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Math Help - Inverse cosine

  1. #1
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    Inverse cosine

    cos2x = 3/4

    So to find x, I would go: (cos-1 (3/4)) / 2

    This answer gives x = 20.7

    But the memo gives x = 0.36 and 2.78.

    How did they get to those 2 values?
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  2. #2
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    Re: Inverse cosine

    The book's answer is given in radians. Yours is in degrees. Always use radians!

    Also, there are infinite number of possible such x:

    2x \in \left\{\pm \cos^{-1}(3/4) +  2 \pi n \ | \ n \in \mathbb{Z} \right\}, so

    x \in \left\{\pm \frac{\cos^{-1}(3/4)}{2} + \pi n \ | \ n \in \mathbb{Z} \right\}.

    The books 2nd answer corresponds to \pi - \frac{\cos^{-1}(3/4)}{2}.

    In a sense, there are just two "different" answers (given by that \pm), because all the rest are just integer mulitples of \pi added to those two.
    Last edited by johnsomeone; October 3rd 2012 at 07:23 AM.
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  3. #3
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    Re: Inverse cosine

    Ah, yes I forgot that the sum includes the limits 0 <= x <= pi.

    So how would I go about working it out?
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    Re: Inverse cosine

    Inverse cosine-cos.png

    Ah, yes I forgot that the sum includes the limits 0 <= x <= pi.
    Which sum?
    Attached Thumbnails Attached Thumbnails Inverse cosine-cos.png  
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  5. #5
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    Re: Inverse cosine

    Did you understand what I wrote?

    Do you know where " 2x \in \left\{\pm \cos^{-1}(3/4) +  2 \pi n \ | \ n \in \mathbb{Z} \right\}" comes from?

    It's algebraically how you get all solutions. (In this case, we had to divide by 2 everywhere from that initial set.) Then to see how many solutions fall within some interval stated by your problem, you simply examine your solution set and algebraically decide which ones are in the desired range. That typically involves "getting an inequality for n" and solving it.

    -----------------------------------------------------------------------------------------------------------------
    Here that would be 0 \le \cos^{-1}(3/4) + \pi n \le \pi and 0 \le -\cos^{-1}(3/4) + \pi m \le \pi.

    Since \cos^{-1} is in [0, \pi], for the first case must have n=0 only, since if you added \pi to something positive you'd jump ahead of \pi, and likewise if you subtracted a \pi, since you're not bigger than \pi, you'd end up negative.

    For simlar reasons, m = 1.

    So you should hopefully just "see it". However, in algebraic detail, solving those inequalities you get:

    -\frac{\cos^{-1}(3/4)}{\pi} \le n \le \frac{\pi - \cos^{-1}(3/4)}{\pi} and \frac{\cos^{-1}(3/4)}{\pi} \le m \le \frac{\pi + \cos^{-1}(3/4)}{\pi}

    Now you could put that into your calculator to get: -.23 \le n \le .77 and .23 \le m \le 1.23, so n = 0 and m = 1.

    Or you could note that, since \cos^{-1}(3/4) is in (0, \pi), it follows that 0 < \frac{\cos^{-1}(3/4)}{\pi} < 1.

    Thus

    -1<-\frac{\cos^{-1}(3/4)}{\pi} \le n \le \frac{\pi - \cos^{-1}(3/4)}{\pi} < 1 and 0< \frac{\cos^{-1}(3/4)}{\pi} \le m \le \frac{\pi + \cos^{-1}(3/4)}{\pi}< 1+1 = 2,

    so again n = 0 and m = 1.
    -----------------------------------------------------------------------------------------------------------------

    I'll give a visual description - maybe it'll make sense, maybe not:
    The full solution is an infinite set of points. Imagine them highlighted on a number line. The solution in a range will just be a "window" on the number line. However, that "2x" means that the final x you get will "bring more solutions into your window", because to go from 2x to x is to multiply by 1/2, which "shrinks the number line".
    Last edited by johnsomeone; October 3rd 2012 at 08:01 AM.
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  6. #6
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    Re: Inverse cosine

    Ok, I understood what you said there. So, my answer of 20.7 is just one of those valid numbers on the line, but it doesn't fall within the interval. So how do I get those answers of x = 0.36 and 2.78?
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    Re: Inverse cosine

    Quote Originally Posted by yorkey View Post
    Ah, yes I forgot that the sum includes the limits 0 <= x <= pi.
    So how would I go about working it out?
    First, the given answer is in radian measure.
    Your answer in the OP is in degrees.

    If the range is really 0\le x\le  \pi .
    The other answer is \pi -\frac{\arccos\left(\frac{3}{4}\right)}{2}.
    Last edited by Plato; October 3rd 2012 at 12:39 PM.
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  8. #8
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    Re: Inverse cosine

    I'm afraid I still don't understand. Here's the question:
    Attached Thumbnails Attached Thumbnails Inverse cosine-question.png  
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  9. #9
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    Re: Inverse cosine

    Quote Originally Posted by yorkey View Post
    I'm afraid I still don't understand. Here's the question:
    Let this be a lesson to you. Always always post the entire question.
    Having seen the whole question, I agree with the first reply. There are two answers in the range 0\le x\le \pi.
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