WEBVTT
00:00:02.710 --> 00:00:14.720
Express cos to the sixth power π in terms of cos six π, cos five π, cos four π, cos three π, cos two π, cos π, and any constant terms.
00:00:15.020 --> 00:00:19.090
In this question, we want to find an expression for the cos to the sixth power of π.
00:00:19.200 --> 00:00:27.050
And this expression needs to consist of terms which only involve cos of integer multiple values of π and constant terms.
00:00:27.160 --> 00:00:37.660
In some questions like this, itβs possible to find expressions for each of these multiple angle formula for cosine and then get our answer to be equal to what we want, in this case cos to the sixth power π.
00:00:37.750 --> 00:00:42.450
However, to do this, we would need to do this for each of the multiple angle formulas in the question.
00:00:42.580 --> 00:00:48.310
So although this method might be possible, it would be very time consuming, and it would take a lot of trial and error.
00:00:48.600 --> 00:00:53.330
Instead, weβre going to need to recall we can find expressions like this by using De Moivreβs theorem.
00:00:53.490 --> 00:00:56.410
So to answer this question, weβll start by recalling De Moivreβs theorem.
00:00:56.450 --> 00:01:07.490
One version of this says, for any integer value of π and real number π, the cos of π plus π sin of π all raised to the πth power is equal to the cos of ππ plus π sin of ππ.
00:01:07.760 --> 00:01:11.970
However, if we were to try and directly use this expression to answer our question, we might struggle.
00:01:12.180 --> 00:01:20.080
Since we want to find an expression for cos to the sixth power of π, weβd need to set π equal to six to get cos to the sixth power to appear in our expression.
00:01:20.200 --> 00:01:28.750
However, when we distribute the πth power over our parentheses, we would end up with fifth power cos terms, fourth power cos terms, all the way down to zeroth power cos terms.
00:01:28.870 --> 00:01:32.300
And this seems like it would be very difficult to get rid of all of these terms.
00:01:32.330 --> 00:01:34.990
So instead, weβre going to need to recall a very useful result.
00:01:35.220 --> 00:01:44.640
If we call the complex number inside our parentheses π§, then De Moivreβs theorem is telling us, π§ to the πth power is equal to the cos of ππ plus π sin of ππ.
00:01:44.840 --> 00:01:57.120
And we can then use De Moivreβs theorem to find two useful results, π§ to the πth power plus one over π§ to the πth power is equal to two cos of ππ and π§ to the πth power minus one over π§ to the πth power is equal to two π sin of ππ.
00:01:57.400 --> 00:02:02.770
And to prove these results, we directly use De Moivreβs theorem, which we remember is true for any integer value of π.
00:02:02.800 --> 00:02:06.850
And we know one over π§ to the πth power is the same as π§ to the power of negative π.
00:02:07.010 --> 00:02:09.910
These results are very useful and theyβre worth committing to memory.
00:02:10.130 --> 00:02:15.790
Weβre going to focus on the top result, which weβre going to use to find an expression for cos to the sixth power of π.
00:02:15.980 --> 00:02:19.560
We start off by setting our value of π equal to one in this expression.
00:02:19.590 --> 00:02:23.030
This gives us two cos of π is equal to π§ plus one over π§.
00:02:23.290 --> 00:02:30.420
And remember, the question wants us to find an expression for the cos to the sixth power π, so weβre going to raise both sides of this equation to the sixth power.
00:02:30.790 --> 00:02:35.890
On the left-hand side of our equation, we can simplify by taking the sixth power of both of our two factors.
00:02:36.120 --> 00:02:39.740
This gives us two to the sixth power times cos to the sixth power of π.
00:02:39.990 --> 00:02:44.100
And we can simplify this since we know two to the sixth power is equal to 64.
00:02:44.470 --> 00:02:48.930
So the left-hand side of our equation simplifies to give us 64 cos to the sixth power of π.
00:02:49.240 --> 00:02:53.420
And on the right-hand side of our equation, we have the sum of two numbers raised to an exponent.
00:02:53.420 --> 00:02:55.520
We can do this by using the binomial formula.
00:02:55.760 --> 00:03:07.640
We recall, this tells us for any positive integer π, π plus π all raised to the πth power is equal to the sum from π equals zero to π of π choose π times π to the power of π multiplied by π to the power of π minus π.
00:03:07.950 --> 00:03:11.910
So we can use this to distribute the exponent of six over the right-hand side of our equation.
00:03:12.120 --> 00:03:21.930
We get that six choose zero π§ to the sixth power plus six choose one times π§ to the fifth power times one over π§ plus six choose two times π§ to the fourth power multiplied by one over π§ all squared.
00:03:22.100 --> 00:03:28.800
And then we keep adding terms of this form all the way up to six choose six multiplied by one over π§ all raised to the sixth power.
00:03:29.160 --> 00:03:33.570
Remember, the question wants us to find an expression for the cos to the sixth power of π.
00:03:33.600 --> 00:03:36.890
But it wants this in terms of cos of multiple angles of π.
00:03:37.010 --> 00:03:39.940
We can rearrange this to get the cos to the sixth power of π.
00:03:39.970 --> 00:03:45.180
However, the right-hand side of this expression is in terms of π§; itβs not yet in terms of cos of angles of π.
00:03:45.510 --> 00:03:48.410
This means weβre going to need to simplify the right-hand side of our equation.
00:03:48.440 --> 00:03:49.750
Weβll do this term by term.
00:03:49.980 --> 00:03:54.430
In our first term, six choose zero is equal to one, so the first term is just π§ to the sixth power.
00:03:54.750 --> 00:04:01.900
In our second term, six choose one is equal to six and π§ to the fifth power multiplied by one over π§ can be simplified by using our laws of exponents.
00:04:01.930 --> 00:04:03.530
Itβs equal to π§ to the fourth power.
00:04:03.790 --> 00:04:06.050
So our second term is just six π§ to the fourth power.
00:04:06.430 --> 00:04:14.970
In our third term, six choose two is 15 and π§ to the fourth power multiplied by one over π§ squared is π§ to the fourth power over π§ squared, which is just equal to π§ squared.
00:04:15.220 --> 00:04:17.240
So our third term is 15π§ squared.
00:04:17.440 --> 00:04:20.660
And we can simplify the rest of the terms in this expansion in the same way.
00:04:20.710 --> 00:04:26.010
We get 20, 15 over π§ squared, six over π§ to the fourth power, and one over π§ to the sixth power.
00:04:26.190 --> 00:04:27.830
And now we can notice something interesting.
00:04:27.860 --> 00:04:30.830
We can simplify this expression by using De Moivreβs theorem.
00:04:30.910 --> 00:04:38.090
For example, we have the first term in this expansion plus the last term in this expansion is π§ to the sixth power plus one over π§ to the sixth power.
00:04:38.400 --> 00:04:44.410
By setting π equal to six in our useful result, we know this is equal to two times the cos of six π.
00:04:44.560 --> 00:04:47.750
And this isnβt the only time this result appears in our expansion.
00:04:47.780 --> 00:04:49.600
We can use this two more times.
00:04:49.690 --> 00:05:01.350
So weβll pair up the terms in our expansion to give us π§ to the sixth power plus one over π§ to the sixth power plus six π§ to the fourth power plus six over π§ to the fourth power plus 15π§ squared plus 15 over π§ squared plus 20.
00:05:01.600 --> 00:05:04.650
And before we apply De Moivreβs theorem, we can notice something interesting.
00:05:04.650 --> 00:05:08.060
In our first set of parentheses, we can take out the shared factor of six.
00:05:08.250 --> 00:05:13.090
And similarly, in our third set of parentheses, we can take out the shared factor of 15.
00:05:13.340 --> 00:05:14.900
This gives us the following expression.
00:05:14.930 --> 00:05:18.960
And now we can simplify all three of our sets of parentheses by using De Moivreβs theorem.
00:05:19.130 --> 00:05:24.920
By setting π equal to six, we know π§ to the sixth power plus one over π§ to the sixth power is two cos of six π.
00:05:25.180 --> 00:05:30.220
With π as four, π§ to the fourth power plus one over π§ to the fourth power is two cos four π.
00:05:30.250 --> 00:05:34.780
And with π as two, π§ squared plus one over π§ squared is two cos of two π.
00:05:35.120 --> 00:05:51.440
So by substituting these expressions in and remembering we need to multiply by our coefficients and we still need to add 20 at the end, we get 64 cos to the sixth power of π is equal to two cos of six π plus 12 cos of four π plus 30 cos of two π plus 20.
00:05:51.640 --> 00:05:53.700
And now this is almost in exactly the form we want.
00:05:53.740 --> 00:05:57.690
All we need now is to rearrange to make cos to the sixth power of π the subject.
00:05:57.890 --> 00:06:01.680
And to do this, weβre going to need to divide both sides of our equation through by 64.
00:06:01.860 --> 00:06:04.090
So to do this, letβs start by clearing some space.
00:06:04.090 --> 00:06:07.680
We have cos to the sixth power of π is equal to the following expression.
00:06:07.990 --> 00:06:12.000
And to evaluate this, weβre going to divide every term in this expression by 64.
00:06:12.360 --> 00:06:22.890
Doing this, we get two cos six π all over 64 plus 12 cos four π all over 64 plus 30 cos two π all over 64 plus 20 divided by 64.
00:06:23.130 --> 00:06:29.310
All thatβs left to do now is cancel all of the shared factors of two in both our numerator and denominator for all four of our terms.
00:06:29.310 --> 00:06:30.540
We get the following expression.
00:06:30.540 --> 00:06:32.320
And this gives us our final answer.
00:06:32.420 --> 00:06:40.320
Therefore, by using De Moivreβs theorem, we were able to express cos to the sixth power of π in terms of cosines of integer multiple angles of π.
00:06:40.690 --> 00:06:53.240
We got cos to the sixth power of π is equal to one over 32 times cos six π plus three over 16 times cos four π plus 15 over 32 times cos of two π plus five over 16.