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I have been asked by a member to comment on this worked example question encountered in an online tigonometry course.

The temperature during the week oscillates daily between 48o and 74o. (presumabably Farenheit).
If the minimum occurs at 5am, at what time is the temperature at 65oF ?

The lecturer chooses a sinusoidal model for the variation, but uses a very overcomplicated method where he arrives at and solves the following equation.

$f\left( \theta \right) = - 13\cos \left( {\frac{\pi }{2}\left( {\theta - 5} \right)} \right) + 61$

The lecturer has chosen to use a cosine wave as the model, using radian measure for the angles, and has come up with a 'grand unified' formula which is fraught with the dnager of miscalculations as will become clear.

It is vey important to solve and understand the Physics as well as the Mathematics of a model.

I have shown the Physics in Fig1, covering both the day in question and part of the day before and part of the day after.
The lecturer failed to note the characteristics of a sinusoidal wave - that it passes throuhg its speial points (its peaks and zeros) at spacings of 90o or 6 hours or 'quadrants'.
This is possibly because he thought to work in radians.
It is worth noting that most scientific instruments that measure angle use the degrees, minutes and seconds units. Almost none read directly in radians.
Maybe this is why he made the incorrect statement that we cannot locate the peaks, troughs and zeros of the model wave, only the minimum at 5am.
I have labelled the important ones A, B and C in green in Fig1.
I have also shown where the target 65oF intersects the model curve at P and Q within the target day and at R on the day before.

It is also worth noting that working in DMS (degrees etc) has advantages in some disciplines since time, latitude and longitude in geography and azimuth and right ascension in astronomy are measured in these.
So 15o corresponds to 1 hour.

The lecturer also points out that there are actually many angles that satisfy the inverse trigonometric equation, but goes on to make very heavy weather of choosing the right one.
He doesn't stress enough that a calculator or set of tables will only provide one of these angles, not necessarily the one you want.

So I have shown in Fig2 and Fig3 the standard way of overcoming this issue.
If positive values are taken the calculator or table will always give you the first quadrant anngle (I have called A to distinguish from his theta).
A mnemonic based on the english word CAST is shown to help remember which trig functions are positive in the other quadrants as well.
Fig3 shows how to establish the angles to be used in each quadrant.

So here are 3 separate ways of working this out, without solving that overblown equation.
We will see how my comments pan out as we work through them.

So starting from point A which is the normal starting point of a cosine wave, where the cosine has a value of +1.
This must be two quadrants of 6 hours ie 2x6 =12 hours back from the minimum at 5am (point B). That is 5pm the previous day.

Counting from point A

Sorry I see I've used A for both the angle and the point, hopefully this will not cause too much confusion.

P is in the 4th quadrant,
Q is in the 5th quadrant, but one day later than R which is in the first quadrant and therefore at the same time of day as Q.

for point R, cos A = 4/13  therefore A = 72o or 72/15 = 4.8 hours later than point A

So R is the point at 1700 + 4.8 hours or 21.8 hours, the previous day.
So Q is the point at 21.8 hours on the target day in question.

P is in the 4th quadrant so is (360-A) = (360- 72) = 288o .
288 degrees is 288/15 = 19.2 hours
So P is the point at 1700 + 19.2 hours or 1700 +6+6+6+1.2 hours = 12.2 hours the following day.

Counting from point B

In relation to point B, P is in the second quadrant and Q is in the third quadrant.
Thus both P and Q will have negative cosines.

Using a calculator to calculate the inverse cosine of -(4/13) will correctly give the correct angle of 108o.
However there is a trap here as this will not give the value for Q directly.

Better to ignore the -ve and follow the standard procedure of Fig3.

For P  Angle  = $180 - {\cos ^{ - 1}}\left( {\frac{4}{{13}}} \right)$  ie (180 - 72) ie 108o or 108/15 = 7.2 hours

For Q Angle =   $180 + {\cos ^{ - 1}}\left( {\frac{4}{{13}}} \right)$ ie (180 + 72) = 252o or 252/15 =  16.8 hours

So P is the point at 0500 + 7.8 = 12.2 hours

and Q is the point at 0500 + 16.8 = 21.8 hours.

Counting form point C

Since C is a zero it would be the natural point to count from if modelling with a sine wave, rather than a cosine wave.

P is then in the first quadrant and Q in the second so both are then positive.

So Angle is  ${\sin ^{ - 1}}\left( {\frac{4}{{13}}} \right) = {18^o}$  ie 18/15 = 1.2 hours
So P is the point 1100 + 1.2 = 12.2 hours

For Q we have Angle  = (180 - 18) = 162o ie 162/15 = 10.8 hours
So Q is the point 1100 + 10.8 = 21.8 hours.

So all three methods come up with the same answer (and the same one the lecturer obtained)

But it shows the value of solving the Physics as well as the fancy trigonometry.

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So all three ways are equal? You are right all 3 should be explained so you know what the equation is doing. Did the instructor plug and chug or was it derived?

Obviously the equation is easier. I think the instructor’s goal was to find the frequency and create an equation that could be reused.

My question is if you graph the sine wave and take the angle between those points, how are you sure the drawn sine wave is the same as the equation wave? I mean did you form a curve that is equal value for value of the equation and do all points align? Would you have to make new “measurements” for each point to “shape” the modified sine wave to the actual modified sine wave?

I said the equation was easier. I meant easier once you know what is what. If I was in that class I would be confused too. But that is why I am asking questions. I never took physics 102 and I know how important this is too electricity and radio waves.

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