Prove-that-e-i-1-e-i-2-e-i-2-

Question Number 128805 by bramlexs22 last updated on 10/Jan/21

Prove that e^{i θ} + 1 = e^{- \frac{i θ}{2}} + e^{\frac{i θ}{2}} ?

Commented by mr W last updated on 10/Jan/21

L H S = (1 + \cos θ) + i \sin θ

R H S = 2 \cos \frac{θ}{2}

L H S \neq R H S!

Answered by benjo_mathlover last updated on 10/Jan/21

e^{i θ} + 1 = \cos θ + i \sin θ + 1

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

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

e^{- \frac{i θ}{2}} + e^{\frac{i θ}{2}} = 2 \cos \frac{θ}{2} \neq 1 + \cos θ + i \sin θ