About this video
This video proves that the integral over zero to two pi of sin(mt) cos(nt) is zero for any integers m and n. It uses product-to-sum identities to rewrite the product as a sum of sines, then applies the earlier result that integrals of sine over whole periods vanish. These facts underpin finding Fourier coefficients.
You'll be able to
- state why the integral of sin(mt) over zero to two pi is zero for integer m
- use product-to-sum identities to rewrite sin(mt) cos(nt)
- show that the integral of sin(mt) cos(nt) over one period is zero
Key words
- Definite integral
- The signed area under a function between two limits, here zero and two pi.
- Product-to-sum identity
- A trig identity that rewrites a product of sine and cosine as a sum of sines.
- Integer multiple
- A whole-number multiple m or n used inside the sine and cosine arguments.
- Fourier coefficient
- A number that measures how much of each sine or cosine frequency a function contains.
Transcript
- [Voiceover] We're in our quest to give ourselves a little bit of a mathematical underpinning of definite integrals of various combinations of trig functions. So, it'll be hopefully straightforward for us to actually find the coefficients, our Fourier coefficients, which we're going to do a few videos from now.
And we've already started going down this path. We've established that the definite integral from zero to two pi of sine of mt dt is equal to zero, and that the definite integral of cosine mt dt is equal to zero for any non-zero integer m. And actually, we can generalize that a little bit.
For sine of mt, it could be for any m actually. And if you don't believe me, I encourage you, so let me write this, for any, for any integer m, this top interval's going to be zero and this second interval for any for non-zero, non-zero integer, integer m. And you can see, if you had zero in this second case, it would be cosine of zero t, so this would just evaluate to one, so you'd just be integrating the value one from zero to two pi, and so that's going to have a non-zero value.
So, with those two out of the way, let's go a little bit deeper, get a little bit more foundations. So, I now want to establish that the definite integral from zero to two pi of sine of mt times cosine of nt dt, that this equals zero for any integers, integers m and n.
And they could even be the same. They don't have to necessarily be different, but they could be different. How do we do this? Well, let's just rewrite this part right over here, leveraging some trig identities. And if this is completely unfamiliar to you, I encourage you to review your trig identities on Khan Academy.
So, this is the same thing, has a definite integral from zero to two pi. This sine of mt times cosine nt, we can rewrite it using the product to sum formulas. So, that is, let me use a different color here. So, this thing right over here that I've underlined in magenta, or that I'm squaring off in magenta, that can be rewritten as one half times sine of m plus n t, sine of m plus n t, plus sine, sine of m minus n, m minus n t.
And then let me just close that with a dt, dt. Now, if we were to just rewrite this using some of our integral properties, we could rewrite it as, so this part over here, we could, and let's assume we distribute the one half, so we're gonna distribute the one half and use some of our integral properties.
And so what are we going to get? So, this part roughly, right over here, we could rewrite as one half times the definite integral from zero to two pi of sine of m plus n t dt and then this part, once you distribute the one half and you use some integration properties, this could be plus one half times the definite integral from zero to two pi of sine of m minus n, m minus n t dt.
Now, what are each of these things going to be equal to? Well, isn't this right over here, isn't that just some integer? If I have to take the sum of two arbitrary integers, that's going to be some integer. So, that's going to be some integer. And this, too, is going to be some integer right over here.
And we've already established that the definite integral of sine of some integer times t dt is zero. So, by this first thing that we already showed, this is going to be equal to zero, that's going to be equal to zero. It doesn't matter that you're multiplying by one half.
One half times zero is zero. One half times zero is zero. This whole thing is going to evaluate to zero. So, there you go; we've proven that as well.
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