Can You Apply Binomial Theorem to Expand (x + y)^5?

[sparsh]

using eulers formula express cos5ø in terms of cosø. Hence show that =cos(pie/10) is a root of the equation 16x^4 - 20x^2 +5 =0 ..

Thanks in advance .

 

[HallsofIvy]

Looks like homework to me so I'm moving this to the homework section. Do you know what Euler's formula is? If so, go ahead and apply it!

 

[sparsh]

Yup i know it

But the problem is that when i convert cos5ø in terms of (cosø + isinø )
then how do i remove the sin term ...

like this complex number has only got its real part. And using eulers formula we can't have angles different for the cos and sine terms..

Moreover we have to use the obtained result in the 2nd part of the question

 

[matt grime]

If you know it then use it, or at least write here what you've obtained by using it, so that you can get help (not just the answer).

You know how to get rid of sin's and turn them into cos's from trig (cos^2+sin^2=1) and the real part only has even powers of sin in it.

 

[Reshma]

I think you will have to consider only the real part here since you have only 'x' terms.
Use the De Moivre's formula for powers and roots of complex numbers.
\left[r(\cos \phi + i\sin \phi)\right]^n = r^n(\cos n\phi + i\sin n\phi)

Hope this helps a bit...

 

[sparsh]

I converted it into something like this ::

cos 5ø = (cosø)^5 + i cos(90-ø)^5 - icos(90-5ø)

Then what should i do

 

[matt grime]

That is not correct. In fact it looks as though you think that (x+y)^5=x^5+y^5.

Do not write sin(x)=cos(90-x), by the way. There is no need to do that.

 

[sparsh]

@ matt

By the eulers formula or De Movier's Theorum (cosø+isinø)^5 is cos 5ø+isin5ø
Is there anything wrong with that.

Actually I got to give the exam after two days . So please it would be really great if you could just give me the solution to this problem. I have spent too much time on it already. Have to study Physics and Chem too

 

[matt grime]

that is right, and nothing to do with what i said: you failed to expand the fifth power as a power properly.

certainly (cosx+isinx)^5=cos5x+isin5x, but you also seem to think that this is the same as cos^5(x)+isin^5(x), which is obviously false.

 

[sparsh]

I think i got the solution to this problem...

Just to share it with you. We can break e^i5ø into e^i3ø e^i2ø and then solve .
Thanks anyways for all your help ...

 

[HallsofIvy]

sparsh, in your original post you said "use Euler's formula". My first response was "Do you know what Euler's formula is? If so, go ahead and apply it!" Your last response was the first one where it looks like you are actually using Euler's formula!

 

[sparsh]

Can i know what is the basic difference between the Eulers Formula and the De moivers theorum. I mean they produce the same result.

 

[Reshma]

Can i know what is the basic difference between the Eulers Formula and the De moivers theorum. I mean they produce the same result.

Euler's formula states:
e^{i\phi} = \cos \phi + i\sin \phi

De Moivre's Formula states:
(\cos \phi + i\sin \phi)^n = \cos(n\phi)+i\sin(n\phi)

De Moivre's formula is useful when you are dealing with powers and roots of complex numbers.

 

[Curious3141]

I think the basic problem is knowledge and application of Binomial Theorem, not De Moivre's or Euler's formulae.

Sparsh, you can expand second and third powers of (x + y).

Can you do the same for the fifth power ?