Graphing a sine

[llessurthick13]

u probably won't answer this or see this but if u do:

is there any way u can send me the points on the graph

[ProfessorElvisZap]

@llessurthick13 Since the graph is of y=2 Sin(3x-pi), I suggest setting the argument, 3x-pi, equal to several of your favorite angles and going with with. For example, 3x-pi= pi/6, gives x=7pi/(18). The value y=2( 3 (7pi/(18)--pi)= 2 Sin(pi/6)=1. Thus (7 pi/18, 1) is a point on the curve. Continue in this fashion with 3x-pi=pi/4, 3x-pi=pi/3, and 3x-pi= pi/2. These are exercises that you should work out on your own :-)

[saracaree]

he is a lefty!

[consumers007]

This could help a lot of people :D

[godlike899]

i'm confused...

it is said there that to get the period, you need to equate the one inside in the parentheses to 2π

but my teacher told be that in order to get the period, you should follow this formula: 2π/b

[ProfessorElvisZap]

@godlike899 A sine curve, y=a sin(bx+c), completes a full period in the interval when the argument (bx+c) moves from 0 to 2pi.

So I set bx+c=0 to see when the wave begins: x=-c/b. I set bx+c=2pi, to see where the wave ends x=(2pi-c)/b.

The period is the difference in these values (2pi-c)/b - (-c/b)= 2pi/b. Your teacher and I agree.

[dirtpile732]

this helped me so much. thanks!

[anjelgiofu]

Awesome. :)