Prove by Induction: cos(6\theta)+isin(6\theta)=(cos\theta +isin\theta )^6

[Mentallic]

Homework Statement

Prove by induction cos(6\theta)+isin(6\theta)=(cos\theta +isin\theta )^6 for all \theta

Homework Equations

by DeMoivres theorem, (cos\theta +isin\theta )^n=cos(n\theta)+isin(n\theta)

The Attempt at a Solution

Well when I first looked at this problem I thought there was an error in the question because from what I know, induction has a flaw in that it can only prove for integers, not ALL values.

Anyway, so I need to prove:

cos6(k+1)+isin6(k+1)=(cos(k+1)+isin(k+1)) ^6

Any help would be appreciated

 

[Mark44]

I think you can use Euler's formula here: e^(ia) = cos(a) + i sin(a). You also need the fact that e^(ab) = (e^a)^b.

 

[Mentallic]

So
cos6(k+1)+isin6(k+1)=e^{i6(k+1)}=(e^{6i})^{k+1}=e^{6i}.(e^{6i})^k

Sorry but I'm unsure what to do from here. And please correct me if I'm wrong.

 

[Mark44]

So
cos6(k+1)+isin6(k+1)=e^{i6(k+1)}=(e^{6i})^{k+1}=e^{6i}.(e^{6i})^k

Sorry but I'm unsure what to do from here. And please correct me if I'm wrong.

Try this instead.
cos6(k+1)+isin6(k+1)=e^{i6(k+1)}=(e^{i(k + 1)})^{6}

Now you're almost there...

 

[gabbagabbahey]

I think you may be misinterpreting the question here. I assume that you are supposed to start by showing cos(n\theta)+isin(n\theta)=(cos\theta +isin\theta )^n for all \theta, and some positive integer n (I'd use n=1 as your starting point) by expanding (cos\theta +isin\theta )^k and using the multiple angle trig identities. Then assuming it is true for n=k show that it is true for n=k+1 and hence by induction that it is true for all n, including n=6

Mark's method isn't really a proof by induction, since you are assuming that Euler's formula holds for all \theta.

 

[Mark44]

I think you may be misinterpreting the question here. I assume that you are supposed to start by showing cos(n\theta)+isin(n\theta)=(cos\theta +isin\theta )^n for all \theta, and some positive integer n (I'd use n=1 as your starting point) by expanding (cos\theta +isin\theta )^k and using the multiple angle trig identities. Then assuming it is true for n=k show that it is true for n=k+1 and hence by induction that it is true for all n, including n=6

Mark's method isn't really a proof by induction, since you are assuming that Euler's formula holds for all \theta.

It seemed odd to me, too, but I'm just going by what mentallic wrote in the first post in this thread:

Prove by induction cos(6\theta)+isin(6\theta)=(cos\theta +isin\theta )^6 for all theta.

and

Well when I first looked at this problem I thought there was an error in the question because from what I know, induction has a flaw in that it can only prove for integers, not ALL values.

 

[gabbagabbahey]

It is a poorly worded question; they may as well have just asked to prove by induction that (x)^6=x^6 for all x.

 

[Mentallic]

Well this is the only question I got wrong in that test (besides a couple sloppy errors). I was unsure how I would prove for all \theta without using DeMoivres theorem, and at the same time, if I were to assume the theorem true, then... I have pretty much the same thing as what you said gabbagabbahey:

It is a poorly worded question; they may as well have just asked to prove by induction that for all x.

So I instead went ahead and proved DeMoivres theorem cos(n\theta)+isin(n\theta)=(cos\theta+isin\theta)^n
And thus, true for n=6 and I probably cheated here when I finished with the statement, hence true for all \theta.

Try this instead.
cos6(k+1)+isin6(k+1)=e^{i6(k+1)}=(e^{i(k + 1)})^{6}

Now you're almost there...

Sorry Mark, but I was never taught Euler's method. We rather used the mod-arg form rcis\theta. But I have heard that there is a clear relationship between these 2 forms, so if we could convert to the mod-arg form, I could proceed

It is a poorly worded question; they may as well have just asked to prove by induction that (x)^6=x^6 for all x.

If it is poorly worded, what should the question have been instead? Because this is exactly what we needed to answer.