How Do You Derive Trigonometric Identities from DeMoivre's Theorem?

[kathrynag]

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.

 

[Dick]

Sure. e^(i*2*theta)=(e^(i*theta))^2. Write down both sides and equate the imaginary parts.

 

[kathrynag]

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

 

[Dick]

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.

 

[kathrynag]

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?

 

[Dick]

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.

 

[kathrynag]

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

 

[Dick]

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?

 

[kathrynag]

Yes. And cos(2t)=cos^2(t)-sin^2(t). Right?

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