An Interesting Trigonometric Expression | Problem 494

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[@scottleung9587]

I used a third method, which was using Euler's number or polar form. After dealing with reciprocals and basic algebra, I got -e^(-i*theta), just like WA.

[@cformanorth]

I got the same as WA's "-e^(-i*theta)". I personally find this "simpler" than "e^i(pi-theta)" - is there a reason that the latter is preferred other than subjective preference?

[@iabervon]

Third method: substitute z=e^(-i theta)=cos theta - i sin theta. Then cos theta + i sin theta = e^(i theta) = 1/z (because the modulus is 1). Then we're trying to simplify (1-z)/(1-1/z). Multiply top and bottom by z to get z(1-z)/(z-1)=-z.

[@shmuelzehavi4940]

My approach:
f⁡(θ) = (1 - cos⁡θ + i sin⁡θ) / (1 - cos⁡θ - i sin⁡θ)
= (1 - e^(-iθ)) / (1 - e^iθ )
= (e^iθ - 1) / (e^iθ (1 - e^iθ))
= - 1/e^iθ
= - e^(-iθ)
= - cos⁡θ + i sin⁡θ

[@Don-Ensley]

problem
Can you simplify
( 1- cos θ + i sin θ ) / ( 1- cos θ - i sin θ )
Call it R for ratio.
R = ( 1- cos θ + i sin θ ) / ( 1- cos θ - i sin θ )
Wrap the cos and sin terms in parentheses.
R = [ 1- ( cos θ - i sin θ ) ] /
[ 1- ( cos θ + i sin θ ) ]
By even cos symmetry and odd symmetry of sin, the numerator may be re-expressed as
R = [ 1- ( cos (-θ) + i sin (-θ) ) ] /
[ 1- ( cos θ + i sin θ ) ]
Use Euler's formula.
R = [ 1- e^(- i θ ) ] / [ 1- e^( i θ ) ]
Multiply by 1 in the form of e^( i θ ) / e^( i θ ) . Leave the denominator unchanged but distribute the e^( i θ ) over the numerator.
R = [ 1/ e^( i θ ) ] [ e^( i θ )- 1 ] / [ 1- e^( i θ ) ]
= - [ 1/ e^( i θ ) ]
= - e^(- i θ )
= - ( cos θ - i sin θ )
answer
- cos θ + i sin θ