While writing the script for the previous video, after showing this limit, I wanted
to say it is impossible to be calculated using algebraic methods, but then I figured that
it is indeed possible using an alternative argument, so I changed the script. The trigonometric
identity, which is the formula for the sum of sines, will be the focus of this video,
and we can use algebra, or geometry to derive the identity, and you can do the judgment
which method is superior.
For the algebraic method, the easiest method is to multiply the entire expression by sin
theta over 2. Then distribute sin theta over 2 to all the terms. Now, we can use the product
to sum formula in trigonometry. This is quite simplified, but we want to further manipulate
it for reasons I will explain later. This is done by considering sum to product formula,
which is basically the reverse of the previous formula. We apply the formula, and then realize
that we can pull out the negative sign to the front, so these two negative signs cancel
out each other. And finally, realize that we have multiplied by sine theta over 2 to
make all these possible, and so we have this formula for the sum of sines. The main reason
not to stop after the telescoping series is that the sum of sines formula bears a very
strong resemblance with this famous formula derived by Gauss, which makes memorizing the
formula easier. That is the algebraic derivation.
The geometric proof for this is a little bit tricky, because there needs to be some slight
adjustments to the argument depending on the range of theta. But for this video, we will
ultimately want a result for the limit representation of the integral, and the angles here are all
acute, at most a right angle, so we only focus on this scenario.
To start us off, remember that in the previous video, we established that these vertical
segments are of length sin k pi over 2n, where k ranges from 1 to n. However, from this point
on, we will use theta to denote pi over 2n to save space, and so these lengths are sin
k theta. What we will do is to move these triangles elsewhere. All we do is just translation
and nothing else. If we put a coordinate system, what we want to
know is the coordinates of this point, specifically the y-coordinate. The reason why we consider
the y-coordinate is that it is simply the sum of sines we are interested in. To get
the formula for it, we should come up with another alternative to calculate the y-coordinate
of the white point. This alternative way is to consider d and phi in the diagram, then
the sum of sines would be equal to d sin phi. We need to figure out what d and phi are.
We would be a little intuitive here because writing down all the rigorous steps might
be daunting and distracting from the main line of reasoning. Firstly, we can see that
the coloured lines form a part of the circle. Then imagine you are on a train from the origin,
initially facing the direction that the red line says, then travel along the coloured
tracks towards the white point. During the process, we originally go from a direction
that measures an angle of theta to the horizontal, ultimately to a direction that is an angle
of n theta to the horizontal. In a sense we can say that the average direction of the
train is n + 1 theta over 2, which should be exactly phi, because this white line should
be the average direction that the train is going.
Now let’s say instead of focusing on the direction that you are going, focus on the
direction of the centre of the circle. This direction would change by n theta throughout
the journey because the direction of the train also changes by n theta. To calculate d using
this fact, we first cut the isosceles triangle along its symmetry, we will then get this
yellow equation. Using a similar argument, replacing the length d with a coloured edge,
we have a similar equation. By simply dividing the first equation by the second one, we finally
obtain d in terms of n and theta. So now that we have d and phi, we can finally substitute
it back to this alternative way of obtaining the y-coordinate of the white point, and we
get this formula that we have seen before using algebraic derivation.
Now that you have seen both derivations of the same formula, you can judge by yourself
which of them is superior. However I would provide some objective points for you to consider.
For the algebraic method, of course you need to know the sum-to-product and product-to-sum
formulae, and that multiplying by sin theta over 2 is a good idea, especially if you don’t
already know the formula. But of course, it is a rather quick process, and just in my
opinion, the cancelling of the terms is very satisfying. For the geometric derivation,
of course you don’t need any advanced knowledge, just simple geometry. But coming up with the
consideration of the y-coordinate of the white dot is not easy, and it is a rather long process
even if you know what you’re doing. So I will leave it up to you to decide which method
wins.
In any case, now that we have derived this formula, we can put it into good use. Remember
that this is still a video in the series Doing Integrals without Integration, so we are still
calculating this integral. Now we can substitute theta to be pi over 2n, and so we finally
have this expression as the limit. Then we can split up the limits, so that the two limits
exist. To tackle the first limit, we use the substitution u equals pi over 4n, which makes
u tend to 0 since n tends to infinity, and so we have this limit, which equals 2. The
second limit is much easier. We just need to see that n+1 over n tends to 1 when n tends
to infinity, and so we have its limit to be sin squared pi over 4, and since sin pi over
4 is 1 over the square root of 2, the integral equals 1.
Whether you like the algebraic or the geometric derivation, or the sum of projections in the
previous video, leave a comment down below. Don’t forget to like and subscribe to Mathemaniac
if you enjoyed this video!