# Thread: Prove using Binomial and DeMoivre's Theorem?

1. ## Prove using Binomial and DeMoivre's Theorem?

If Z = cos θ + i sin θ, (θ belongs to set of real numbers, i = √-1), then according to DeMoivre's theorem, Z^n = cos nθ + i sin nθ, (n belongs to set of natural numbers). Using this theorem alongwith binomial expansion, prove that:

(i) cos 2nθ =( r = 0 to n) (2n C 2r) cos^(2(n - r)) θ sin^(2r) θ
(ii) cos 2nθ = (r = 0 to n) (2n C (2r + 1))cos^(2(n-r) - 1) θ sin^(2r + 1) θ

NOTE: 2n C 2r is 2n choose 2r.

2. Hi

$\cos(2n\theta) = Re(\cos(2n\theta) + i \sin(2n\theta))$

$\cos(2n\theta) = Re\left((\cos \theta + i \sin \theta)^{2n}\right)$

$\cos(2n\theta) = Re\left(\sum_{k=0}^{2n} \binom{2n}{k} \cos^{2n-k} \theta +(i \sin \theta)^{k}\right)$

$(i \sin \theta)^{k} = i^k \sin^k \theta$ is pure complex if k is odd and real if k is even

Set k=2r

3. Originally Posted by running-gag
Hi

$\cos(2n\theta) = Re\left(\sum_{k=0}^{2n} \binom{2n}{k} \cos^{2n-k} \theta + (i \sin \theta)^{k}\right)$
Instead of +, shoudn't it be . ?

So, the question appears to have a small mistake, doesn't it? As in the R.H.S, it says in the question that the summation is from r = 0 to n instead of r = 0 to 2n as you proved. Thanks for the help!

4. Originally Posted by fardeen_gen
Instead of +, shoudn't it be . ?
Yes sorry

Originally Posted by fardeen_gen
So, the question appears to have a small mistake, doesn't it? As in the R.H.S, it says in the question that the summation is from r = 0 to n instead of r = 0 to 2n as you proved. Thanks for the help!
No
k is from 0 to 2n and k=2r therefore r is from 0 to n

5. No
k is from 0 to 2n and k=2r therefore r is from 0 to n
Oops. My bad. Thanks once again!

6. But then in (ii), the sin term will disappear entirely(setting k = 2r + 1?). And also, the summation would be from r = -1/2(possible?) to r = (2n - 1)/2.
What am I missing?

7. Originally Posted by fardeen_gen
(i) cos 2nθ =( r = 0 to n) (2n C 2r) cos^(2(n - r)) θ sin^(2r) θ
(ii) cos 2nθ = (r = 0 to n) (2n C (2r + 1))cos^(2(n-r) - 1) θ sin^(2r + 1) θ

NOTE: 2n C 2r is 2n choose 2r.
To be clear
(i) is false since $(i \sin \theta)^{2r} = i^{2r} \sin^{2r} \theta = (-1)^r \sin^{2r} \theta$
(ii) is false because
- it is $\sin(2n\theta)$
- r cannot go up to n because you would get $\binom{2n}{2n+1}$
Use the imaginary part instead of the real part to get the second relation