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.