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I tried to continue my homework but ran into another problem I couldn't do, literally can't continue now.

I have to graph $\displaystyle y=\frac 32 \sin2\left(x+ \frac{\pi}{4}\right)$

What do I do with $\sin 2$? Is it $\sin(2x + \pi/2)$, $\sin4(x/2 + \pi/8)$ or what?

Edit: I've got that the period of the function is $\pi$.

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What do you mean by sintwo? This makes it less clear, not more. You need to distinguish between the two readings (of the original) in Robert Israel's answer. –  Ross Millikan Jun 13 '11 at 23:03
    
Adam: I take it you mean $y = \frac 32 \sin(2(x + \frac{\pi}{4}))$, is that right? that is $\frac 32 \times \sin(2\alpha)$ where $\alpha = (x + \frac {\pi}{4})$? –  amWhy Jun 13 '11 at 23:12
    
As @Ross noted below, (if my comment is correct in that you were wondering what to do about sin(2 times an angle), then your expansions are both correct, and your period is correct (as you wrote in reply to Ross. You're on your way...From your answer to Didier, it seems clear that you are not dealing with sin squared...I'll edit your question to make this clear, but if you don't like my edit, feel free to change it. –  amWhy Jun 13 '11 at 23:29
    
Check if your answer is correct with Wolfram. It's useful to note that $\sin 2(x + \pi/4) = \sin (2x + \pi/2) = \cos 2x$. –  Luke Jun 13 '11 at 23:48
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3 Answers 3

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Based on what you give lower down, you want to graph $y=\frac {3}{2} \sin (2(x+\frac{\pi}{4}))$. Both your expansions are correct. To determine the period, how much does $x$ have to increase to make the argument of the sine function (the stuff in the parentheses) increase by $2\pi$? Both your expansions should give the same answer to that question.

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Usually to find the period I just divide 2pi by the number behind sin or whatever. So I guess it should be pi? –  Adam Jun 13 '11 at 22:29
    
If I interpreted it correctly, you are right that $\pi$ is the period. The period is also $\pi$ if it is $\frac{3}{2} \sin^2 (x + \frac{\pi}{4})$, but the graph looks different. –  Ross Millikan Jun 13 '11 at 22:49
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Since the sine function has period $2\pi$ and anti-period $\pi$, the function $\sin^2$ has period $\pi$. Another approach is to use the double angle formula $\cos(2u)=1-2\sin^2(u)$, hence your function is $y=(3/4)(1-\cos(2x+\pi/2))=(3/4)(1+\sin(2x))$.

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Eh, I don't know what any of that means and I don't think we have covered any of that in class. –  Adam Jun 13 '11 at 22:29
    
Any of that? Really? Surely you covered the sine function, didn't you? You also might know what a period is... Anyway, if you want to get some further explanations (which I am quite willing to write) about some steps of this solution, I suggest you try a different approach than your last comment... –  Did Jun 13 '11 at 22:36
    
I don't know what anti-period pi is, function sin squared, double angle formula cos(2u) or u. –  Adam Jun 13 '11 at 22:38
    
An anti-period of a function $f$ is any number $T$ such that $f(x+T)=-f(x)$ for every $x$. The function $\sin^2$ is the function $f$ defined by $f(x)=(\sin(x))^2$. The double angle formula is the formula written in my post, which holds for any real number $u$ (hence $u$ is a dummy variable here) and which you could use for $u=x+\pi/4$. Anything else unclear? –  Did Jun 13 '11 at 22:42
    
I understand that but I can't wrap my head around how that works or how to use it all. I have never seen sin^2 on a graph. –  Adam Jun 13 '11 at 22:45
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If the question is not clear, you should ask your teacher to clarify it. We can't read the teacher's mind - we can't even see the original question, just your transcription of it. I rather suspect that it should be $\frac{3}{2} \sin^2 (x + \frac{\pi}{4})$, that is, (3/2) * (sin(x+pi/4))^2, but that's only a guess. It might be $\frac{3}{2} \sin(2(x+\frac{\pi}{4}))$.

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It is exactly what I put down. –  Adam Jun 13 '11 at 22:27
    
Also I do ask the teacher questions after class but the class moves really fast and I have 4 sections of homework to cover, I can't wait until tomorrow to finish 3 more sections of homework, because then I will have 7 sections to do. –  Adam Jun 13 '11 at 22:30
    
@Adam: You are unlikely to get the right answer unless you work on the right problem. We had this issue on your last question as well-you need to know whether you have $\sin 2\theta$ or $\sin^2\theta$. The way you write it is not clear. If it is not clear in your notes, you are in trouble. –  Ross Millikan Jun 13 '11 at 22:47
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