prove-that-e-i-cos-isin-

Question Number 170110 by mathlove last updated on 16/May/22

p r o v e t h a t

e^{i θ} = c o s θ + i s i n θ

Commented by ajfour last updated on 16/May/22

1 \times e^{i θ} = z i s

1 r o t a t e d b y θ a n t i c l o c k w i s e .

h e n c e z = \cos θ + i \sin θ = e^{i θ}

d r a w a d i a g r a m n u k n o w .

Answered by floor(10²Eta[1]) last updated on 16/May/22

using the mclaurin series :

f (x) = \underset{n = 0}{\sum^{\infty}} \frac{f^{(n)} (0)}{n!} x^{n}

for e^{x}, sinx, cosx, we have :

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

sinx = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots

cosx = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \dots

\Rightarrow e^{i θ} = 1 + i θ + \frac{{(i θ)}^{2}}{2!} + \frac{{(i θ)}^{3}}{3!} + \dots

= 1 + i θ - \frac{θ^{2}}{2!} - i \frac{θ^{3}}{3!} + \frac{θ^{4}}{4!} + \frac{i θ^{5}}{5!} - \frac{θ^{6}}{6!} - \frac{i θ^{7}}{7!} + \dots

= (1 - \frac{θ^{2}}{2!} + \frac{θ^{4}}{4!} - \frac{θ^{6}}{6!} + \dots) + i (θ - \frac{θ^{3}}{3!} + \frac{θ^{5}}{5!} - \frac{θ^{7}}{7!} + \dots)

= \cos θ + isin θ

Commented by peter frank last updated on 16/May/22

thank you

Commented by mathlove last updated on 16/May/22

t h a n k s

Answered by MJS_new last updated on 18/May/22

\cos θ = \frac{e^{i θ} + e^{- i θ}}{2}

\sin θ = \frac{e^{i θ} - e^{- i θ}}{2 i}

\cos θ + i \sin θ = \frac{e^{i θ} + e^{- i θ}}{2} + i \frac{e^{i θ} - e^{- i θ}}{2 i} = e^{i θ}

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