Graphing sin curve?

Question

I am trying real hard to figure this out and I know the what it should be but I just can't get it.

y = 5 sin 2x

So the period is 2pi/2 which is pi. So I know that this has double the frequency of sinx.

I know the amplitude is 5 so it goes from -5,5 on y.

Now the book says take my period and divide by 4 to get my points. Each time I do this I get only the upper half of my graph? Please can someone give me a step by step of the little equations used and explain, divide this by this and check unit circle for this and then multiply that etc etc. For some reason I'm missing something real easy here that this book just assumes we know because it skips right through getting the points and goes on to phase shifting without showing how to get points...

Please help me!!!

Answer 1

the sin x curve is periodic, so it is enough to know shape over one period and repeat the shape.
sin x is 0 for x = 0, pi, 2pi.
exactly between 0 and pi, you get max = 1
exactly between pi and 2 pi you have minimum = -1.
now to plot sin 2x plot the sin x curve
instead of the x intercepts being 0, pi , 2pi
you will divide each by 2
your maximum and minimum are exactly between your intercepts. the 5 multiplies the amplitude by 5, so just relabel the highest y as 5 instead of 1 and minimum as -5 instead of -1.

to draw sin 3x, draw sinx curve.
divide each x - intercept by three and you have the sin 3x curve.
for: 2sin3x just change scale on y axis so that maximum is 2 insteead of 1 and minimum is -2 instead of -1

Answer 2

The "main points" of the graph are at quarter cycles.

Sin is 0 at 0. It is positive max at 1/4 of the cycle, zero again at 2/4 of cycle, maximum negative (i.e., minimum) at 3/4 and zero again at 4/4.

So the book means take the period that you have found (pi) and divide it by 4 to find the critical points:

0 ... 0
pi/4 ... +5
2pi/4 ... 0
3pi/4 ... -5
pi ... 0
etc. for ever.

Next,
divide cycle by 12 (here, we get pi/12) and measure this from the roots (where y = 0); this gives you the half values:

0 + pi/12= pi/12 ... +2.5

2pi/4 = 6pi/12

6pi/12 - pi/12 = 5pi/12 ... +2.5
6pi/12 + pi/12 = 7pi/12 ... - 2.5
pi - pi/12 = 11pi/12 ... -2.5

(In the normal cycle -- 2pi -- the angle 2pi/12 = pi/6 has sine of 0.5)

The missing "twelfths" are where sin = half of root(3) = √3/2 = 0.866...

0 ... 0
1pi/12 ... +2.5
2pi/12 ... +4.33 (= 5 * 0.866)
3pi/12 ... +5
4pi/12 ... +4.33
5pi/12 ... +2.5
6pi/12 ... 0
7pi/12 ... -2.5
8pi/12 ... -4.33
9pi/12 ... -5
10pi/12 ... -4.33
11pi/12 ... -2.5
12pi/12 ... 0

This is about as far as you can go without anything more than a pencil and paper (unless you have memorized other values of sine)

---

Unless you are having calculator problem

In radians:
Do "3*pi/4" press "=" (this is x)
*2 = (this is 2x)
"Sin", then press "=" (This is Sin(2x) )
*5 = (This is 5 Sin(2x) )

Answer 3

Yes, you are all correct. Your period is indeed pi. One-fourth of that is pi/4. You would evaluate y(0) = 0, y(pi/4) = 5, y(pi/2) = 0, y(3pi/4) = -5, y(pi) = 0.

Answer 4

I don't really know if this website will work, but on my College Blackboard account they posted Falcon, the E-Learning Resources Indexer on there and this is a result from there.

http://falcon.tynemet.ac.uk/frameset.php?url=Sciences%20and%20Mathematics/Mathematics/granada_RB42_Sine,%20Cosine%20and%20Tangent/


And this website is for when you have a graphical calculator.

http://www.shodor.org/unchem/math/calc/index.html