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Math Help - Trigonometric functions and exact values

  1. #1
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    Trigonometric functions and exact values

    How should I go about solving this??

    If 0 is the same or less than x, which is the same as or less than 360,
    find:

    {x:cos2x=1}

    I cant seem to work out how to express this in degrees, rather than exact form in radians

    Thanks in advance,
    Steve
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    Quote Originally Posted by Stevo_Evo_22 View Post
    How should I go about solving this??

    If 0 is the same or less than x, which is the same as or less than 360,
    find:

    {x:cos2x=1}

    I cant seem to work out how to express this in degrees, rather than exact form in radians

    Thanks in advance,
    Steve
    Well, one full circle is 2\pi radians, so 2\pi\text{ rad} = 360^\circ or \pi\text{ rad} = 180^\circ

    You should become familiar enough with the common trig angles (0, 30, 45, 60, 90, etc) that you can convert without much thought. If you aren't there yet, just keep practicing.

    For \cos2x=1, we must ask, what angle gives a cosine of 1? This should come quickly if you think of the unit circle: \cos0^\circ = 1 (and again, if you aren't familiar with your unit circle, practice until you get good with it). In fact, any angle that is a multiple of 360 will have a cosine of 1, but since we are limited to the interval 0^\circ\leq x\leq 360^\circ\Rightarrow0^\circ\leq2x\leq 720^\circ, we get

    \cos2x = 1\Leftrightarrow 2x = 0^\circ\text{ or }2x = 360^\circ\text{ or }2x = 720^\circ

    \Rightarrow x = 0^\circ,\;x = 180^\circ,\text{ or }x = 360^\circ

    So, using your set notation,

    \left\{x\in\mathbb{R}\;\big|\;\cos2x = 1,\;0\leq x\leq 2\pi\right\} = \left\{0,\;\pi,\;2\pi\right\}
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  3. #3
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    But how do you find out that 2x=0, 2x=180 and 2x=360?
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    Quote Originally Posted by Stevo_Evo_22 View Post
    But how do you find out that 2x=0, 2x=180 and 2x=360?
    As I said, you need to know your unit circle, and you should know it well. I gave you this link; there is a very useful picture there, and you can find others easily.

    A point on a unit circle centered at the origin will be located at (\cos\theta,\;\sin\theta), where \theta is the angle between the positive x-axis and the line passing through the point and the origin (see my drawing below).

    Now, we want an angle that will give a cosine of 1. So we look on the unit circle for those angles that give an x coordinate of 1, and this will clearly be for

    \theta = 0^\circ, \pm360^\circ, \pm720^\circ,\ldots = 0 + 2n\pi = 0^\circ + 360^\circ n,\;n\in\mathbb{Z}

    Ideally, you should know (as in, have memorized) the sines and cosines of 0,\;\frac{\pi}6,\;\frac{\pi}4,\;\frac{\pi}3,\;\fra  c{\pi}2. Most other common angles can be found by comparison with a reference angle on the unit circle. Similarly, if presented with the reverse problem (finding an angle that will give a particular value for a certain function--that is, evaluating an inverse trig function) you should be able to do it without difficulty. I know having to memorize things like this can be unpleasant, but being able to quickly evaluate trigonometric functions at common angles (and in this case, the reverse) is important.

    One thing though: memorizing the sines and cosines of those angles is made much easier if you are familiar with the properties of 45-45-90 and 30-60-90 triangles--look it up if you like.
    Attached Thumbnails Attached Thumbnails Trigonometric functions and exact values-ucircle.png  
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  5. #5
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    AHAAAA!!!!

    Thankyou soooo much-your explanation has made the unit circle so much clearer to me you have no idea. I have all the exact values for sin, cos and tan 30, 45 and 60 but we weren't told to memorise them-it seems like this whole unit is a blur and none of it really made sense until now. I understand now how logically the angle must be 0, 180 or 360, but I still don't see the way that one can work it out-I will give you the way my teacher has done a worked example and could you please tell me what she did? Thanks.

    If 0 is the same as or less than x which is the same as or less than 360, find {x: cosx/2=(square root 2)/2}

    positive---1st or 4th quadrant

    inverse cosine((square-root 2)/2)=45degrees

    0 is the same as or less than x/2 which is the same as or less than 180

    x/2 = 45

    x=90
    Last edited by Stevo_Evo_22; June 30th 2008 at 02:30 AM.
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    Quote Originally Posted by Stevo_Evo_22 View Post
    I will give you the way my teacher has done a worked example and could you please tell me what she did? Thanks.

    If 0 is the same as or less than x which is the same as or less than 360, find {x: cosx/2=(square root 2)/2}
    Again, you just need to be familiar with your unit circle. You should know that \cos\frac\pi4=\frac{\sqrt2}2. If you do not have it memorized, consult your notes or reference material and see which basic angle gives a cosine of \frac{\sqrt2}2 (which can also be written \frac1{\sqrt2}) and you should find that one such angle is \frac\pi4.

    So, if \cos u = \frac{\sqrt2}2, then one possible solution for u is \frac\pi4. But, looking at the unit circle, you can see that there is another point with that x-coordinate, namely at \theta = -\frac\pi4 = \frac{7\pi}4. Then you also need to account for the fact that both of these angles can be written in infinitely many ways, by adding multiples of 2\pi.


    Going back to your worked example, it should hopefully make a bit more sense:

    We want to find x such that \cos\frac x2=\frac{\sqrt2}2,x\in[0,2\pi]

    positive---1st or 4th quadrant
    The cosine we were given is a positive value, \frac{\sqrt{2}}2, so it must lie either in the first or fourth quadrant (think about your unit circle: the cosine is the x-coordinate, so it will be positive in quadrants I and IV). This means that our angle must lie either in \left[0,\frac\pi2\right] or in \left[\frac{3\pi}2,2\pi\right].

    inverse cosine((square-root 2)/2)=45degrees
    If you are familiar with your unit circle, you should be able to determine that \arccos\frac{\sqrt2}2 = \frac\pi4 or 45^\circ. This is the angle in the first quadrant, and looking at the unit circle you should be able to find the fourth quadrant angle as \frac{7\pi}4 = 315^\circ. Note, however, that our condition for x means that 0\leq\frac x2\leq\pi, so 45^\circ is the only angle we need consider.

    x/2 = 45
    We found that our angle was 45, but our angle is \frac x2, so we have

    \frac x2 = 45^\circ = \frac\pi4

    \Rightarrow x = 90^\circ = \frac\pi2
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    Note, however, that our condition for x means that 0\leq\frac x2\leq\pi, so 45^\circ is the only angle we need consider

    Why is 7pi/4 not possible? Isn't it smaller than 2pi?
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    Quote Originally Posted by Stevo_Evo_22 View Post
    Note, however, that our condition for x means that 0\leq\frac x2\leq\pi, so 45^\circ is the only angle we need consider

    Why is 7pi/4 not possible? Isn't it smaller than 2pi?
    Yes, but our angle is x divided by 2. So if 0^\circ\leq x\leq360^\circ, then 0^\circ\leq\frac x2\leq180^\circ (dividing through by 2), so we only consider angles between 0 and 180 (or between 0 and \pi, if you like).

    Edit: In other words, if we considered \frac{7\pi}4, we would have

    \frac x2 = \frac{7\pi}4\Rightarrow x = \frac{7\pi}2, but this value of x is not in [0,2\pi]
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