Express cos(θ₁ + θ₂ + θ₃) using Euler's formula

[frankR]

The problem states:

Express Cos( Θ₁ + Θ₂ + Θ₃) in terms of Sin(Θ_k) and Cos(Θ_k), k = 1, 2, 3, using the relation e^+/-i*Θ = Cos(Θ) +/- i*Sin(Θ). [Hint: Use the product property of the exponential e.g., e^{(Θ₁ + Θ₂)} = e^i*Θ₁e^i*Θ₂.]




I'm really confused by in terms of Sin(Θ_k) and Cos(Θ_k), k = 1, 2, 3, how does this apply to the problem?

I'm really lost, someone please steer me in the right direction.

Thanks,

Frank

Edit: Not sure what's wrong with my &theta ?

 

[Claude Bile]

You need to put: - (Using A's instead of thetas)

cos(A1+A2+A3) + isin(A1+A2+A3)

into exponential form:

e^-i(A1+A2+A3)

then using the property of exponentials, express the above exponent as:

(e^-iA1)(e^-iA2)(e^-iA3)

Then put each exponential back into trig form and multiply everything out. separate the real and imaginary terms and equate cos(A1+A2+A3) to the real parts. This will contain both cos and sin terms.

Claude.

 

[krab]

Originally posted by frankR
The problem states:

Express Cos( θ₁ + θ₂ + θ₃) in terms of Sin(θ_k) and Cos(θ_k), k = 1, 2, 3, using the relation e^+/-i*θ = Cos(θ) +/- Sinh(θ). [Hint: Use the product property of the exponential e.g., e^{(θ₁ + θ₂)} = e^i*θ₁e^i*θ₂.]




I'm really confused by in terms of Sin(θ_k) and Cos(θ_k), k = 1, 2, 3, how does this apply to the problem?

I'm really lost, someone please steer me in the right direction.

Thanks,

Frank

Edit: Not sure what's wrong with my &theta ?

You forgot the ;

 

[frankR]

I actually trired that method, but was unsure if it was correct since the terms were so messy. I'll continue with that method and post my solution.

Thanks.

 

[frankR]

Is this how it's expressed:

±Cos{Θ_k} = e ^{± (Θ_k)} ± Sin{Θ_k}, where k=1,2,3.

Thanks.

 

[krab]

±Cos{?k} = e ± (?k) ± Sin{?k}

No.

Cos(?) = (e^(i?)+e^(-i?))/2

 

[Claude Bile]

Originally posted by frankR
I actually trired that method, but was unsure if it was correct since the terms were so messy. I'll continue with that method and post my solution.

Thanks.

Recall the formula for the double angle formula for cos:

cos(A+B) = cos(A)cos(B)-sin(A)sin(B)

Not the neatest answer, and that is just for the two angle case. Based on this alone, you should probably expect your answer to be a bit messy.

Claude.