Video: Using Inverse Functions to Solve Trigonometric Equations Modeling Real-Life Situations

The temperature fluctuation on a cold winters day (in degrees Celsius) is modeled by 𝑇 = 3 cos ((πœ‹/12)(𝑑 βˆ’ 14)) + 2, where 𝑑 is the time of day expressed in hours after Midnight. At what times of the day was the temperature 0℃?

06:20

Video Transcript

The temperature fluctuation on a cold winter’s day, in degrees Celsius, is modeled by 𝑇 is equal to three cos of πœ‹ by 12 times 𝑑 minus 14 plus two, where 𝑑 is the time of day expressed in hours after midnight. At what times of the day was the temperature zero degrees Celsius?

What this question is asking us to do is to find little 𝑑 when big 𝑇 is equal to zero. First, let’s note the range of values that little 𝑑 can take. Since 𝑑 is the time after midnight in hours and there are 24 hours in a day, this means that little 𝑑 can take values between nought and 24. And so we get nought is less than or equal to 𝑑, which is less than 24.

Now when we look at the function for the temperature, we see that we have cos of πœ‹ by 12 times 𝑑 minus 14. And so to make calculating these values easier, we need to find the range of values for the part inside the cos function. So that’s πœ‹ by 12 times 𝑑 minus 14. In order to find this range, we just need to adapt the range of values of little 𝑑 that we’ve already found.

So let’s start by subtracting 14 from each part of the inequality. And this gives us that minus 14 is less than or equal to 𝑑 minus 14, which is less than 10. Now we can multiply each part of the inequality by πœ‹ over 12. And this leaves us with this. And we can simplify this to minus seven πœ‹ by six is less than or equal to πœ‹ by 12 times 𝑑 minus 14 is less than five πœ‹ by six. And so now we have found our range of values for πœ‹ by 12 lots of 𝑑 minus 14.

Now let’s set big 𝑇 equal to zero. This gives us that zero is equal to three cos of πœ‹ by 12 times 𝑑 minus 14 plus two. And we can rearrange this to make the cos part the subject, which gives us that cos of πœ‹ by 12 times by 𝑑 minus 14 is equal to minus two-thirds. In order to make this look a little nicer, we can substitute in πœƒ for πœ‹ by 12 times 𝑑 minus 14. So now all we are solving is cos of πœƒ is equal to minus two-thirds. And we can use the range that we found earlier to help us solve this.

And this range is for πœ‹ by 12 times 𝑑 minus 14, just the same as πœƒ. So we are finding values of πœƒ such that minus seven πœ‹ by six is less than or equal to πœƒ which is less than five πœ‹ by six. Now we are ready to solve cos of πœƒ is equal to minus two- thirds. Let’s draw a cos graph to help us work this out. Here, we have our graph for 𝑦 equals cos πœƒ. Now let’s mark on the range for πœƒ. We have that minus seven πœ‹ by six is less than or equal to πœƒ.

And so we can draw a solid line in at minus seven πœ‹ by six. Then we also have that πœƒ is less than five πœ‹ by six. So we draw a dashed line in at five πœ‹ by six. Next, let’s add the line 𝑦 equals minus two-thirds. And so the solutions to our equation where the two lines 𝑦 equals minus two-thirds and 𝑦 equals cos πœƒ intersect. As we can see, we have two solutions.

Typing cos inverse of minus two over three into our calculator gives us a value of 2.300523983. And this value lies between πœ‹ by two and πœ‹. So therefore, this is the solution on the right of our graph. In order to find the other solution, we can use the fact that the graph is symmetric about the 𝑦-axis. So that means that the second solution will simply be the negative of the first solution. This gives us negative 2.300523983.

We could round here. However, this could cause us to lose accuracy in our answer. So let’s keep these both as cos inverse of minus two-thirds for now. We have that πœƒ is equal to cos inverse of minus two-thirds. Now we can substitute back in πœƒ which we had before. So we have that πœ‹ by 12 times by 𝑑 minus 14 is equal to cos inverse of minus two-thirds. Let’s rearrange this for 𝑑.

We can start by multiplying both sides by 12 over πœ‹. And now, we simply add 14 to both sides. And we get that 𝑑 is equal to 14 plus 12 over πœ‹ cos inverse of minus two-thirds. Now all we need to do is substitute in the two values for cos inverse of minus two-thirds that we found earlier. Substituting in the positive solution gives us 𝑑 is equal to 22.78735433 hours. And substituting in the negative solution gives us that 𝑑 is equal to 5.212645674 hours.

Now since we cannot have a decimal point of an hour. We need to calculate these decimal points in minutes. In order to do this, we can take the decimal point and multiply it by 60. So we get an answer of 47 minutes. It is okay to round here since we don’t need to go any more accurate than minutes. For the second time that we calculated, we now need to multiply the part after the decimal point by 60. And this gives us a time of 13 minutes. So therefore, it was zero degrees at both five hours and 13 minutes after midnight and 22 hours and 47 minutes after midnight.

Writing these as times of the day we get a solution of 5:13 AM and 10:47 PM.

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