Deriving DeMoivre's Formula?

In summary: Thanks!In summary, we derived the identities cos2theta=(costheta)^2-(sintheta)^2 and sin2theta=2sin theta costheta from Demoivre's theorem by equating the imaginary parts of both sides. We also used the identity cos(2t)=cos^2(t)-sin^2(t) to simplify our calculations.
  • #1
kathrynag
598
0

Homework Statement


Derive from Demoivre's theorem that
cos2theta=(costheta)^2-(sintheta)^2

sin2theta=2sin theta costheta

Homework Equations





The Attempt at a Solution


I really am clueless on this.
I considered using (costheta +isintheta)^n, but don't know where to go with this idea.
 
Last edited:
Physics news on Phys.org
  • #2
Sure. e^(i*2*theta)=(e^(i*theta))^2. Write down both sides and equate the imaginary parts.
 
  • #3
Dick said:
Sure. e^(i*2*theta)=(e^(i*theta))^2. Write down both sides and equate the imaginary parts.

cos2theta+isin2theta=(costheta)^2+2isintheta-(sintheta)^2
cos2theta=costheta^2-sintheta^2




cos2theta+isin2theta=(costheta)^2+2isintheta-(sintheta)^2
sin2theta=2sintheta-costheta-sintheta^2+costheta^2
I get stuck here
 
  • #4
Let's just call theta t, ok? On one side you have cos(2t)+i*sin(2t). On the other side you have (cos(t)+i*sin(t))*(cos(t)+i*sin(t)). Multiply that out carefully. You are getting things all mushed up.
 
  • #5
Dick said:
Let's just call theta t, ok? On one side you have cos(2t)+i*sin(2t). On the other side you have (cos(t)+i*sin(t))*(cos(t)+i*sin(t)). Multiply that out carefully. You are getting things all mushed up.

cos(2t)+isin(2t)=(cost+isint)^2
cos(2t)+isin(2t)=(cost)^2+2isint-(sint)^2
isin2t=(cost)^2+2isint-(sint)^2-cos(2t)
Is there something I'm missing on what to do here?
 
  • #6
On your second line the imaginary part is wrong. (a+bi)*(a+bi)=a^2-b^2+2abi. Now if the two sides are equal then the real part of one side equals the real part of the other side. Same for the imaginary parts.
 
  • #7
kathrynag said:
cos(2t)+isin(2t)=(cost+isint)^2
cos(2t)+isin(2t)=(cost)^2+2isint-(sint)^2
isin2t=(cost)^2+2isint-(sint)^2-cos(2t)
Is there something I'm missing on what to do here?


cos(2t)+isin(2t)=(cost+isint)^2
cos(2t)+isin(2t)=(cost)^2+2isintcost-(sint)^2
sin(2t)=2sintcost
 
  • #8
kathrynag said:
cos(2t)+isin(2t)=(cost+isint)^2
cos(2t)+isin(2t)=(cost)^2+2isintcost-(sint)^2
sin(2t)=2sintcost

Yes. And cos(2t)=cos^2(t)-sin^2(t). Right?
 
  • #9
Dick said:
Yes. And cos(2t)=cos^2(t)-sin^2(t). Right?

Yep, that makes sense. i was having a brain fart
 

1. What is DeMoivre's formula?

DeMoivre's formula is a mathematical formula that relates complex numbers to trigonometric functions. It states that for any complex number z = r(cos θ + i sin θ), where r is the magnitude and θ is the angle, the nth power of z can be expressed as rn(cos nθ + i sin nθ).

2. How is DeMoivre's formula derived?

DeMoivre's formula can be derived using Euler's formula, which states that eix = cos x + i sin x. By substituting x = nθ into Euler's formula and raising both sides to the nth power, we can obtain DeMoivre's formula.

3. What is the significance of DeMoivre's formula?

DeMoivre's formula is significant because it allows us to easily calculate the powers of complex numbers, which are essential in many mathematical and scientific fields. It also has applications in physics, engineering, and signal processing.

4. Can DeMoivre's formula be used to find roots of complex numbers?

Yes, DeMoivre's formula can be used to find n th roots of a complex number. By setting r to the nth root of the original complex number and solving for θ, we can find the n distinct roots of the original number.

5. Are there any limitations to DeMoivre's formula?

DeMoivre's formula is limited to calculating powers and roots of complex numbers. It cannot be used for other operations such as addition, subtraction, multiplication, or division. Additionally, it only applies to complex numbers in polar form.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
30
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
805
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
336
Back
Top