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Math Help - Extremely Urgent Trig Help Needed

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    Extremely Urgent Trig Help Needed

    I have an assignment to turn in tomorrow that will be a large part of my grade. I have had virtually no time to work over break and many other things to work on. Yada yada yada everyone rationalizes why they didn't get stuff done so it doesn't really matter why. I need to know how to do these problems by tomorrow morning. If someone was kind enough to type out the explanations I could finish it during lunch easily, I used to know how but forgot over a couple weeks, guess that goes to show my learning wasn't very deep.

    So here are the problems I will need to know by tomorrow. I appreciate anyone who can help me out right away.

    1.  \sin 2 \theta \ = 1/2

    2. <br />
3 \sqrt{2} \sin \theta + 2 = -1

    3.  2 \sin^2 \ theta\ - 3 \cos \theta +1 = 0

    4.  \cos 2\ theta = cos \ theta \

    Any and all help right away is very much appreciated. As I said this is really important and urgent.
    Last edited by Gerbilkit; January 2nd 2008 at 08:51 PM.
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    Quote Originally Posted by Gerbilkit View Post
    I have an assignment to turn in tomorrow that will be a large part of my grade. I have had virtually no time to work over break and many other things to work on. Yada yada yada everyone rationalizes why they didn't get stuff done so it doesn't really matter why. I need to know how to do these problems by tomorrow morning. If someone was kind enough to type out the explanations I could finish it during lunch easily, I used to know how but forgot over a couple weeks, guess that goes to show my learning wasn't very deep.

    So here are the problems I will need to know by tomorrow. I appreciate anyone who can help me out right away.

    1. Sin(2 theta) = 1/2
    what range for \theta do you want to solve for?

    you should know that (in general) if \sin x = \frac 12 then x = \frac {\pi}6 + 2k \pi or x = \frac {5 \pi}6 + 2n \pi for k,n \in \mathbb{Z}

    so, \sin 2 \theta = \frac 12 \implies 2 \theta = \frac {\pi}6 + 2k \pi or 2 \theta = \frac {5 \pi}6 + 2n \pi for k,n \in \mathbb{Z}

    now find \theta
    Last edited by Jhevon; January 2nd 2008 at 08:32 PM.
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    Quote Originally Posted by Gerbilkit View Post
    2. 3 square root of 2 sin(theta) + 2 = -1
    kinda confused here... did you mean 3 \sqrt{2 \sin \theta} + 2 = -1 ?
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    Sorry forgot to add these should all be solved on the interval of 0 <= theta <= 2pi.

    Could you elaborate slightly? My memory of the periodic functions is a bit hazy at the moment. Although I am looking at some more examples currently. Also my brain is a bit fried at the moment.

    And yes that was exactly what I meant, sorry I'm not great with the math tags having just started.
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    Quote Originally Posted by Gerbilkit View Post
    3. 2sin squared theta - 3 cos(theta) +1 = 0
    recall that \sin^2 \theta = 1 - \cos^2 \theta, so

    2 \sin^2 \theta - 3 \cos \theta + 1 = 0

    \Rightarrow 2 - 2 \cos^2 \theta - 3 \cos \theta + 1 = 0

    \Rightarrow 2 \cos^2 \theta + 3 \cos \theta - 3 = 0

    now this is quadratic in \cos \theta, if we say replace \cos \theta with x, we get:

    2x^2 + 3x - 3 = 0

    we can solve for x here, once we get x, just replace it with \cos \theta and solve for \theta
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    Quote Originally Posted by Gerbilkit View Post
    4. cos (2theta) = cos (theta)
    here, use the fact that \cos 2 \theta = 2 \cos^2 \theta - 1. plug that in for \cos 2 \theta and then you would solve it exactly like number 3
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    Quote Originally Posted by Gerbilkit View Post
    Sorry forgot to add these should all be solved on the interval of 0 <= theta <= 2pi.
    in that case, 2 \theta = \frac {\pi}6 or \frac {5 \pi}6

    Could you elaborate slightly? My memory of the periodic functions is a bit hazy at the moment. Although I am looking at some more examples currently. Also my brain is a bit fried at the moment.
    hmm, that will be a pain. how is your knowledge of reference angles or of the graphs of sine and cosine?

    And yes that was exactly what I meant, sorry I'm not great with the math tags having just started.
    ok then,

    3 \sqrt{2 \sin \theta} + 2 = -1 .............we are going to solve for sine. subtract 2 from both sides

    \Rightarrow 3 \sqrt{2 \sin \theta } = -3 ...........divide by 3 on both sides

    \Rightarrow \sqrt{2 \sin \theta } = -1 .............square both sides

    \Rightarrow 2 \sin \theta = 1 .............divide by 2 on both sides

    \Rightarrow \sin \theta = \frac 12

    which is the same as question 1


    (by the way, be sure to check your solutions in the original equation. sometimes when you square things and such, you introduce erroneous solutions).

    of course we can tell that this equation has no solution from the get-go. so doing all of this, while you would want to show your work, is a waste of time. why is that?
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    Quote Originally Posted by Jhevon View Post
    here, use the fact that \cos 2 \theta = 2 \cos^2 \theta - 1. plug that in for \cos 2 \theta and then you would solve it exactly like number 3
    I solved that out and factored it, and I got cos = -1/2 and cos = 1. Right now I'm thinking cos = 1 or 2 pi is the answer. Is that right?

    I'm sorry, I messed up when I wrote that to you. The sin value is not actually under the square root, only the two. So it doesn't quite work out the way you did it.
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    Quote Originally Posted by Gerbilkit View Post
    I solved that out and factored it, and I got cos = -1/2 and cos = 1. Right now I'm thinking cos = 1 or 2 pi is the answer. Is that right?
    yes, \cos \theta = - \frac 12 or \cos \theta = 1. but after that, you are wrong. you are supposed to find \theta, for what values of \theta do the above happen?

    Quote Originally Posted by Gerbilkit View Post
    I'm sorry, I messed up when I wrote that to you. The sin value is not actually under the square root, only the two. So it doesn't quite work out the way you did it.
    ok, then you tell me where to go with it this time. the plan of attack is the same. we are going to isolate the \sin \theta. so what's the equation? what does \sin \theta = ?
    Last edited by ThePerfectHacker; January 12th 2008 at 09:30 PM.
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    The true equation is.

    <br />
3 \sqrt{2} \sin \theta + 2 = -1<br />

    All the others should be correct.
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    Quote Originally Posted by Gerbilkit View Post
    The true equation is.

    <br />
3 \sqrt{2} \sin \theta + 2 = -1<br />

    All the others should be correct.
    yes. now solve for \sin \theta
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    I worked out everything now for the most part except for these two. The sin of 2 theta = 1/2 one. Please just work this out someone, I really tried to remember and work out periodic properties but I'm failing atm.

    Also I tried that one that worked out to a quadratic and didn't factor.


    <br />
2x^2 + 3x - 3 = 0<br />

    I put it through the quadratic equation and came up with.

     - 3 +- \sqrt{33}/ \ 4

    Did I do it wrong? I can't come up with an exact value of theta for this junk.
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    Quote Originally Posted by Gerbilkit View Post
    I worked out everything now for the most part except for these two. The sin of 2 theta = 1/2 one. Please just work this out someone, I really tried to remember and work out periodic properties but I'm failing atm.
    sin(2\theta)=\frac{1}{2}

    Let y = 2\theta

    sin(y)=\frac{1}{2}

    y = \frac{\pi}{6}

    2\theta = \frac{\pi}{6}

    \theta = \frac{\pi}{12}
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    Quote Originally Posted by Gerbilkit View Post
    Also I tried that one that worked out to a quadratic and didn't factor.


    <br />
2x^2 + 3x - 3 = 0<br />

    I put it through the quadratic equation and came up with.

     - 3 +- \sqrt{33}/ \ 4

    Did I do it wrong? I can't come up with an exact value of theta for this junk.
    Your solution to the quadratic looks fine. You could not have factored it, and were forced to use the quadratic formula.
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    Quote Originally Posted by colby2152 View Post
    sin(2\theta)=\frac{1}{2}

    Let y = 2\theta

    sin(y)=\frac{1}{2}

    y = \frac{\pi}{6}

    2\theta = \frac{\pi}{6}

    \theta = \frac{\pi}{12}
    we want 0 \le \theta \le 2 \pi so \theta = \frac {5 \pi}{12} is the other solution
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