<p>wait nevermind you’re right</p>
<p>int’l plug in the absolute value of cos (x/2). The same y value appears every 2pi, which is the period. 4pi works too but only because it is a multiple of 2pi. Absolute value matters.</p>
<p>I remember that trig problem too…
The equation indicated not only an absolute value, but a shift up the y-axis (at least, that’s what I had in my calculator afterwards)…when I plugged it into my calculator, I saw that there were no negative values, even without the abs value…</p>
<p>I might be wrong, but I do believe it’s 4pi…</p>
<p>Well I put 4 pi but I think it’s wrong. You probably already know this, but the period is how long it takes before you get the same y value. A shift in the y axis will not change the period, only the y values. Consider sinx. The period is 2pi for sinx and for sinx+2. Just sinx goes from 1 to 3 with sinx+2 and -1 to 1 with sinx.</p>
<p>^ false. It’s not when you get the same y value. For instance, the graph of sin(theta) crosses the x-axis three times total to complete its period. A period is not one arch. It’s one full wave (an arch and its inverse arch).</p>
<p>To illustrate my point: <a href=“http://www.analyzemath.com/Graphing/sine2.gif[/url]”>http://www.analyzemath.com/Graphing/sine2.gif</a>
That’s one full period. Note that y=0 more than twice.</p>
![http://www.analyzemath.com/Graphing/sine2.gif[/url]](http://www.analyzemath.com/Graphing/sine2.gif%5B/url%5D){kind=link}
<p>fro any of you who have read the stranger by albert camus, you will realize that argument over the period of a function is absurd. what is arguing going to change…nothing, what is knowing that your answer is right going to do…nothing, just drop it.</p>
<p>yeah you’re right.
so what was the answer?</p>
<p>What TheMan66 meant was probably
f(x) = f(x+T) for ANY x.</p>
<p>Sketch
y1=sin(x),
y2=abs(sin(x/2)) and
y3=abs(sin(x/2)) + 1
on [Graphing</a> Calculator](<a href=“Title of the document”>Title of the document).
All three functions y1, y2, and y3 have T=2pi.</p>
<p>I plugged the trig period equation into my calculator (and cleared it during the test, so I don’t have it anymore…) with abs(), so it was entered exactly as asked.</p>
<p>From memory, I think the equation was y=abs(…+3). Since the equation is moved up 3 units from the x-axis, it is unaffected by the absolute value function.</p>
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<p>The period, in radians, is 4π.</p>
<p>I had my calculator in degrees mode. The graph spanned the whole width of the screen, which was 720 degrees, i.e. 4π. I’m pretty sure the answer is 4π.</p>
<p>===============
Edit: Unless the +3 was outside abs()? I sure hope not :(</p>
<p>Wow this one problem is a big fuss. Graph abs(…(x/))+3.
You will see that the wave function repeats itself every 2pi.</p>
<p>I’m pretty sure it was 2Pi.
Though its true that just because, lets say the maximum point, is the same every 2pi, it doesn’t mean it’s the period. However, the graph of it was sin ( I think), and that would mean no matter what shifts there are, an absolute value would divide its normal period by 2.</p>
<p>Graph it yourself, the bottom part of the graph, the part that is negative in the Y, will be flipped, giving it the same exact graph every 2 pi</p>
<p>Edit: was it sin or cos
EditEdit: Doesn’t matter if its sin or cos. abs will make both graphs have periods half the original.
EditEditEdit: To above poster, are you sure that’s the equation? Looks weird.</p>
<p>
If Angelo and ichiruki remember the question correctly, then the answer is indeed
4pi.</p>
<p>sorry meant abs(…(x/2))+3
I put 4pi so that would be awesome if that was the answer.</p>
<p>ok guys, its official, a friend of mine has so kindly violated the CB rules and wrote that question on his registration ticket (such a useful bit of paper!) and confirmed that the +3 did in fact exist and is inside the absolute value. when i graphed it, i looked exactly as i remember it and there is no question that the period is 4pi.</p>
<p>Indeed, if that is correct, 4Pi would be correct. But the question is, is HE correct.</p>
<p>:/
Committee, period, deep water (maybe) and 1 omit.
This is so nerve racking :]</p>
<p>im sure that my friend is right 200% ;)</p>
<p>Thanks to the unknown hero, int’lstudent’12’s friend, this question is settled:</p>
<p>the period of the “weird” trig function ……………………………… 4pi</p>
<p>I’ve refrained from arguing, because I thought it was pointless, but it should be obvious to anyone who’s taken precal that the period is 4pi, right off the bat, no graphing necessary. The absolute value has “absolutely” (teehee pun) nothing to do with it. It’s the vital “x/2” that counts. That’s equivalent to (abbreviated equation, disregarding the stuff that doesn’t matter): sin(1/2)x. That number in the middle is always known as “b”, and the period of any trigonometric function is 2pi/b…ALWAYS. Since “b” is 1/2, the period is 4pi. Easy.</p>
<p>so why in my calc it was only cos(x/2)? I then reflected and got 2pi… I have to be blind not to see +3. Yeah I have -2.5 shortseeing, but does it mean I can’t read properly? So in this case, I made a stupid mistake/mistype/misthink :)</p>