Question 1183251
<br>
I will strongly disagree with the comments from tutor @ikleyn about the relative merits of her response and the response from the other tutor.<br>
The other tutor's response, although it is not a particularly good straightforward presentation, contains a lot of information for a student who is trying to learn how to work problems like this.<br>
On the other hand, tutor @ikelyn's statement early in her response that the equation is "obviously" _____ is of absolutely no use to a student who is trying to learn how to work the problem.  The equation might be obvious to her; but I very much doubt it is obvious to a beginning student.<br>
Here is what I would do with this problem.<br>
The statement of the problem implies that the period of the sinusoidal function is 24 hours.  With t=0 at midnight and the minimum temperature at 6am, the minimum temperature occurs exactly one-quarter of the way through the cycle.  That means at t=0 the temperature is at the midline and is decreasing.<br>
A basic sine function at t=0 is at the midline and increasing; a basic cosine function at t=0 is at its maximum. So a sinusoidal function that is at the midline and decreasing at t=0 is most easily represented by a negative sine function with no phase shift:<br>
{{{T(t) = -a*sin(b(t-0))+d}}}<br>
a is the amplitude; b is 2pi divided by the period; the 0 represents no phase shift; and d is the midline.<br>
The minimum and maximum temperatures are 80 and 90; that makes the midline 85 and the amplitude 5.  And the period is 24 hours.  So the function is<br>
{{{T(t) = -5sin((t/24)2pi)+85}}}<br>
A graph, showing the temperature over a little more than one period....<br>
{{{graph(400,400,-2,26,75,95,80,90,-5*sin((x/24)2pi)+85)}}}<br>