Step by step solution to "Prove that e−ix=cos⁡(x)−isin⁡(x)"

Agohofidov6

Answered question

2021-12-11

Prove that

e^{- i x} = \cos (x) - i \sin (x)

Pansdorfp6

Beginner2021-12-12Added 27 answers

\sin (x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots

\cos (x) = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots

e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{P} {4}}{4!} + \dots

Multiply

e^{x}

- i

1 + (- i x) + \frac{{(- i x)}^{2}}{2!} + \frac{{(- i x)}^{3}}{3!} + \frac{{(- i x)}^{4}}{4!} + \dots

Expand

1 - i x - \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \dots

Factorise it:

(1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots) - i (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots)

And the first part of the equation is equal to

\cos (x)

and the second part to

\sin (x)

. Thus, we can replace them.

(\cos (x)) - i (\sin (x))

\cos (x) - i \sin (x)

Annie Gonzalez

Beginner2021-12-13Added 41 answers

You can prove this using Taylor's Series.

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