Question 1194135
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The starting point for the function (t=0) is at low tide.  Since the maximum of the basic cosine function is at t=0, we can use a negative cosine function with no horizontal shift.<br>
High tide is 52 inches and low tide is 12 inches.  The difference is 40 inches, so the amplitude is 20; the midline is 32.  So the function is of the form<br>
{{{y=-20*cos(bx)+32}}}<br>
The only missing part is the coefficient b, which is (2pi) divided by the period.<br>
Half the period is between low tide and high tide, which is 6.25 hours; so the period is 12.5 hours.  The desired function is then<br>
{{{y=-20*cos((2pi/12.5)x)+32}}}<br>
A graph, showing low tide of 12 inches at t=0 (9:15am) and high tide of 52 inches at t=6.25 (3:30pm) -- and the next low tide 12.5 hours after the first one (9:45pm):<br>
{{{graph(400,400,-pi/2,9pi/2,-10,70,-20*cos((2pi/12.5)x)+32)}}}<br>
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Commenting on tutor @ikleyn's comments about the other responses, note that the response from alan3354 is in fact correct, because he used a horizontal shift of pi radians, making the positive coefficient on the cosine function correct.<br>
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To the student, who asked a completely different question in a "thank you" note regarding this problem....<br>
We generally can't answer questions that you ask in that way.  You need to post your questions in the normal way.<br>
However, there is a quick answer to the question you asked this time.<br>
The angle t goes from 0 to 2pi; but the angle in the problem is 2t, so 2t goes from 0 to 4pi.  The two answers you didn't find are each simply 2pi more than the answers you DID find, because the range is 0 to 4pi instead of 0 to 2pi.<br>