WEBVTT
00:00:00.226 --> 00:00:12.716
Use Eulerβs formula to express sin cubed π cos squared π in the form π sin π plus π sin three π plus π sin five π, where π, π, and π are constants to be found.
00:00:13.446 --> 00:00:21.766
Hence, find the solutions of sin five π minus sin three π equals zero in the interval π is greater than or equal to zero and less than π.
00:00:22.166 --> 00:00:23.526
Give your answer in exact form.
00:00:24.196 --> 00:00:32.136
We begin by recalling the fact that sin π is equal to one over two π times π to the ππ minus π to the negative ππ.
00:00:32.386 --> 00:00:37.246
And cos π is equal to a half times π to the ππ plus π to the negative ππ.
00:00:37.836 --> 00:00:42.736
This means we can find the product of sin cubed π and cos squared π.
00:00:43.016 --> 00:00:53.066
We can write it as one over two π times π to the ππ minus π to the negative ππ cubed times a half times π to the ππ plus π to the negative ππ squared.
00:00:53.496 --> 00:00:56.856
One over two π cubed is negative one over eight π.
00:00:57.366 --> 00:00:59.786
And a half squared is one-quarter.
00:01:00.116 --> 00:01:02.866
So we can rewrite our expression a little bit further.
00:01:02.936 --> 00:01:06.546
We find the product of negative one over eight π and a quarter.
00:01:06.546 --> 00:01:08.416
And we get negative one over 32π.
00:01:09.016 --> 00:01:11.546
And we can rewrite the rest of our expression as shown.
00:01:11.996 --> 00:01:16.326
Weβre now going to use the binomial theorem to expand each of the sets of parentheses.
00:01:16.936 --> 00:01:23.916
The first part becomes π to three ππ plus three choose one π to the two ππ times negative π to the negative ππ and so on.
00:01:24.516 --> 00:01:33.356
And this simplifies to π to the three ππ minus three π to the ππ plus three π to the negative ππ minus π to the negative three ππ.
00:01:33.626 --> 00:01:37.756
Letβs repeat this process for π to the ππ plus π to the negative ππ squared.
00:01:38.086 --> 00:01:45.166
When we do, we get π to the two ππ plus two π to the zero which is just two plus π to the negative two ππ.
00:01:45.436 --> 00:01:48.066
Weβre going to need to find the product of these two expressions.
00:01:48.166 --> 00:01:49.726
Weβll need to do that really carefully.
00:01:49.946 --> 00:01:55.086
Weβll need to ensure that each term in the first expression is multiplied by each term in the second expression.
00:01:55.636 --> 00:01:58.976
And we can write sin cubed π cos squared π as shown.
00:01:59.476 --> 00:02:00.766
Now, thereβs quite a lot going on here.
00:02:00.766 --> 00:02:04.536
So you might wish to pause the video and double-check your answer against mine.
00:02:04.926 --> 00:02:07.556
Weβre going to gather the corresponding powers of π together.
00:02:07.796 --> 00:02:11.386
Weβll gather π to the five ππ and π to the negative five ππ.
00:02:11.716 --> 00:02:15.336
Weβll collect π to the plus and minus three ππ.
00:02:15.626 --> 00:02:18.666
And weβll gather π the ππ and π to the negative ππ.
00:02:19.036 --> 00:02:20.736
Letβs neaten things up somewhat.
00:02:20.736 --> 00:02:34.826
We end up with negative one over 32π times π to the five ππ minus π to the negative five ππ minus π to the three ππ minus π to the negative three ππ minus two times π to the ππ minus π to the negative ππ.
00:02:35.116 --> 00:02:38.256
And now, you might be able to spot why we chose to do this.
00:02:38.366 --> 00:02:40.616
We can now go back to the given formulae.
00:02:41.036 --> 00:02:42.956
Letβs clear some space for the next step.
00:02:43.636 --> 00:02:45.896
We kind of unfactorize a little.
00:02:46.026 --> 00:02:49.426
And we can rewrite sin cubed π cos squared π as shown.
00:02:49.736 --> 00:02:55.566
And we can therefore replace π to the ππ plus π to the negative ππ with sin π and so on.
00:02:56.036 --> 00:03:03.736
And we can see that sin cubed π cos squared π is equal to a 16th times two sin π plus sin three π minus sin five π.
00:03:04.316 --> 00:03:09.626
Since π, π, and π are constants to be found, we can say that π, the coefficient of sin π, is one-eighth.
00:03:09.816 --> 00:03:12.496
π, the coefficient of sin three π, is a 16th.
00:03:12.816 --> 00:03:16.466
And π, the coefficient sin five π, is negative one 16th.
00:03:16.786 --> 00:03:18.836
Letβs now consider part two of this question.
00:03:19.126 --> 00:03:24.146
We begin by using our answer to part one and multiplying both sides by 16.
00:03:24.596 --> 00:03:30.036
We then subtract two sin π from both sides and multiply through by negative one.
00:03:30.396 --> 00:03:35.846
And we can now see that weβve got an equation in sin five π minus sin three π.
00:03:36.066 --> 00:03:39.646
Weβre told that sin five π minus sin three π is equal to zero.
00:03:39.876 --> 00:03:45.176
So we let two sin π minus 16 sin cubed π cos squared π be equal to zero.
00:03:45.536 --> 00:03:48.186
And then, we factor by two sin π.
00:03:48.636 --> 00:03:54.056
Since the product of these two terms is equal to zero, this means that either of these terms must be equal to zero.
00:03:54.436 --> 00:04:03.366
So either two sin π is equal to zero and dividing by two, we could see that sin π is equal to zero or one minus eight sin squared π cos squared π is equal to zero.
00:04:03.876 --> 00:04:10.496
Given the interval π is greater than or equal to zero and less than π, we can see that one of our solutions is when π is equal to zero.
00:04:10.816 --> 00:04:13.286
Weβre going to rewrite our other equations somewhat.
00:04:13.286 --> 00:04:16.506
We know that sin two π is equal to two sin π cos π.
00:04:16.906 --> 00:04:21.646
Squaring this, we get sin squared two π equals four sin squared π cos squared π.
00:04:22.076 --> 00:04:27.636
And thatβs in turn means that our equation is one minus two sin squared two π equals zero.
00:04:28.096 --> 00:04:34.736
Rearranging to make sin two π the subject, we see that sin two π is equal to plus or minus one over root two.
00:04:35.226 --> 00:04:44.266
Starting with the positive square root for π in the interval given, we know that sin two π is equal to one over root two when π is equal to π by eight or three π by eight.
00:04:44.656 --> 00:04:47.046
Similarly, we can solve for the negative square root.
00:04:47.076 --> 00:04:49.636
And we get five π by eight and seven π by eight.
00:04:50.086 --> 00:04:59.466
And there are therefore five solutions to the equation sin five π minus sin three π equals zero in the interval π is greater than or equal to zero and less than π.
00:04:59.856 --> 00:05:05.276
They are zero π by eight, three π by eight, five π by eight, and seven π by eight.