SOLUTION: The graph represents a sine function in the form y = A sin B(t + C) + D or a cosine function in the form y = A cos B(t + C) + D. Write both equations for each. Answer key: y =

Algebra ->  Trigonometry-basics -> SOLUTION: The graph represents a sine function in the form y = A sin B(t + C) + D or a cosine function in the form y = A cos B(t + C) + D. Write both equations for each. Answer key: y =       Log On


   

Question 1205033: The graph represents a sine function in the form
y = A sin B(t + C) + D or a cosine function in the form y = A cos B(t + C) + D. Write both equations for each.
Answer key: y = 4 sin πt, y = 4 cos π(t - 1/2)


By the way, is there an edit button or similar function to make changes to a post after I've already posted it in case I discover later on I've made a typo?
Found 2 solutions by MathLover1, math_tutor2020:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

The graph represents a sine function in the form y+=+A+sin+B%28t+%2B+C%29+%2B+D
or a cosine function in the form y+=+A+cos+B%28t+%2B+C%29+%2B+D.
Write both equations for each.

from the graph we see that amplitude is 4
we also see that the mid line is x-axis, so the graph is not shifted vertically and D=0
there is also no horizontal shift, so+C=0
since the graph passes through the origin, the graph of a sine function always intersect the mid line (which is x-axis in your case) at x=0, we have sine function
so, we will use
y+=+A+sin+B%28t+%2B+C%29+%2B+D
since A=4 , C=0, and D=0, we have
y+=+4sin%28+B%28t+%29%29

one full cycle is 2 units, so+B=2
period is 2pi%2FB=2pi%2F2=pi

and your function is:
y+=+4sin%28+pi%2At%29

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Students aren't able to edit, but tutors can. So please be sure to review your questions carefully. If you do notice a mistake after the fact, then you can mention it in the thank you note to the tutor. A possibly better alternative is to make a new post.

The highest and lowest points occur when y = 4 and y = -4 respectively.
Use the midpoint formula to find that y = 0 is the midpoint.
This leads us to D = 0 which represents the midline.
This applies to both sine and cosine templates.

The height of this curve is 8 units.
Half of which is 4, which is the value of A.
A = 4
This applies to both sine and cosine templates.
Technically we could go for A = -4, but I'll stick with the positive version since that's what your teacher picked.

The neighboring peak points occur when x = 0.5 and x = 2.5
The gap is 2.5 - 0.5 = 2 units which is the period.
The curve repeats itself every 2 units along the x axis.

T = period = 2
T = 2pi/B
B = 2pi/T
B = 2pi/2
B = pi
This applies to both sine and cosine templates.

So far we found that:
A = 4
B = pi
D = 0
and they work for both sine and cosine templates mentioned.
Side note: Cosine is a phase shifted version of sine.

Let's plug those values in. We'll also plug in one of the points on the curve.
I'll plug in (0.5, 4)
This will allow us to solve for C.

So,
y = A*sin( B(x+C) ) + D
4 = 4*sin( pi(0.5+C) ) + 0
4 = 4*sin( pi(0.5+C) )
sin( pi(0.5+C) ) = 4/4
sin( pi(0.5+C) ) = 1
pi(0.5+C) = arcsin(1)
pi(0.5+C) = pi/2
0.5+C = 1/2
1/2+C = 1/2
C = 1/2 - 1/2
C = 0

Summary for the sine template
A = 4
B = pi
C = 0
D = 0
This will mean we go from y = A*sin(B(t+C)) + D to y = 4*sin(pi*t)

Cosine will look almost identical in terms of A,B,C,D values.
A,B,D will be the same.
C is going to be different.

Let's plug the known A,B,D values into the cosine template.
Also let's plug in (0.5, 4) so we can solve for C.
y = A*cos(B(x+C)) + D
4 = 4*cos(pi(0.5+C)) + 0
4 = 4*cos(pi(0.5+C))
cos(pi(0.5+C)) = 1
pi(0.5+C) = arccos(1)
pi(0.5+C) = 0
0.5+C = 0
C = -0.5
C = -1/2

For cosine we have
A = 4
B = pi
C = -1/2
D = 0
which explains how we arrive at y = 4*cos(pi(t - 1/2))

Confirmation using Desmos graph
https://www.desmos.com/calculator/uatjjocx1m
I'm using x in place of t much of the time, but you get the idea.