How do I express cos(theta) in terms of e^i(theta) and e^-i(theta)?

Question

This one has me stumped...if you could tell me how you got to the answer so I understand the proof that would be great!

Thanks for your help :)

Answer 1

e^(iθ) = cos θ + i sin θ
e^(-iθ) = cos (-θ) + i sin (-θ) = cos θ - i sin θ

Now, add:

e^(iθ) + e^(-iθ) = 2 cos θ.

Divide by 2:

[e^(iθ) + e^(-iθ)] / 2 = cos θ.

To get sin θ, multiply the second equation by -1, then do the same thing.

Answer 2

E Itheta

Answer 3

1. e^(i theta) = cos(theta) + i sin(theta)
2. e^-(i theta) = cos(-theta) + i sin(-theta)

Now, equation number 2 can be put as:
e^-(i theta) = cos(theta) - i sin(theta)

Adding (1) and (2) we get:

e^(i theta) + e^-i(theta) = 2cos(theta) // divide by 2

cos*theta) = [e^(i theta) + e^-i(theta) ] / 2

Answer 4

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