Transforming Trigonometric Equation

[Musa Ali]

Homework Statement

In 2001, Windsor, Ontario received its maximum amount of sunlight,
15.28 hrs, on June 21, and its least amount of sunlight, 9.08 hrs, on
December 21

1. Due to the Earth's revolution about the sun, the hours of daylight function is periodic. Determine an equation that can model the hours of daylight function for Windsor, Ontario.

Homework Equations

y=a sin[b(x-c)] + d

The Attempt at a Solution

a=(max-min)/2=(15.28-9.08)/2=3.1

b=2π/period=2π/365=0.02

d=(max+min)/2=(15.28+9.08)/2=12.18

To find , we must substitute y and x for 15.28 and 172, which is June 21, respectively.

y=a sin⁡〖[b(x-c)]〗+d

15.28=3.1 sin⁡〖[0.02(172-c)]〗+12.18

15.28-12.18=3.1 sin⁡〖[170.28-0.99c]〗

3.1=3.1 sin⁡〖[170.28-0.99c]〗

1=sin⁡〖[3.44-0.02c]〗

sin^(-1)⁡〖(1)〗=3.44-0.02c

90=3.44-0.02c

90-3.44=-0.02c

86.56=-0.02c

4328=c

 

[SammyS]

Homework Statement

In 2001, Windsor, Ontario received its maximum amount of sunlight,
15.28 hrs, on June 21, and its least amount of sunlight, 9.08 hrs, on
December 21

1. Due to the Earth's revolution about the sun, the hours of daylight function is periodic. Determine an equation that can model the hours of daylight function for Windsor, Ontario.

Homework Equations

y=a sin[b(x-c)] + d

The Attempt at a Solution

a=(max-min)/2=(15.28-9.08)/2=3.1

b=2π/period=2π/365=0.02

d=(max+min)/2=(15.28+9.08)/2=12.18

To find , we must substitute y and x for 15.28 and 172, which is June 21, respectively.

y=a sin⁡〖[b(x-c)]〗+d

15.28=3.1 sin⁡〖[0.02(172-c)]〗+12.18

15.28-12.18=3.1 sin⁡〖[170.28-0.99c]〗

3.1=3.1 sin⁡〖[170.28-0.99c]〗

1=sin⁡〖[3.44-0.02c]〗

sin^(-1)⁡〖(1)〗=3.44-0.02c

90=3.44-0.02c

90-3.44=-0.02c

86.56=-0.02c

4328=c

Do you have a question?

That's a huge round-off error for b.

Plug in 355 & see what the answer is for that day.

 

[Musa Ali]

Do you have a question?

That's a huge round-off error for b.

Plug in 355 & see what the answer is for that day.

I am aware of the fact that the value I get for c is horribly skewed. I would like to know where exactly I have gone wrong.

 

[SammyS]

I am aware of the fact that the value I get for c is horribly skewed. I would like to know where exactly I have gone wrong.

The value of 0.2 for b is at least 16% larger than the correct value of 2π/365 . I think that's a significant error.

How do you know that you've gone wrong?

 

[SammyS]

I am aware of the fact that the value I get for c is horribly skewed. I would like to know where exactly I have gone wrong.

Do you want to do this in degrees or in radians ?

This appears to be the biggest issue. Using radians for b, then using degrees when evaluating the inverse sine will cause a BIG problem.