The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

De moivres theorem binomial expansion

find cos3a in terms of cosa
cos3a=(z^3+z^-3)/2
where z = cosa+jsina
so z^3 = (cosa+jsina)^3 and z^-3=(cosa-jsina)^3 <<is this right?
let c=cosa and s=jsina
cos3a=((c3+3c2s+3cs2+s3)+(c3+3c2(−s)+3c(−s)2+1(−s)3))/2 ((c^3+3c^2s+3cs^2+s^3)+(c^3+3c^2(-s)+3c(-s)^2+1(-s)^3))/2
=(2c3+6cs2)/2=c3+3cs2 (2c^3+6cs^2)/2=c^3+3cs^2
this is wrong according to the book where the last line is meant to be c^3-3cs^2
wheres the silly mistake?
Reply 1
Avatar for RichE
RichE
15
Original post by Daiblain
find cos3a in terms of cosa
cos3a=(z^3+z^-3)/2
where z = cosa+jsina
so z^3 = (cosa+jsina)^3 and z^-3=(cosa-jsina)^3 <<is this right?
let c=cosa and s=jsina
cos3a=((c3+3c2s+3cs2+s3)+(c3+3c2(−s)+3c(−s)2+1(−s)3))/2 ((c^3+3c^2s+3cs^2+s^3)+(c^3+3c^2(-s)+3c(-s)^2+1(-s)^3))/2
=(2c3+6cs2)/2=c3+3cs2 (2c^3+6cs^2)/2=c^3+3cs^2
this is wrong according to the book where the last line is meant to be c^3-3cs^2
wheres the silly mistake?


What happened to j?

You've worked out (c+s)^3+(c-s)^3

It's easier to work it out at the real part of (c+js)^3

Edit: I see you've included j in your s, so your s is probably different from the book's.
(edited 7 years ago)
Reply 2
Avatar for B_9710
B_9710
18
It's much easier if you use De Moivre's theorem in the following way
cos⁡3θ=ℜ(z3)=ℜ((cos⁡θ+isin⁡θ)3) \displaystyle \cos 3\theta =\Re (z^3)=\Re ((\cos \theta +i\sin \theta )^3) .
Just expand (cos⁡θ+isin⁡θ)3 (\cos \theta +i\sin \theta )^3 and equate the real parts.
(edited 7 years ago)
Reply 3
Avatar for Daiblain
Daiblain
OP
1
thanks

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you