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Math Help - [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:

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

    Now, using the half angle identity I get

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

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

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

    \cos^2 \alpha - 1 = 0

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

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

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

    The answers to the problem are:

    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 \cos^2 \alpha - 1 = 0 you get \cos^2 \alpha = 1, from which you have the solutions

    \cos \alpha = \pm 1.

    Thus, \alpha = k \pi.

    The only way you could get three solutions would be to have an equation with \cos^3 (\alpha + \delta) or \cos (3\alpha + \delta) in it, for some \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 \alpha=2t and your equation becomes 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; July 13th 2009 at 09:57 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 \cos^2 \alpha - 1 = 0 you get \cos^2 \alpha = 1, from which you have the solutions

    \cos \alpha = \pm 1.

    Thus, \alpha = k \pi.

    The only way you could get three solutions would be to have an equation with \cos^3 (\alpha + \delta) or \cos (3\alpha + \delta) in it, for some \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( 4 \pi). Instead of k \pi<br />
    Because of the original 2 \cos( \alpha /2 ) the period is 4 \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 \cos^2 \alpha - 1 = 0 you get \cos^2 \alpha = 1, from which you have the solutions

    \cos \alpha = \pm 1.

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

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


    So, Krizalid's method is probably the best one:
    Let \alpha = 2t, then
    \begin{aligned}<br />
2\cos t &= \cos (2t) + 1 \\<br />
2\cos t &= 2\cos^2(2t) - 1 + 1 \\<br />
2\cos t &= 2\cos^2 t \\<br />
2\cos^2 t - 2\cos t &= 0 \\<br />
2\cos t(cos t - 1) &= 0 \\<br />
\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; July 13th 2009 at 09:58 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 \alpha = k \pi implies that \alpha = 2\pi would be a solution, but plugging in that solution for alpha doesn't work:
    2 \cos( \alpha /2 ) = \cos \alpha + 1
    2 \cos( 2\pi /2 ) = \cos 2\pi + 1
    2(-1) = 1 + 1
    -2 = 2 ????

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


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