I have been trying to simplify this expression:
cos(θ)−1cos(θ)+isin(θ)
into:
isin(θ).
However I can't find what steps to take to get to this simplification.
Euler's formula states:
eiθ=cos(θ)+isin(θ)
It is linked to this formula however I am not sure how to go about this.
Answer
cos(θ)−1cos(θ)+isin(θ)=cos(θ)−1eiθ= cos(θ)−e−iθ=12eiθ+12e−iθ−e−iθ= =12eiθ−12e−iθ=i(12ieiθ−12ie−iθ)=isin(θ)
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