Question

Trig problem - Could someone show me the steps for solving for t?

720 = 11.895+2.545sin [(2pi/366)(t-80.5)]

Answer 1

first, isolate the sin by subtracting 11.895 on both sides and dividing by 2.545.
Then you need to find out wich angle (in radians first) gives that sin.

Answer 2

i've actually done that... i've gotten as far as 278.233 = sin [2pi/366(t-80.5)], but I don't understand what to do after that....

Answer 3

Oh, wait. That is impossible. Sin and Cos cant be greater than 1. Is your equation correct?

Answer 4

i think so.... my problem is supposed to be modelling daylight hours. i'm trying to solve for t if the hours of daylight is 720

Answer 5

Write the original problem then please.

Answer 6

a) Find a formula for the daylight function using constants determined from the data;
b) In the context of daylight duration, discuss solving the inequality y(t) ≥ 720 for t

the maximum amount of daylight is 14.44 and the minimum is 9.35

this is my formula:
y=11.895 + 2.545 sin [2pi/366(t-80.5)]

Answer 7

amplitude:
A=1/2 (14.44-9.35)
=2.545

wavelength
B= 2pi/366

center line
1/2 (14.44+9.35)
=11.895

C= 80.5

Answer 8

What is C, and what units are you using on B?

Answer 9

C is the phaseshift .... I found a similar solution/problem here:
In Philadelphia the number of hours of daylight on day t (where t is the number of days after Jan. 1) is modeled by the function
L(t)= 12+2.83sin(2pi/365(t-365))

A) Which days of the year have about 10 hours of daylight?
B) Which days of the year have more than 10 hours of daylight?
math - drwls, Saturday, August 13, 2011 at 10:13pm
(A) Solve
10 = 12 + 2.83 sin[(2*pi/365*(t-365)]

-2 = 2.83 sin[(2*pi*/365)(t-365)]
-0.7067 = sin[(2*pi*/365)(t-365)]
sin[(2*pi*/365)(t-365)] = -0.7848
Solve for t.

[(2*pi*/365)(t-365)] = -0.90244
Use the first value

t-365 = -52
t = 313 days
Oct 1 is day 303. So Oct 11 is one answer. The other day will be 52 days after the winter solstice, or about Feb 11.

Answer 10

Ok, I understand your formula, except for the phase shift. That must be related to the day when there is more or less sunlight in the year, is that how you found it?
Now the problem here if you found the correct phase shift is with the interpretation of the problem. What does that y(t)>720 means? It cannot be your function, thats for sure. Do they give you anymore information?

Answer 11

i assumed the question was asking if the amount of daylight is about 720. the data from where i got the max/min was in hours and minutes, by the way

Answer 12

Ok, that is probably the origin of the problem. You cant have 720 hours of daylight, so maybe thats minutes. Actually, minutes are consistent with this number. Try using minutes and remember to change your values acordingly

Answer 13

Ah, okay! Thanks!

So, to make it into minutes.... \[y=60(11.895)+60(2.545) \sin [(2\pi/366)(t-80.5)]\]

then let y =720?

Answer 14

I assume I can follow the steps that this problem has? although, i don't understand how to do step 3 in order to take the arcsin of both sides (inverse sine)

10 = 12 + 2.83 sin[(2*pi/365*(t-365)]

Step 1: -2 = 2.83 sin[(2*pi*/365)(t-365)]
Step 2: -0.7067 = sin[(2*pi*/365)(t-365)]
Step 3: sin[(2*pi*/365)(t-365)] = -0.7848
Solve for t.

[(2*pi*/365)(t-365)] = -0.90244
Use the first value

t-365 = -52
t = 313 days

Answer 15

I only took a quik look, but that seems ok. The arcsin you can only calculate using a calculator I guess. The phase shift I didn't tried to see if its correct, but -365 will change nothing in the function.

Answer 16

Ah, well thanks for your help anyhow!