Question Video: Finding the Frequency of a Trigonometric Function

Video Transcript

What is the frequency of the function 𝑓 of 𝑡 equals 𝑎 cos 𝑏𝑡 minus 𝑐 plus 𝐾?

We have a function of 𝑡. We can see 𝑡 here. What does this function 𝑓 do to the input 𝑡? Well, it multiplies it by 𝑏 to get 𝑏𝑡. And then, to this product we subtract 𝑐. We find the cosine of this difference. We multiply that by 𝑎. And then, finally, we add 𝐾. 𝑎, 𝑏, 𝑐, and 𝐾 are just some numbers. And to see what role they play in this function, it’s best to just try graphing this function for different values of 𝑎, 𝑏, 𝑐, and 𝐾.

You should find that whatever values of 𝑎, 𝑏, 𝑐, and 𝐾 you pick, the graph looks a bit like a sine curve. We say it is sinusoidal. But if you increase the value of 𝑎 while keeping 𝑏, 𝑐, and 𝐾 the same, you should find that the wave gets bigger. The oscillations are more pronounced. This value 𝑎 is called the amplitude of the wave. This amplitude is the height of the wave when measured from the average value of the wave. It’s half the height of the wave if you measure the height from the highest point of the wave to the lowest point. So, that’s 𝑎. What about to 𝑏?

Well, let’s see what happens if we increase 𝑏 while keeping 𝑎, 𝑐, and 𝐾 the same. As we increase 𝑏, we see that the distance between successive peaks, that’s the highest points on the graph, decreases. The peaks get closer together as we increase 𝑏. And the same is true of the troughs, or lowest points on the wave. This function is periodic, and we see that its period is decreasing.

Remember we have time 𝑡 on the 𝑥-axis. And so, the peaks looking closer together on the graph actually tells us that the peaks are becoming more frequent. So, we can say that as 𝑏 increases, frequency increases. Notice that, unlike before when I said 𝑎 is the amplitude, I haven’t said that 𝑏 is the frequency. Just that as 𝑏 increases, frequency increases. This is because the frequency is not simply 𝑏. We’ll see why later.

Moving on, we can see that increasing 𝑐, the graph is translated to the right. 𝑐 is related to something called the phase of the wave. And increasing 𝐾, translates the graph up. 𝐾 is the average value of the function. The equation of the midline of the graph is 𝑦 equals 𝐾. Let’s get back to our question, which is what is the frequency of the function 𝑓 of 𝑡 equals 𝑎 cos 𝑏𝑡 minus 𝑐 plus 𝐾. Well, we’ve seen that only the value of 𝑏 affects the frequency, but I’ve said that the frequency is not 𝑏.

Consider a slightly different function, the function 𝑔 defined by 𝑔 of 𝑡 equals 𝐴 times cos two 𝜋𝑓𝑡 plus 𝜑 plus 𝐾. And just for a moment, compare this to the function 𝑓. We can see that the lowercase 𝑎 in the function 𝑓 has been replaced by the uppercase, or capital, 𝐴 in the function 𝑔. 𝑏 in the function 𝑓 has been replaced by two 𝜋𝑓 in the function 𝑔. The minus 𝑐 has been replaced by a plus 𝜑. We can also think of this as replacing 𝑐 by negative 𝜑. And 𝐾 has kept its role.

Now I can tell you the amplitude, frequency, phase, and average value of this function 𝑔. Capital 𝐴 is its amplitude. And this isn’t surprising, as this capital 𝐴 plays the same role in 𝑔 that the lower case 𝑎 plays in 𝑓. And we said that the lowercase 𝑎 was the amplitude of our function 𝑓. 𝑓 is the frequency of the function 𝑔.

And note here that 𝑓 is not simply equal to 𝑏. 𝑏 equals two 𝜋𝑓. We were right to say that the frequency is not 𝑏, although it is related to 𝑏. If we divide by two 𝜋 on both sides, we find that the frequency is in fact 𝑏 over two 𝜋. This is the frequency of the function 𝑓 of 𝑡 equals 𝐴 cos 𝑏𝑡 minus 𝑐 plus 𝐾 that we’re looking for. So, we found the answer to our question. It’s 𝑏 divided by two 𝜋.

But for completeness, I should say that 𝜑 is the phase of the function 𝑔. And so, negative 𝑐 is the phase of 𝑓. And 𝐾 is, of course, the average of both functions 𝑔 and 𝑓. Now, of course, you could ask why two 𝜋 is involved in the definition of the function 𝑔. Well, let capital 𝑇 be the period of the function 𝑔, that’s the time it takes before the function repeats. So, 𝑔 of 𝑡 plus capital 𝑇 is 𝑔 of 𝑡.

We use the definition of the function 𝑔. And we can subtract 𝐾 from both sides and then divide through on both sides by 𝐴. And so, we see that cos of two 𝜋𝑓 times 𝑡 plus capital 𝑇 plus 𝜑 equals cos of two 𝜋𝑓𝑡 plus 𝜑.

Now, one way for this equation to hold is that the inputs to cos are both the same. But we could also add or subtract any multiple of two 𝜋 and their cosines would be the same. And by the definition of the period, exactly one oscillation has occurred. And so, we just have one two 𝜋 to add. After one period of the function, we’ve moved by two 𝜋 radians to repeat again. Now it’s just some more algebra.

We cancel the 𝜑s, distribute two 𝜋𝑓 over the terms in parentheses on left-hand side, which allows us to cancel further. We can also divide through by two 𝜋 now to get 𝑓𝑡 on the left-hand side and just one on the right-hand side. And if we divide through by the period capital 𝑇 of our function 𝑔, we find that 𝑓 is one over this period capital 𝑇.

And so, by definition, 𝑓 is the frequency of our function 𝑔. This has justified our claim that 𝑓 is indeed the frequency of the function 𝑔. And we could perform basically the same procedures to show that the frequency of the original function 𝑓 is 𝑏 over two 𝜋.