Expressing a trig function as a complex expodential (HELP)

In summary, the problem involves expressing Cos( Θ1 + Θ2 + Θ3) in terms of Sin(Θk) and Cos(Θk), using the relation e+/-i*Θ = Cos(Θ) +/- i*Sin(Θ) and the product property of exponential. Using the formula for the double angle formula for Cos, the solution may be a bit messy but can be obtained by expressing the exponential form as (e^-iA1)(e^-iA2)(e^-iA3) and then converting each exponential back into trigonometric form and multiplying everything out.
  • #1
frankR
91
0
The problem states:

Express Cos( Θ1 + Θ2 + Θ3) in terms of Sin(Θk) and Cos(Θk), 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) = ei*Θ1ei*Θ2.]




I'm really confused by in terms of Sin(Θk) and Cos(Θk), 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 ?
 
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  • #2
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. Seperate the real and imaginary terms and equate cos(A1+A2+A3) to the real parts. This will contain both cos and sin terms.

Claude.
 
  • #3
Originally posted by frankR
The problem states:

Express Cos( θ1 + θ2 + θ3) in terms of Sin(θk) and Cos(θk), 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) = ei*θ1ei*θ2.]




I'm really confused by in terms of Sin(θk) and Cos(θk), 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 ;
 
  • #4
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.
 
  • #5
Is this how it's expressed:

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

Thanks.
 
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  • #6
±Cos{?k} = e ± (?k) ± Sin{?k}

No.

Cos(?) = (e^(i?)+e^(-i?))/2
 
  • #7


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.
 

FAQ: Expressing a trig function as a complex expodential (HELP)

What is a complex exponential?

A complex exponential is a mathematical expression of the form ex, where e is the base of the natural logarithm and x is a complex number.

Why would I want to express a trig function as a complex exponential?

Expressing a trig function as a complex exponential can make it easier to solve certain mathematical problems, as complex exponentials have special properties that can simplify calculations.

How do I express a trig function as a complex exponential?

To express a trig function as a complex exponential, you can use Euler's formula eix = cos(x) + i sin(x), where i is the imaginary unit.

What are the advantages of expressing a trig function as a complex exponential?

Expressing a trig function as a complex exponential can make it easier to manipulate and solve equations involving trigonometric functions. It can also help in understanding the underlying mathematical concepts and relationships.

Are there any limitations to expressing a trig function as a complex exponential?

While expressing a trig function as a complex exponential can be useful in certain situations, it may not be necessary or practical in all cases. It may also make calculations more complex in some cases.

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