How Do You Find the Equation for Sine and Cosine from a Graph?

[Ronka]

Find the equation for Sine and cosine from a graph

Homework Statement

The Attempt at a Solution

Hello people, i have had a little problem solving this graph. It looks really easy, but I am still couldn't solve it ?!

a x Sin (bx + c) + d

a = amplitude
b = period
c = phase shift
d = vertical shift

thx in advances.

Edit:

I have solved the equation for the Sine ( thank you folks )
Equation for Sine:
2 sin(1(x-2))

Trying to find the equation for Cosine:

 

[micaele]

edited out my answer because i wasn't aware of this rule. way to botch the first post in the forum, eh?

well, to be more helpful without giving an answer:

when you're looking at a graph, it really is rather simple... that purple one sure looks pretty basic, don't you think? pay attention to the places where the line/curve touches the x-axis and where the inflection points are so that you can better figure the stuff out.

 

[catkin]

As you say, it's two graphs of the form Y = a sin(bX + c) + d

So the answer is giving two sets of values for a, b, c and d.

Can you spot any of them? The a and d values are the easiest ones to start with.

 

[catkin]

As it says in the FAQ (https://www.physicsforums.com/showthread.php?t=94379) "Please DO NOT do someone's homework for them or post complete solutions to problems. Please give all the help you can, but DO NOT simply do the problem yourself and post the solution (at least not until the original poster has tried his/her very best)."

 

[Ronka]

a = 2
d = 0

b = period remain the same = 2 phi
c = ...

since the graph is shifted to the right, means that it should be -c not +c
c = -2 ?!?

 

[catkin]

Is that the green or the pink you're working on, Ronka?

 

[Ronka]

Im trying to find the equation of the green one. the pink one is just Sin(x).

Green ( one period or 2 phi ) :

2 sin(1(x - 2) ?! is this right?

 

[catkin]

Yes to a = 2 and d = 0.

It's pi not phi.

If pink is sin(X) then X is in radians.

Check:
sin(0) = 0
sin(π/2) = 1
sin(π) = 0
OK.

2 sin(1(x - 2) can't be right because the parentheses don't balance.
If you meant 2 sin(1(x - 2)) then it's simpler to write 2 sin(x - 2)
sin(x - 2) is good; it shifts the positive-going zero-crossing of sin(x) to the right.
Yes -- 2 sin(x - 2) looks right.

 

[Ronka]

Thanks:

Equation for Sine:
2 sin(1(x-2))

Now i finally understand how to find the equation for Sine
Im now trying to find the equation for Cosine

I'll give it a try first. i'll post when I'm stuck :)

thx!

 

[catkin]

Now I don't understand!

Why are you choosing "2 sin(1(x-2))" in preference to the simpler "2 sin(x-2)"?

Are you wanting to show that b is 1 in "Y = a Sin (bx + c) + d"?

 

[Ronka]

Just to make myself clear that that the period has been tretched ( i know you could clearly see from the graph ) but it just help me rember :)

 

[catkin]

The period hasn't been stretched. If you count the X distance between corresponding points on the green graph (like the maxima, the minima or the positive- and negative-going zero crossings) they are the same as the pink graph.

a = amplitude is stretched by times 2
b = period is unchanged
c = phase shift is -2
d = vertical shift is zero