The expected low temperature T( in ^∘ F) in Fairbanks, Alaska, may be approximated by T=36 sin[(

step 1 All right. So we are looking at an equation right here. This equation represents the expected lo

The expected low temperature T( in ^∘ F) in Fairbanks,...

Key Concepts

Trigonometric Inequalities

Trigonometric inequalities involve finding the range of values for which a trigonometric function satisfies a given condition. Solving these inequalities often requires determining the critical points, typically using inverse trigonometric functions, and then identifying the intervals within one period where the inequality holds.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as inverse sine, are used to find the angle corresponding to a given sine value. This is especially important when solving equations or inequalities involving sine functions, as they allow one to determine the boundary angles where the function achieves specific values.

Sinusoidal Functions

Sinusoidal functions, such as the sine function, are used to model periodic phenomena. They have the general form a sin(b(t - c)) + d where 'a' represents the amplitude, 'b' affects the period, 'c' is the phase shift, and 'd' is the vertical shift. This type of function helps in understanding periodic behavior and fluctuations over regular intervals.

Amplitude, Period, and Phase Shift

The amplitude of a sinusoidal function indicates the maximum deviation from its midline, while the period determines the length of one complete cycle. The phase shift represents a horizontal translation of the function. Recognizing these characteristics is crucial when interpreting or predicting behavior in models, such as temperature variations that follow a seasonal pattern.

Transcript

00:03 All right.

00:04 So we are looking at an equation right here.

00:11 This equation represents the expected low temperature in fairbanks, alaska, in fahrenheit, where t is days.

00:24 And in this case, we are going to say that t equals zero, is january 1st.

00:39 Okay.

00:41 And they want to know how many days during the year.

00:46 Will the low temperature be below or less than negative four? okay.

00:55 So i'm actually going to set this up as an inequality.

00:59 There's a couple ways you could do this.

01:01 You could just graph it, but looking at this particular problem in the book, it doesn't look like that's what you're being asked to do.

01:09 Looks like you're actually being asked to solve it.

01:12 So i'm going to go ahead and walk through how to solve this.

01:19 So i'm setting my equation so that, oh my gosh, i mess that up.

01:27 One moment.

01:29 That should be the other way.

01:36 There we go, because negative four is the, all the way up to negative four, not quite negative four, right? so we want a greater than there.

01:45 And then t minus 101.

01:49 Okay, plus 14.

01:50 So i'm going to go ahead and start solving this.

01:54 Subtract 14 from both sides to give me negative 18.

02:01 And then i'll go ahead and divide both sides.

02:12 365, not 36, 21.

02:18 Divide both sides by 36.

02:20 So this is going to end up being negative one half greater than the sign of 2 pi over 365 times t minus 101.

02:36 Okay, so now i'm going to take the sign inverse of both sides.

02:40 I'll change colors here so you can kind of see.

02:44 So sign inverse of both sides.

02:51 And of all that mess and all this mess.

02:56 Okay.

02:56 So when i do that, the sign inverse of negative one half in radiance is approximately negative 0 .5236.

03:16 Now, the thing about this, though, is that i'm looking at my unit circle here.

03:26 Okay, here's this little angle, right? so negative 0 .5, 2, 3, 6.

03:38 Okay? and actually, sign is also negative in the third quadrant.

03:45 Okay.

03:46 So i need to figure out what this angle here.

03:48 Here is in order to determine my second equation, because there's actually going to be two equations here that we have to solve in order to figure out the solution to this.

04:08 So i'm actually going to find the positive version of this.

04:14 If i were to turn the negative 0 .5 -236 into a positive, i'm actually going to find.

04:21 Going to get a number that's outside of the of the year because remember we're going from zero to 364 since zero is january 1st.

04:33 So to find this, what i do is if you remember this reference angle right here has to be the same as the angle in the fourth quadrant.

04:46 So i'm just going to add pie to the positive version.

04:54 Of 0 .526.

04:58 And that is going to give me 3 .6652.

05:08 So that'll be the second equation that we solve.

05:13 So i will multiply both sides by 365 divided by 2 pi...