The problem:

    $\displaystyle 2 \cos( \alpha /2 ) = \cos \alpha + 1$ Solve for all possible

    Now, using the half angle identity I get

    $\displaystyle 2 (\pm \sqrt \frac{1+\cos \alpha}{2}) = \cos \alpha + 1$

    $\displaystyle 4 \frac{1 + \cos \alpha}{2} = (\cos \alpha + 1)^2$

    $\displaystyle 2 + 2 \cos \alpha = \cos^2 \alpha + 2 \cos \alpha + 1$ simplify

    $\displaystyle \cos^2 \alpha - 1 = 0 $

    This is where I'm at. The only thing I can think of is to factor out a $\displaystyle \cos \alpha$

    but then I'm left with $\displaystyle \cos \alpha ( \cos \alpha + \frac{1}{\cos \alpha}) = 0 $

    or $\displaystyle \cos \alpha ( \cos \alpha + \sec \alpha)= 0$

    The answers to the problem are:

    $\displaystyle 180 + k (720), 540 + k(720), k(720)$

    So I'm pretty sure I'm missing something here.

    thank you!
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  2. #2
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    From $\displaystyle \cos^2 \alpha - 1 = 0$ you get $\displaystyle \cos^2 \alpha = 1$, from which you have the solutions

    $\displaystyle \cos \alpha = \pm 1$.

    Thus, $\displaystyle \alpha = k \pi$.

    The only way you could get three solutions would be to have an equation with $\displaystyle \cos^3 (\alpha + \delta)$ or $\displaystyle \cos (3\alpha + \delta)$ in it, for some $\displaystyle \delta$. Also, the solutions would be equally spaced, which your given solution is not.
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  3. #3
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    Put $\displaystyle \alpha=2t$ and your equation becomes $\displaystyle 2\cos t=\cos(2t)+1=2\cos^2t,$ which is not hard to solve.
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  4. #4
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    (Post removed because I can't read... )


    01
    Last edited by yeongil; Jul 13th 2009 at 08:57 PM. Reason: Post removed because I can't read...
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    From $\displaystyle \cos^2 \alpha - 1 = 0$ you get $\displaystyle \cos^2 \alpha = 1$, from which you have the solutions

    $\displaystyle \cos \alpha = \pm 1$.

    Thus, $\displaystyle \alpha = k \pi$.

    The only way you could get three solutions would be to have an equation with $\displaystyle \cos^3 (\alpha + \delta)$ or $\displaystyle \cos (3\alpha + \delta)$ in it, for some $\displaystyle \delta$. Also, the solutions would be equally spaced, which your given solution is not.
    The answer in book says it's it's k(720) or k($\displaystyle 4 \pi$). Instead of $\displaystyle k \pi
    $
    Because of the original $\displaystyle 2 \cos( \alpha /2 )$ the period is 4 $\displaystyle \pi$.
    Would that explain why there are three separate answers?
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  6. #6
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    Now that I looked at the problem more closely, the problem with your method, which icemanfan and I expanded on...
    Quote Originally Posted by icemanfan View Post
    From $\displaystyle \cos^2 \alpha - 1 = 0$ you get $\displaystyle \cos^2 \alpha = 1$, from which you have the solutions

    $\displaystyle \cos \alpha = \pm 1$.

    Thus, $\displaystyle \alpha = k \pi$.
    ... is that by squaring both sides you introduced an extraneous roots. Saying that the solution is $\displaystyle \alpha = k \pi$ implies that $\displaystyle \alpha = 2\pi$ would be a solution, but plugging in that solution for alpha doesn't work:
    $\displaystyle 2 \cos( \alpha /2 ) = \cos \alpha + 1$
    $\displaystyle 2 \cos( 2\pi /2 ) = \cos 2\pi + 1$
    $\displaystyle 2(-1) = 1 + 1$
    $\displaystyle -2 = 2 ????$

    (This is what I get for reading too fast. )


    So, Krizalid's method is probably the best one:
    Let $\displaystyle \alpha = 2t$, then
    $\displaystyle \begin{aligned}
    2\cos t &= \cos (2t) + 1 \\
    2\cos t &= 2\cos^2(2t) - 1 + 1 \\
    2\cos t &= 2\cos^2 t \\
    2\cos^2 t - 2\cos t &= 0 \\
    2\cos t(cos t - 1) &= 0 \\
    \end{aligned}$

    So
    2 cos t = 0 or cos t = 0, which means that
    t = π/2 + 2πk and t = 3π/2 + 2πk
    But we said α = 2t, so
    α = π + 4πk and α = 3π + 4πk
    [or α = 180 + (720)k or α = 540 + (720)k].

    cos t - 1 = 0 or cos = 1, which means that
    t = 0 + 2πk or t = 2πk
    But again, α = 2t, so
    α = 4πk
    [or α = (720)k].


    01
    Last edited by yeongil; Jul 13th 2009 at 08:58 PM.
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  7. #7
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    Quote Originally Posted by yeongil View Post
    Now that I looked at the problem more closely, the problem with your method, which icemanfan and I expanded on...

    ... is that by squaring both sides you introduced an extraneous roots. Saying that the solution is $\displaystyle \alpha = k \pi$ implies that $\displaystyle \alpha = 2\pi$ would be a solution, but plugging in that solution for alpha doesn't work:
    $\displaystyle 2 \cos( \alpha /2 ) = \cos \alpha + 1$
    $\displaystyle 2 \cos( 2\pi /2 ) = \cos 2\pi + 1$
    $\displaystyle 2(-1) = 1 + 1$
    $\displaystyle -2 = 2 ????$

    (This is what I get for reading too fast. )


    So, Krizalid's method is probably the best one:
    Let $\displaystyle \alpha = 2t$, then
    $\displaystyle \begin{aligned}
    2\cos t &= \cos (2t) + 1 \\
    2\cos t &= 2\cos^2(2t) - 1 + 1 \\
    2\cos t &= 2\cos^2 t \\
    2\cos^2 t - 2\cos t &= 0 \\
    2\cos t(cos t - 1) &= 0 \\
    \end{aligned}$

    So
    2 cos t = 0 or cos t = 0, which means that
    t = π/2 + 2πk and t = 3π/2 + 2πk
    But we said α = 2t, so
    α = π + 4πk and α = 3π + 4πk
    [or α = 180 + (720)k or α = 540 + (720)k].

    cos t - 1 = 0 or cos = 1, which means that
    t = 0 + 2πk or t = 2πk
    But again, α = 2t, so
    α = 4πk
    [or α = (720)k].


    01
    Fantastic! Thank you everyone for the help
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