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

• Mentallic
In summary, the question asks to prove by induction that cos(6\theta)+isin(6\theta)=(cos\theta +isin\theta )^6 for all \theta. The approach involves using DeMoivres theorem to show that cos(n\theta)+isin(n\theta)=(cos\theta+isin\theta)^n, and then using induction to show that this holds for n=6. The question may be better phrased as proving that (x)^6=x^6 for all x using induction.
Mentallic
Homework Helper

## 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

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.

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.

Mentallic said:
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...

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 $\theta$.

gabbagabbahey said:
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.

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.

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:
gabbagabbahey said:
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 $$\theta$$.

Mark44 said:
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

gabbagabbahey said:
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.

## 1. What is the purpose of using induction in this proof?

The purpose of using induction in this proof is to show that the given statement is true for all natural numbers. This allows for a more efficient and concise way of proving the statement, rather than having to prove it for each individual case.

## 2. How do you begin a proof by induction?

To begin a proof by induction, you must first establish a base case, which is typically the smallest value of the natural numbers. Then, you assume that the statement holds for some arbitrary value of the natural numbers, and use that assumption to prove that it also holds for the next value. This process is repeated until the desired statement is proven for all natural numbers.

## 3. What is the role of the cosine and sine functions in this proof?

The cosine and sine functions are used because the given statement involves complex numbers, and the cosine and sine functions are essential in representing complex numbers in polar form. By utilizing these functions, we can easily manipulate and simplify the given statement to prove its validity.

## 4. Can this proof be used for other trigonometric functions?

Yes, this proof can be used for other trigonometric functions, as long as the statement being proved involves complex numbers and can be written in terms of cosine and sine functions. However, the specific steps and calculations may vary depending on the trigonometric function being used.

## 5. Are there any limitations to using induction in this proof?

One limitation of using induction in this proof is that it can only be used for proving statements that hold for all natural numbers. If the statement does not hold for all natural numbers, then induction cannot be used. Additionally, the proof may become more complex if the statement involves more than one variable or condition.

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