Products and Sums of Series

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Cos pi over seven, times cos two pi over seven, times cos four pi over seven. None of these angles is in our table of standard values, right? Still, by the end of this video, you will find this product exactly, with no calculator. Products like this come up often in JEE questions. So, let's begin.

0:24Angles that double: the product formula

So, here is our product: cos A, times cos two A, times cos four A, and so on, where the three dots mean the pattern continues. The last factor is cos of, two raised to the power, n minus one, all times A, so the powers of two run from zero up to n minus one, there are n factors, and each angle is double the one before. Okay? We already know a double angle formula. So tell me, what is two sine A cos A? It is sine two A, yes, exactly! So multiply the whole product by two sine A, and two sine A times cos A collapses into sine two A, right? Then two sine two A cos two A gives sine four A, and two sine four A cos four A gives sine eight A, all the way down the chain. With me? Each step needs one more two, so with n factors, we use two to the power n in all. So the product is, on top, sine of two to the power n times A, and at the bottom, two to the power n times sine A, okay? Multiply and divide by two sine A, that is the whole idea. But careful, this formula needs two conditions. We divide by sine A, so sine A must not be zero; the crossed equal sign means not equal to. And the angles must double, so for angles like A, two A, three A, the formula does not apply directly. That is the classic trap. Got it?

1:52Angles in equal steps: sums of sines and cosines

Now let us add sines of angles that go up in equal steps, each angle inside round brackets. The Greek letter alpha is the first angle, the Greek letter beta is the step, and the big sigma means add up, for k from zero to n minus one. The sum of the sines is sine of n beta over two, over sine of beta over two, times the sine of the angle, alpha plus n minus one times beta over two. Okay? For cosines, the same fraction stays in front, and only the last sine becomes a cos. Right? Now look at that last angle, alpha plus n minus one times beta over two. The first angle is alpha, and the last is alpha plus n minus one times beta. So what is this angle, in simple words? It is the average of the first and the last angle, yes! That is the topper's way to remember it: the fraction in front, times the sine or the cos of the average angle. With me? There is one condition: beta must not be two k pi, with k an integer, where pi, the Greek letter, is the radian measure of one hundred and eighty degrees. If beta were two k pi, then sine of beta over two, at the bottom, would be zero. Clear?

3:08Telescoping: turn each term into a difference

The third tool is telescoping. Capital A and capital B are angles, and cot is cos over sine. One over sine A times sine B equals cot A minus cot B, over sine of B minus A, so a product becomes a difference, okay? For cosines, one over cos A times cos B equals tan B minus tan A, over sine of B minus A; notice that tan B comes first. Both hold wherever both sides are defined, so nothing at the bottom may be zero. Right? And one more: tan theta equals cot theta, minus two cot two theta, where theta, the Greek letter, is any angle with sine two theta not zero. Got it? Here is why differences help: in a long sum, the end of one term cancels the start of the next, and only the first and the last pieces survive, like a telescope folding up. Clear?

4:02Example 1: cos pi/7 cos 2pi/7 cos 4pi/7, and a sine product

Now example one, our puzzle from the start: cos pi over seven, times cos two pi over seven, times cos four pi over seven. Then sine pi over fourteen, times sine three pi over fourteen, times sine five pi over fourteen. Pause the video and try it first. Done? Okay, let's check. These angles double, so capital A is pi over seven, and n is three, so two to the power three is eight. So the product is sine of eight pi over seven, over eight times sine pi over seven. With me? Eight pi over seven is pi plus pi over seven, and adding pi flips the sign of the sine. So the top is minus sine pi over seven, which cancels with the sine pi over seven at the bottom, and the first product is minus one over eight. So our puzzle is solved, right? Now the second product. How can we turn these sines into cosines? We use complements, yes! The sine of an angle is the cos of pi over two minus that angle. So sine pi over fourteen is cos three pi over seven, sine three pi over fourteen is cos two pi over seven, and sine five pi over fourteen is cos pi over seven. Okay? But three pi over seven is not in our chain, so write it as cos of pi minus four pi over seven, which is minus cos four pi over seven. Got it? So the second product is minus our first product: minus of minus one over eight, which is plus one over eight. Same cosines, opposite sign. Beautiful, isn't it?

5:43Example 2: prove the pi/11 cosine product equals 1/32

Now a harder one, JEE level. Prove that cos pi over eleven, times cos two pi over eleven, and so on up to cos five pi over eleven, equals one over thirty two. Pause the video and try it first. Tried it? Good, let's check. Five cosines, and two to the power five is thirty two, so we use our formula straight away, right? No, no, wait, look at the angles first! Pi, two pi, three pi, four pi and five pi, over eleven, do not double, and that is exactly the trap. So we fix them: cos eight pi over eleven is cos of pi minus three pi over eleven, which is minus cos three pi over eleven. Clear? And sixteen pi over eleven is two pi minus six pi over eleven, so its cos is cos six pi over eleven, which is minus cos five pi over eleven. With me? So swap them in, and the two minus signs multiply to give a plus. Now the product is cos pi over eleven, times cos two pi, four pi, eight pi and sixteen pi, all over eleven, and the angles double! Okay? So capital A is pi over eleven and n is five, and we get sine of thirty two pi over eleven, over thirty two times sine pi over eleven. Right? Now the last step. Can you simplify sine of thirty two pi over eleven? Take away two pi to get sine ten pi over eleven, and ten pi over eleven is pi minus pi over eleven, so that is just sine pi over eleven. It cancels, and the product is one over thirty two, which proves it. Got it?

7:26Quick revision: doubling product, AP sums, telescoping

So, quick revision. Angles that double: sine of two to the power n times A, over two to the power n times sine A, and if the angles do not double, fix them first. Angles in equal steps: sine of n beta over two, over sine of beta over two, times the sine or cos of the average angle. Telescoping: turn each term into a difference, so the middle terms cancel. Now go and try the practice problems on this page. All the best!