How would you go about to attack this question??? (1 Viewer)

[Michaelmoo]

Express Cos^7(theta) [i.e. Cos theta to the power of 7] in terms of cosines of multiples of theta.

The text book goes through a solution, although it doesnt really explain the reason for the approach. Or is this just one of those questions where that you MUST have done before to realise?

Please explain with reasoning.

Thanks in advance?

 

[Timothy.Siu]

get a pen and stab at it

 

[Michaelmoo]

get a pen and stab at it

Yes!!! Thanks!!!

 

[Timothy.Siu]

do u need someone to do it?

 

[Michaelmoo]

do u need someone to do it?

well not necesarily do it. Just a brief explanantion on how to approach it/ and how would you know how to approach it.

 

[Timothy.Siu]

does cos⁷A=(cos 7A-7cos 5A+21cos 3A-35cos A)/64

 

[Michaelmoo]

does cos⁷A=(cos 7A-7cos 5A+21cos 3A-35cos A)/64

Yes. But without the minusses, just plusses throughout i.e.

cos⁷A=(cos 7A+7cos 5A+21cos 3A+35cos A)/64

How and why?

 

[anon09]

Use De Moivre's Theorem:

cos7@ + isin7@ = (cos@ + isin@) ^7

Expand RHS

Gather real terms, which equal cos7@

And, using trig identities, eliminate sin@'s to get the answer in terms of cos

 

[Michaelmoo]

Use De Moivre's Theorem:

cos7@ + isin7@ = (cos@ + isin@) ^7

Expand RHS

Gather real terms, which equal cos7@

And, using trig identities, eliminate sin@'s to get the answer in terms of cos

But Im after Cos(theta) to the power of 7 not cos(7theta)

 

[Timothy.Siu]

oh crap, yeah my bad,

well u do z=cos A+isin A

z+z^-1=2cos A

(2cos A)⁷=128cos⁷A

(z+z^-1)⁷=z⁷+7z⁵+21³+35z+35z^-1+21z^-3+7z^-5+z^-7

=(z⁷+z^-7)+7(z⁵+z^-5)+21(z³+z^-3)+35(z+z^-1)=2cos 7A-14cos 5A+42cos 3A-70cos A=128 cos⁷A (from above)

used de moirves theorem for z=cos A+isin A
zⁿ=cos nA+isin nA

therefore cos⁷A=(cos 7A+7cos 5A+21cos 3A+35cos A)/64

EDIT: done i think

 

[lyounamu]

Use De Moivre's Theorem:

cos7@ + isin7@ = (cos@ + isin@) ^7

Expand RHS

Gather real terms, which equal cos7@

And, using trig identities, eliminate sin@'s to get the answer in terms of cos

not really.

That method is not really right for this question.

timothy's solution is more correct.

 

[Michaelmoo]

oh crap, yeah my bad,

well u do z=cos A+isin A

z+z^-1=2cos A

(2cos A)⁷=128cos⁷A

(z+z^-1)⁷=z⁷+7z⁵+21³+35z+35z^-1+21z^-3+7z^-5+z^-7

=(z⁷+z^-7)+7(z⁵+z^-5)+21(z³+z^-3)+35(z+z^-1)=2cos 7A-14cos 5A+42cos 3A-70cos A=128 cos⁷A (from above)

used de moirves theorem for z=cos A+isin A
zⁿ=cos nA+isin nA

therefore cos⁷A=(cos 7A+7cos 5A+21cos 3A+35cos A)/64

EDIT: done i think

Ok thanks. Is that something you simply think of on sight of this question, or it is a method expected to be known???

 

[Timothy.Siu]

Ok thanks. Is that something you simply think of on sight of this question, or it is a method expected to be known???

well, u know it now

 

[lyounamu]

Ok thanks. Is that something you simply think of on sight of this question, or it is a method expected to be known???

Well, it's not really expected to know. You can easily think of that on spot if you can see that z + z^-1 = 2cos@ and z - z^-1 = 2isin@ given that z = cis@

But it's from Cambridge as far as I am aware.

 

[anon09]

Yep, my mistake, I didn't read the question properly.

I hope I don't do that in a test!

 

[kaz1]

Ok thanks. Is that something you simply think of on sight of this question, or it is a method expected to be known???

My teacher said you should be able to derive zⁿ+z^-n=2cos nA and zⁿ-z^-n=2isin nA as you cannot just state them in an exam.

 

[lyounamu]

My teacher said you should be able to derive zⁿ+z<SUP>-n</SUP>=2cos nA and zⁿ-z<SUP>-n</SUP>=2isin nA as you cannot just state them in an exam.

Yeah. True.

But in my own 4 unit exam, people just wrote that "formula" and still got a mark for it...I am the one who actually proved it.

 

[youngminii]

But why.. That formula is extremely simple to derive, should be a given.. Better safe than sorry I suppose

 

[Timothy.Siu]

derive that formula? its like 2 steps...
if z=cos a+isin a
zⁿ=cos na+isin na
z^-n=cos -na+isin na=cos na-sin na

therefore zⁿ+z^-n=2cos na

 

[lolokay]

would you get OP's question in a test without the lead in?