graphing sin functions

**netbook** · August 19th 2009, 07:35 PM

ok so i have been asked to find the maximum y of y = 2sin 3x

I am not sure where to begin really or how to graph sin functions, anyone with an idea or two would be greatly appreciated.

**TKHunny** · August 19th 2009, 08:46 PM

This should be easy. You WILL BE expected to know this on an exam.

What does y = sin(x) look like? For this problem, it is most important to knwo the RANGE, [-1,1].

What does y = 2 sin(x) look like? You MUST see immediately that it has RAnge [-2,2]. You tell me why and then solve the problem.

**Amer** · August 19th 2009, 08:50 PM

Originally Posted by netbook

ok so i have been asked to find the maximum y of y = 2sin 3x

I am not sure where to begin really or how to graph sin functions, anyone with an idea or two would be greatly appreciated.

the functions of sin is bounded $-1\leq \sin ax \leq 1$ for any $a\in R$

**netbook** · August 19th 2009, 11:41 PM

ok so then 2 is the maximum y similarly 3sin x would be 3 or asin x a correct?

thanks

**pacman** · August 23rd 2009, 04:03 AM

why not see this graph? It may help you a bit, just click the picture

**HallsofIvy** · August 23rd 2009, 06:23 AM

Originally Posted by netbook

ok so then 2 is the maximum y similarly 3sin x would be 3 or asin x a correct?

thanks

I don't know what you mean by "or asin x a correct"! But yes, y= 2 sin(x) has range from y= -2 to 2.

Now, what about the "3" in sin(3x)? You should know that the basic function sin(x) has period $2\pi$ : one single period goes from x= 0 to $x= 2\pi$ . But here, instead of 'x' we have '3x'. That goes through one single period from 3x= 0 to [tex]3x= 2\pi[/itex] or x= 0 to $x= \frac{2\pi}{3}$ .

Here's what I would do: First draw two (faint) horizontal lines at y= -2 and y= 2. Now, mark the x-axis at x= 0 and $x= \frac{2\pi}{3}$ . Now mark the midpoint ( $x= \frac{\pi}{3}$ but on the actual graph you can do it by "eye"). Finally mark the two midpoints of those two intervals, $x= \frac{\pi}{6}$ and $\frac{\pi}{2}$ .
Starting at x= 0, go up to the upper line at the first mark, $x= \frac{\pi}{6}$ , back down to 0 at the middle, $x= \frac{\pi}{3}$ , down to the lower line at the third mark, $\frac{\pi}{2}$ , and finally, up to 0 at $x= \frac{2\pi}{3}$ . That is one period of the graph. It continues, copying that, on both sides.

**pacman** · August 23rd 2009, 07:51 AM

HallsofIvy, well done

**bandedkrait** · August 23rd 2009, 08:37 AM

I'd just like to add something to all everyone else has said.

1)the maximum value of asinbx where a and b are non-zero constants is always modulus(a).

2) The range of any sinusoidal function is affected only by the value of "a"

3)The period of the function is affected by the value of "b"

So if b>1, then the graph of asinbx will be somewhat compressed as the function attains its extrema very quickly compared to the original y=sinx.
If b<1, the period increases and the graph kind of expands.

4)Using a graphing software is an excellent way to understand all this. It worked really well for me.

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